Mixing
Precalc Trig Identity, verify: $1 + cos(x) + cos(2x) = frac 12 + frac{sin(5x/2)}{2sin(x/2)}$
$begingroup$
Working with LHS:
I've tried using the sum to product trig ID to get:
$1 + 2cos(3x/2)cos(x/2)$ from here I've tried a couple of things, but can't seem to get closer. I've tried changing the $(3x/2)$ into $(5x/2 - x)$ and using sum identity, but this just makes things even messier.
I also tried working the RHS. I'm only allowed to use the basic trig ID's: pythag, double and half angle, and sum to product and product to sum.
trigonometry
edited Jan 16 at 3:47
Andrei
12.3k21128
asked Jan 16 at 3:32
McMathMcMath
456
$endgroup$
add a comment |
$begingroup$
Working with LHS:
I've tried using the sum to product trig ID to get:
$1 + 2cos(3x/2)cos(x/2)$ from here I've tried a couple of things, but can't seem to get closer. I've tried changing the $(3x/2)$ into $(5x/2 - x)$ and using sum identity, but this just makes things even messier.
I also tried working the RHS. I'm only allowed to use the basic trig ID's: pythag, double and half angle, and sum to product and product to sum.
trigonometry
edited Jan 16 at 3:47
Andrei
12.3k21128
asked Jan 16 at 3:32
McMathMcMath
456
$endgroup$
add a comment |
$begingroup$
Working with LHS:
I've tried using the sum to product trig ID to get:
$1 + 2cos(3x/2)cos(x/2)$ from here I've tried a couple of things, but can't seem to get closer. I've tried changing the $(3x/2)$ into $(5x/2 - x)$ and using sum identity, but this just makes things even messier.
I also tried working the RHS. I'm only allowed to use the basic trig ID's: pythag, double and half angle, and sum to product and product to sum.
trigonometry
edited Jan 16 at 3:47
Andrei
12.3k21128
asked Jan 16 at 3:32
McMathMcMath
456
$endgroup$
Working with LHS:
I've tried using the sum to product trig ID to get:
$1 + 2cos(3x/2)cos(x/2)$ from here I've tried a couple of things, but can't seem to get closer. I've tried changing the $(3x/2)$ into $(5x/2 - x)$ and using sum identity, but this just makes things even messier.
I also tried working the RHS. I'm only allowed to use the basic trig ID's: pythag, double and half angle, and sum to product and product to sum.
trigonometry
trigonometry
edited Jan 16 at 3:47
Andrei
12.3k21128
asked Jan 16 at 3:32
McMathMcMath
456
edited Jan 16 at 3:47
Andrei
12.3k21128
asked Jan 16 at 3:32
McMathMcMath
456
edited Jan 16 at 3:47
Andrei
12.3k21128
edited Jan 16 at 3:47
Andrei
12.3k21128
edited Jan 16 at 3:47
Andrei
12.3k21128
12.3k21128
asked Jan 16 at 3:32
McMathMcMath
456
asked Jan 16 at 3:32
McMathMcMath
456
asked Jan 16 at 3:32
McMathMcMath
456
456
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
For $sinfrac{x}{2}neq0$ we obtain:
$$1+cos{x}+cos2x=frac{2sinfrac{x}{2}+2sinfrac{x}{2}cos{x}+2sinfrac{x}{2}cos2x}{2sinfrac{x}{2}}=$$
$$=frac{2sinfrac{x}{2}+sinfrac{3x}{2}-sinfrac{x}{2}+sinfrac{5x}{2}-sinfrac{3x}{2}}{2sinfrac{x}{2}}=frac{1}{2}+frac{sinfrac{5x}{2}}{2sinfrac{x}{2}}.$$
I used the following formula.
$$sinalphacosbeta=frac{1}{2}(sin(alpha+beta)+sin(alpha-beta)).$$
For example,$$2sinfrac{x}{2}cos{x}=2cdotfrac{1}{2}left(sinleft(frac{x}{2}+xright)+sinleft(frac{x}{2}-xright)right)=sinfrac{3x}{2}-sinfrac{x}{2}.$$
answered Jan 16 at 6:17
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
Can you please explain how you get from step to step. I'm only alloweed to use basic trig IDs. thx
$endgroup$
– McMath
Jan 16 at 14:42
$begingroup$
@McMath I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:47
$begingroup$
oh, and then you're putting all over common denom? I'll give it a shot
$endgroup$
– McMath
Jan 16 at 14:57
$begingroup$
By the same way: $cos{x}=frac{2sinfrac{x}{2}cos{x}}{2sinfrac{x}{2}}$. With $cos2x$ we make the similar thing.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:58
$begingroup$
Finally - understand. Thanks! Do you just play around with the identities or is there a general strategy that lead you to use that identity in the first place?
$endgroup$
– McMath
Jan 16 at 15:06
|
show 1 more comment
$begingroup$
Start with the RHS. Notice that
$$ sin(5x/2) = sin(2x + x/2) = sin 2x cos (x/2) + cos 2x sin (x/2) $$
Also, notice that
$$ sin 2x = 2 sin x cos x = 4 sin (x/2) cos (x/2) cos x $$
Therefore,
$$ frac{ sin (5x/2) }{2 sin(x/2) } = frac{4 sin (x/2) cos^2 (x/2) cos x + cos 2x sin (x/2)}{2 sin(x/2)} $$
$$ = 2 cos^2 (x/2) cos x + frac{ cos 2x }{2} $$
Also, we have that $cos^2 (x/2) = frac{ cos x + 1 }{2}$ and so
$$ = cos^2 x + cos x + frac{ cos 2x }{2} $$
$$ = frac{1+cos 2x }{2} + cos x + frac{ cos 2x }{2} $$
$$ cos x + cos 2x + frac{1}{2} $$$
add the missing $1/2$ from the RHS and you have the LHS
answered Jan 16 at 4:03
Jimmy SabaterJimmy Sabater
2,666323
$endgroup$
$begingroup$
How did you get that cos^2(x/2) = (cosx + 1)/2
$endgroup$
– McMath
Jan 16 at 4:22
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double angle formula $ x = 2 cdot frac{x}{2} $
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– Jimmy Sabater
Jan 16 at 4:24
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hmm I don't know that double angle formula, I'm aware of cos(2x) = 2cos^2(x) - 1, but you're missing the 1? If I didn't know your version, can I get there from mine? Also how can you just change the 1/2 into a 1?
$endgroup$
– McMath
Jan 16 at 4:29
add a comment |
$begingroup$
Hint:
Use
http://mathworld.wolfram.com/WernerFormulas.html
$2sindfrac x2cos mx=sinleft(m+dfrac12right)x-sinleft(m-dfrac12right)x$
Set $m=1,2$ and add to find $$2sindfrac x2(cos x+cos2x)=sindfrac{5x}2-sindfrac x2$$
Assuming $sindfrac x2ne0,$ divide both sides by $2sindfrac x2$
See also: How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?
answered Jan 16 at 3:48
lab bhattacharjeelab bhattacharjee
226k15157275
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To be honest, I have no idea how to fit that in... those formulas are listed in my Trig IDs as Product to sum formulas. But it would be more helpful if you give me some step by step guidance.. thx
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– McMath
Jan 16 at 3:57
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McMath: see my answer.
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– Jimmy Sabater
Jan 16 at 4:04
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@McMath, what if $m=1$ and $m=2?$
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– lab bhattacharjee
Jan 16 at 4:16
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if m = 1 I get sin(3x/2) - sin (x/2).... if m =2 I get sin(5x/2) - sin(3x/2) ... but I don't see where this fits in. What are the previous steps???
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– McMath
Jan 16 at 4:38
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@McMath Then $cos x+cos2x=?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:43
|
show 2 more comments
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The left-hand side is $$Re(1+e^{ix}+e^{2ix})=Refrac{e^{3ix}-1}{e^{ix}-1}=Refrac{2ie^{3ix/2}sinfrac{3x}{2}}{2ie^{ix/2}sinfrac{x}{2}}=Refrac{2e^{ix}sinfrac{3x}{2}}{2sinfrac{x}{2}}\=frac{2cos xsinfrac{3x}{2}}{2sinfrac{x}{2}}=frac{sinfrac{x}{2}+sinfrac{5x}{2}}{2sinfrac{x}{2}}=frac12+frac{sinfrac{5x}{2}}{2sinfrac{x}{2}}.$$
answered Jan 16 at 6:29
J.G.J.G.
27.5k22843
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$begingroup$
I've never seen this math formulas. Please explain using only basic trig identities.
$endgroup$
– McMath
Jan 16 at 14:43
add a comment |
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4 Answers
4
active
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4 Answers
4
active
oldest
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$begingroup$
For $sinfrac{x}{2}neq0$ we obtain:
$$1+cos{x}+cos2x=frac{2sinfrac{x}{2}+2sinfrac{x}{2}cos{x}+2sinfrac{x}{2}cos2x}{2sinfrac{x}{2}}=$$
$$=frac{2sinfrac{x}{2}+sinfrac{3x}{2}-sinfrac{x}{2}+sinfrac{5x}{2}-sinfrac{3x}{2}}{2sinfrac{x}{2}}=frac{1}{2}+frac{sinfrac{5x}{2}}{2sinfrac{x}{2}}.$$
I used the following formula.
$$sinalphacosbeta=frac{1}{2}(sin(alpha+beta)+sin(alpha-beta)).$$
For example,$$2sinfrac{x}{2}cos{x}=2cdotfrac{1}{2}left(sinleft(frac{x}{2}+xright)+sinleft(frac{x}{2}-xright)right)=sinfrac{3x}{2}-sinfrac{x}{2}.$$
answered Jan 16 at 6:17
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
Can you please explain how you get from step to step. I'm only alloweed to use basic trig IDs. thx
$endgroup$
– McMath
Jan 16 at 14:42
$begingroup$
@McMath I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:47
$begingroup$
oh, and then you're putting all over common denom? I'll give it a shot
$endgroup$
– McMath
Jan 16 at 14:57
$begingroup$
By the same way: $cos{x}=frac{2sinfrac{x}{2}cos{x}}{2sinfrac{x}{2}}$. With $cos2x$ we make the similar thing.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:58
$begingroup$
Finally - understand. Thanks! Do you just play around with the identities or is there a general strategy that lead you to use that identity in the first place?
$endgroup$
– McMath
Jan 16 at 15:06
|
show 1 more comment
$begingroup$
For $sinfrac{x}{2}neq0$ we obtain:
$$1+cos{x}+cos2x=frac{2sinfrac{x}{2}+2sinfrac{x}{2}cos{x}+2sinfrac{x}{2}cos2x}{2sinfrac{x}{2}}=$$
$$=frac{2sinfrac{x}{2}+sinfrac{3x}{2}-sinfrac{x}{2}+sinfrac{5x}{2}-sinfrac{3x}{2}}{2sinfrac{x}{2}}=frac{1}{2}+frac{sinfrac{5x}{2}}{2sinfrac{x}{2}}.$$
I used the following formula.
$$sinalphacosbeta=frac{1}{2}(sin(alpha+beta)+sin(alpha-beta)).$$
For example,$$2sinfrac{x}{2}cos{x}=2cdotfrac{1}{2}left(sinleft(frac{x}{2}+xright)+sinleft(frac{x}{2}-xright)right)=sinfrac{3x}{2}-sinfrac{x}{2}.$$
answered Jan 16 at 6:17
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
$begingroup$
Can you please explain how you get from step to step. I'm only alloweed to use basic trig IDs. thx
$endgroup$
– McMath
Jan 16 at 14:42
$begingroup$
@McMath I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:47
$begingroup$
oh, and then you're putting all over common denom? I'll give it a shot
$endgroup$
– McMath
Jan 16 at 14:57
$begingroup$
By the same way: $cos{x}=frac{2sinfrac{x}{2}cos{x}}{2sinfrac{x}{2}}$. With $cos2x$ we make the similar thing.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:58
$begingroup$
Finally - understand. Thanks! Do you just play around with the identities or is there a general strategy that lead you to use that identity in the first place?
$endgroup$
– McMath
Jan 16 at 15:06
|
show 1 more comment
$begingroup$
For $sinfrac{x}{2}neq0$ we obtain:
$$1+cos{x}+cos2x=frac{2sinfrac{x}{2}+2sinfrac{x}{2}cos{x}+2sinfrac{x}{2}cos2x}{2sinfrac{x}{2}}=$$
$$=frac{2sinfrac{x}{2}+sinfrac{3x}{2}-sinfrac{x}{2}+sinfrac{5x}{2}-sinfrac{3x}{2}}{2sinfrac{x}{2}}=frac{1}{2}+frac{sinfrac{5x}{2}}{2sinfrac{x}{2}}.$$
I used the following formula.
$$sinalphacosbeta=frac{1}{2}(sin(alpha+beta)+sin(alpha-beta)).$$
For example,$$2sinfrac{x}{2}cos{x}=2cdotfrac{1}{2}left(sinleft(frac{x}{2}+xright)+sinleft(frac{x}{2}-xright)right)=sinfrac{3x}{2}-sinfrac{x}{2}.$$
answered Jan 16 at 6:17
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
For $sinfrac{x}{2}neq0$ we obtain:
$$1+cos{x}+cos2x=frac{2sinfrac{x}{2}+2sinfrac{x}{2}cos{x}+2sinfrac{x}{2}cos2x}{2sinfrac{x}{2}}=$$
$$=frac{2sinfrac{x}{2}+sinfrac{3x}{2}-sinfrac{x}{2}+sinfrac{5x}{2}-sinfrac{3x}{2}}{2sinfrac{x}{2}}=frac{1}{2}+frac{sinfrac{5x}{2}}{2sinfrac{x}{2}}.$$
I used the following formula.
$$sinalphacosbeta=frac{1}{2}(sin(alpha+beta)+sin(alpha-beta)).$$
For example,$$2sinfrac{x}{2}cos{x}=2cdotfrac{1}{2}left(sinleft(frac{x}{2}+xright)+sinleft(frac{x}{2}-xright)right)=sinfrac{3x}{2}-sinfrac{x}{2}.$$
answered Jan 16 at 6:17
Michael RozenbergMichael Rozenberg
104k1892197
edited Jan 16 at 14:46
answered Jan 16 at 6:17
Michael RozenbergMichael Rozenberg
104k1892197
answered Jan 16 at 6:17
Michael RozenbergMichael Rozenberg
104k1892197
answered Jan 16 at 6:17
Michael RozenbergMichael Rozenberg
104k1892197
104k1892197
$begingroup$
Can you please explain how you get from step to step. I'm only alloweed to use basic trig IDs. thx
$endgroup$
– McMath
Jan 16 at 14:42
$begingroup$
@McMath I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:47
$begingroup$
oh, and then you're putting all over common denom? I'll give it a shot
$endgroup$
– McMath
Jan 16 at 14:57
$begingroup$
By the same way: $cos{x}=frac{2sinfrac{x}{2}cos{x}}{2sinfrac{x}{2}}$. With $cos2x$ we make the similar thing.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:58
$begingroup$
Finally - understand. Thanks! Do you just play around with the identities or is there a general strategy that lead you to use that identity in the first place?
$endgroup$
– McMath
Jan 16 at 15:06
|
show 1 more comment
$begingroup$
Can you please explain how you get from step to step. I'm only alloweed to use basic trig IDs. thx
$endgroup$
– McMath
Jan 16 at 14:42
$begingroup$
@McMath I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:47
$begingroup$
oh, and then you're putting all over common denom? I'll give it a shot
$endgroup$
– McMath
Jan 16 at 14:57
$begingroup$
By the same way: $cos{x}=frac{2sinfrac{x}{2}cos{x}}{2sinfrac{x}{2}}$. With $cos2x$ we make the similar thing.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:58
$begingroup$
Finally - understand. Thanks! Do you just play around with the identities or is there a general strategy that lead you to use that identity in the first place?
$endgroup$
– McMath
Jan 16 at 15:06
$begingroup$
Can you please explain how you get from step to step. I'm only alloweed to use basic trig IDs. thx
$endgroup$
– McMath
Jan 16 at 14:42
$begingroup$
Can you please explain how you get from step to step. I'm only alloweed to use basic trig IDs. thx
$endgroup$
– McMath
Jan 16 at 14:42
$begingroup$
@McMath I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:47
$begingroup$
@McMath I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:47
$begingroup$
oh, and then you're putting all over common denom? I'll give it a shot
$endgroup$
– McMath
Jan 16 at 14:57
$begingroup$
oh, and then you're putting all over common denom? I'll give it a shot
$endgroup$
– McMath
Jan 16 at 14:57
$begingroup$
By the same way: $cos{x}=frac{2sinfrac{x}{2}cos{x}}{2sinfrac{x}{2}}$. With $cos2x$ we make the similar thing.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:58
$begingroup$
By the same way: $cos{x}=frac{2sinfrac{x}{2}cos{x}}{2sinfrac{x}{2}}$. With $cos2x$ we make the similar thing.
$endgroup$
– Michael Rozenberg
Jan 16 at 14:58
$begingroup$
Finally - understand. Thanks! Do you just play around with the identities or is there a general strategy that lead you to use that identity in the first place?
$endgroup$
– McMath
Jan 16 at 15:06
$begingroup$
Finally - understand. Thanks! Do you just play around with the identities or is there a general strategy that lead you to use that identity in the first place?
$endgroup$
– McMath
Jan 16 at 15:06
|
show 1 more comment
$begingroup$
Start with the RHS. Notice that
$$ sin(5x/2) = sin(2x + x/2) = sin 2x cos (x/2) + cos 2x sin (x/2) $$
Also, notice that
$$ sin 2x = 2 sin x cos x = 4 sin (x/2) cos (x/2) cos x $$
Therefore,
$$ frac{ sin (5x/2) }{2 sin(x/2) } = frac{4 sin (x/2) cos^2 (x/2) cos x + cos 2x sin (x/2)}{2 sin(x/2)} $$
$$ = 2 cos^2 (x/2) cos x + frac{ cos 2x }{2} $$
Also, we have that $cos^2 (x/2) = frac{ cos x + 1 }{2}$ and so
$$ = cos^2 x + cos x + frac{ cos 2x }{2} $$
$$ = frac{1+cos 2x }{2} + cos x + frac{ cos 2x }{2} $$
$$ cos x + cos 2x + frac{1}{2} $$$
add the missing $1/2$ from the RHS and you have the LHS
answered Jan 16 at 4:03
Jimmy SabaterJimmy Sabater
2,666323
$endgroup$
$begingroup$
How did you get that cos^2(x/2) = (cosx + 1)/2
$endgroup$
– McMath
Jan 16 at 4:22
$begingroup$
double angle formula $ x = 2 cdot frac{x}{2} $
$endgroup$
– Jimmy Sabater
Jan 16 at 4:24
$begingroup$
hmm I don't know that double angle formula, I'm aware of cos(2x) = 2cos^2(x) - 1, but you're missing the 1? If I didn't know your version, can I get there from mine? Also how can you just change the 1/2 into a 1?
$endgroup$
– McMath
Jan 16 at 4:29
add a comment |
$begingroup$
Start with the RHS. Notice that
$$ sin(5x/2) = sin(2x + x/2) = sin 2x cos (x/2) + cos 2x sin (x/2) $$
Also, notice that
$$ sin 2x = 2 sin x cos x = 4 sin (x/2) cos (x/2) cos x $$
Therefore,
$$ frac{ sin (5x/2) }{2 sin(x/2) } = frac{4 sin (x/2) cos^2 (x/2) cos x + cos 2x sin (x/2)}{2 sin(x/2)} $$
$$ = 2 cos^2 (x/2) cos x + frac{ cos 2x }{2} $$
Also, we have that $cos^2 (x/2) = frac{ cos x + 1 }{2}$ and so
$$ = cos^2 x + cos x + frac{ cos 2x }{2} $$
$$ = frac{1+cos 2x }{2} + cos x + frac{ cos 2x }{2} $$
$$ cos x + cos 2x + frac{1}{2} $$$
add the missing $1/2$ from the RHS and you have the LHS
answered Jan 16 at 4:03
Jimmy SabaterJimmy Sabater
2,666323
$endgroup$
$begingroup$
How did you get that cos^2(x/2) = (cosx + 1)/2
$endgroup$
– McMath
Jan 16 at 4:22
$begingroup$
double angle formula $ x = 2 cdot frac{x}{2} $
$endgroup$
– Jimmy Sabater
Jan 16 at 4:24
$begingroup$
hmm I don't know that double angle formula, I'm aware of cos(2x) = 2cos^2(x) - 1, but you're missing the 1? If I didn't know your version, can I get there from mine? Also how can you just change the 1/2 into a 1?
$endgroup$
– McMath
Jan 16 at 4:29
add a comment |
$begingroup$
Start with the RHS. Notice that
$$ sin(5x/2) = sin(2x + x/2) = sin 2x cos (x/2) + cos 2x sin (x/2) $$
Also, notice that
$$ sin 2x = 2 sin x cos x = 4 sin (x/2) cos (x/2) cos x $$
Therefore,
$$ frac{ sin (5x/2) }{2 sin(x/2) } = frac{4 sin (x/2) cos^2 (x/2) cos x + cos 2x sin (x/2)}{2 sin(x/2)} $$
$$ = 2 cos^2 (x/2) cos x + frac{ cos 2x }{2} $$
Also, we have that $cos^2 (x/2) = frac{ cos x + 1 }{2}$ and so
$$ = cos^2 x + cos x + frac{ cos 2x }{2} $$
$$ = frac{1+cos 2x }{2} + cos x + frac{ cos 2x }{2} $$
$$ cos x + cos 2x + frac{1}{2} $$$
add the missing $1/2$ from the RHS and you have the LHS
answered Jan 16 at 4:03
Jimmy SabaterJimmy Sabater
2,666323
$endgroup$
Start with the RHS. Notice that
$$ sin(5x/2) = sin(2x + x/2) = sin 2x cos (x/2) + cos 2x sin (x/2) $$
Also, notice that
$$ sin 2x = 2 sin x cos x = 4 sin (x/2) cos (x/2) cos x $$
Therefore,
$$ frac{ sin (5x/2) }{2 sin(x/2) } = frac{4 sin (x/2) cos^2 (x/2) cos x + cos 2x sin (x/2)}{2 sin(x/2)} $$
$$ = 2 cos^2 (x/2) cos x + frac{ cos 2x }{2} $$
Also, we have that $cos^2 (x/2) = frac{ cos x + 1 }{2}$ and so
$$ = cos^2 x + cos x + frac{ cos 2x }{2} $$
$$ = frac{1+cos 2x }{2} + cos x + frac{ cos 2x }{2} $$
$$ cos x + cos 2x + frac{1}{2} $$$
add the missing $1/2$ from the RHS and you have the LHS
answered Jan 16 at 4:03
Jimmy SabaterJimmy Sabater
2,666323
answered Jan 16 at 4:03
Jimmy SabaterJimmy Sabater
2,666323
answered Jan 16 at 4:03
Jimmy SabaterJimmy Sabater
2,666323
answered Jan 16 at 4:03
Jimmy SabaterJimmy Sabater
2,666323
2,666323
$begingroup$
How did you get that cos^2(x/2) = (cosx + 1)/2
$endgroup$
– McMath
Jan 16 at 4:22
$begingroup$
double angle formula $ x = 2 cdot frac{x}{2} $
$endgroup$
– Jimmy Sabater
Jan 16 at 4:24
$begingroup$
hmm I don't know that double angle formula, I'm aware of cos(2x) = 2cos^2(x) - 1, but you're missing the 1? If I didn't know your version, can I get there from mine? Also how can you just change the 1/2 into a 1?
$endgroup$
– McMath
Jan 16 at 4:29
add a comment |
$begingroup$
How did you get that cos^2(x/2) = (cosx + 1)/2
$endgroup$
– McMath
Jan 16 at 4:22
$begingroup$
double angle formula $ x = 2 cdot frac{x}{2} $
$endgroup$
– Jimmy Sabater
Jan 16 at 4:24
$begingroup$
hmm I don't know that double angle formula, I'm aware of cos(2x) = 2cos^2(x) - 1, but you're missing the 1? If I didn't know your version, can I get there from mine? Also how can you just change the 1/2 into a 1?
$endgroup$
– McMath
Jan 16 at 4:29
$begingroup$
How did you get that cos^2(x/2) = (cosx + 1)/2
$endgroup$
– McMath
Jan 16 at 4:22
$begingroup$
How did you get that cos^2(x/2) = (cosx + 1)/2
$endgroup$
– McMath
Jan 16 at 4:22
$begingroup$
double angle formula $ x = 2 cdot frac{x}{2} $
$endgroup$
– Jimmy Sabater
Jan 16 at 4:24
$begingroup$
double angle formula $ x = 2 cdot frac{x}{2} $
$endgroup$
– Jimmy Sabater
Jan 16 at 4:24
$begingroup$
hmm I don't know that double angle formula, I'm aware of cos(2x) = 2cos^2(x) - 1, but you're missing the 1? If I didn't know your version, can I get there from mine? Also how can you just change the 1/2 into a 1?
$endgroup$
– McMath
Jan 16 at 4:29
$begingroup$
hmm I don't know that double angle formula, I'm aware of cos(2x) = 2cos^2(x) - 1, but you're missing the 1? If I didn't know your version, can I get there from mine? Also how can you just change the 1/2 into a 1?
$endgroup$
– McMath
Jan 16 at 4:29
add a comment |
$begingroup$
Hint:
Use
http://mathworld.wolfram.com/WernerFormulas.html
$2sindfrac x2cos mx=sinleft(m+dfrac12right)x-sinleft(m-dfrac12right)x$
Set $m=1,2$ and add to find $$2sindfrac x2(cos x+cos2x)=sindfrac{5x}2-sindfrac x2$$
Assuming $sindfrac x2ne0,$ divide both sides by $2sindfrac x2$
See also: How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?
answered Jan 16 at 3:48
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
To be honest, I have no idea how to fit that in... those formulas are listed in my Trig IDs as Product to sum formulas. But it would be more helpful if you give me some step by step guidance.. thx
$endgroup$
– McMath
Jan 16 at 3:57
$begingroup$
McMath: see my answer.
$endgroup$
– Jimmy Sabater
Jan 16 at 4:04
$begingroup$
@McMath, what if $m=1$ and $m=2?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:16
$begingroup$
if m = 1 I get sin(3x/2) - sin (x/2).... if m =2 I get sin(5x/2) - sin(3x/2) ... but I don't see where this fits in. What are the previous steps???
$endgroup$
– McMath
Jan 16 at 4:38
$begingroup$
@McMath Then $cos x+cos2x=?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:43
|
show 2 more comments
$begingroup$
Hint:
Use
http://mathworld.wolfram.com/WernerFormulas.html
$2sindfrac x2cos mx=sinleft(m+dfrac12right)x-sinleft(m-dfrac12right)x$
Set $m=1,2$ and add to find $$2sindfrac x2(cos x+cos2x)=sindfrac{5x}2-sindfrac x2$$
Assuming $sindfrac x2ne0,$ divide both sides by $2sindfrac x2$
See also: How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?
answered Jan 16 at 3:48
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
$begingroup$
To be honest, I have no idea how to fit that in... those formulas are listed in my Trig IDs as Product to sum formulas. But it would be more helpful if you give me some step by step guidance.. thx
$endgroup$
– McMath
Jan 16 at 3:57
$begingroup$
McMath: see my answer.
$endgroup$
– Jimmy Sabater
Jan 16 at 4:04
$begingroup$
@McMath, what if $m=1$ and $m=2?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:16
$begingroup$
if m = 1 I get sin(3x/2) - sin (x/2).... if m =2 I get sin(5x/2) - sin(3x/2) ... but I don't see where this fits in. What are the previous steps???
$endgroup$
– McMath
Jan 16 at 4:38
$begingroup$
@McMath Then $cos x+cos2x=?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:43
|
show 2 more comments
$begingroup$
Hint:
Use
http://mathworld.wolfram.com/WernerFormulas.html
$2sindfrac x2cos mx=sinleft(m+dfrac12right)x-sinleft(m-dfrac12right)x$
Set $m=1,2$ and add to find $$2sindfrac x2(cos x+cos2x)=sindfrac{5x}2-sindfrac x2$$
Assuming $sindfrac x2ne0,$ divide both sides by $2sindfrac x2$
See also: How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?
answered Jan 16 at 3:48
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Hint:
Use
http://mathworld.wolfram.com/WernerFormulas.html
$2sindfrac x2cos mx=sinleft(m+dfrac12right)x-sinleft(m-dfrac12right)x$
Set $m=1,2$ and add to find $$2sindfrac x2(cos x+cos2x)=sindfrac{5x}2-sindfrac x2$$
Assuming $sindfrac x2ne0,$ divide both sides by $2sindfrac x2$
See also: How can we sum up $sin$ and $cos$ series when the angles are in arithmetic progression?
answered Jan 16 at 3:48
lab bhattacharjeelab bhattacharjee
226k15157275
edited Jan 16 at 5:28
answered Jan 16 at 3:48
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 16 at 3:48
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 16 at 3:48
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
$begingroup$
To be honest, I have no idea how to fit that in... those formulas are listed in my Trig IDs as Product to sum formulas. But it would be more helpful if you give me some step by step guidance.. thx
$endgroup$
– McMath
Jan 16 at 3:57
$begingroup$
McMath: see my answer.
$endgroup$
– Jimmy Sabater
Jan 16 at 4:04
$begingroup$
@McMath, what if $m=1$ and $m=2?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:16
$begingroup$
if m = 1 I get sin(3x/2) - sin (x/2).... if m =2 I get sin(5x/2) - sin(3x/2) ... but I don't see where this fits in. What are the previous steps???
$endgroup$
– McMath
Jan 16 at 4:38
$begingroup$
@McMath Then $cos x+cos2x=?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:43
|
show 2 more comments
$begingroup$
To be honest, I have no idea how to fit that in... those formulas are listed in my Trig IDs as Product to sum formulas. But it would be more helpful if you give me some step by step guidance.. thx
$endgroup$
– McMath
Jan 16 at 3:57
$begingroup$
McMath: see my answer.
$endgroup$
– Jimmy Sabater
Jan 16 at 4:04
$begingroup$
@McMath, what if $m=1$ and $m=2?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:16
$begingroup$
if m = 1 I get sin(3x/2) - sin (x/2).... if m =2 I get sin(5x/2) - sin(3x/2) ... but I don't see where this fits in. What are the previous steps???
$endgroup$
– McMath
Jan 16 at 4:38
$begingroup$
@McMath Then $cos x+cos2x=?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:43
$begingroup$
To be honest, I have no idea how to fit that in... those formulas are listed in my Trig IDs as Product to sum formulas. But it would be more helpful if you give me some step by step guidance.. thx
$endgroup$
– McMath
Jan 16 at 3:57
$begingroup$
To be honest, I have no idea how to fit that in... those formulas are listed in my Trig IDs as Product to sum formulas. But it would be more helpful if you give me some step by step guidance.. thx
$endgroup$
– McMath
Jan 16 at 3:57
$begingroup$
McMath: see my answer.
$endgroup$
– Jimmy Sabater
Jan 16 at 4:04
$begingroup$
McMath: see my answer.
$endgroup$
– Jimmy Sabater
Jan 16 at 4:04
$begingroup$
@McMath, what if $m=1$ and $m=2?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:16
$begingroup$
@McMath, what if $m=1$ and $m=2?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:16
$begingroup$
if m = 1 I get sin(3x/2) - sin (x/2).... if m =2 I get sin(5x/2) - sin(3x/2) ... but I don't see where this fits in. What are the previous steps???
$endgroup$
– McMath
Jan 16 at 4:38
$begingroup$
if m = 1 I get sin(3x/2) - sin (x/2).... if m =2 I get sin(5x/2) - sin(3x/2) ... but I don't see where this fits in. What are the previous steps???
$endgroup$
– McMath
Jan 16 at 4:38
$begingroup$
@McMath Then $cos x+cos2x=?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:43
$begingroup$
@McMath Then $cos x+cos2x=?$
$endgroup$
– lab bhattacharjee
Jan 16 at 4:43
|
show 2 more comments
$begingroup$
The left-hand side is $$Re(1+e^{ix}+e^{2ix})=Refrac{e^{3ix}-1}{e^{ix}-1}=Refrac{2ie^{3ix/2}sinfrac{3x}{2}}{2ie^{ix/2}sinfrac{x}{2}}=Refrac{2e^{ix}sinfrac{3x}{2}}{2sinfrac{x}{2}}\=frac{2cos xsinfrac{3x}{2}}{2sinfrac{x}{2}}=frac{sinfrac{x}{2}+sinfrac{5x}{2}}{2sinfrac{x}{2}}=frac12+frac{sinfrac{5x}{2}}{2sinfrac{x}{2}}.$$
answered Jan 16 at 6:29
J.G.J.G.
27.5k22843
$endgroup$
$begingroup$
I've never seen this math formulas. Please explain using only basic trig identities.
$endgroup$
– McMath
Jan 16 at 14:43
add a comment |
$begingroup$
The left-hand side is $$Re(1+e^{ix}+e^{2ix})=Refrac{e^{3ix}-1}{e^{ix}-1}=Refrac{2ie^{3ix/2}sinfrac{3x}{2}}{2ie^{ix/2}sinfrac{x}{2}}=Refrac{2e^{ix}sinfrac{3x}{2}}{2sinfrac{x}{2}}\=frac{2cos xsinfrac{3x}{2}}{2sinfrac{x}{2}}=frac{sinfrac{x}{2}+sinfrac{5x}{2}}{2sinfrac{x}{2}}=frac12+frac{sinfrac{5x}{2}}{2sinfrac{x}{2}}.$$
answered Jan 16 at 6:29
J.G.J.G.
27.5k22843
$endgroup$
$begingroup$
I've never seen this math formulas. Please explain using only basic trig identities.
$endgroup$
– McMath
Jan 16 at 14:43
add a comment |
$begingroup$
The left-hand side is $$Re(1+e^{ix}+e^{2ix})=Refrac{e^{3ix}-1}{e^{ix}-1}=Refrac{2ie^{3ix/2}sinfrac{3x}{2}}{2ie^{ix/2}sinfrac{x}{2}}=Refrac{2e^{ix}sinfrac{3x}{2}}{2sinfrac{x}{2}}\=frac{2cos xsinfrac{3x}{2}}{2sinfrac{x}{2}}=frac{sinfrac{x}{2}+sinfrac{5x}{2}}{2sinfrac{x}{2}}=frac12+frac{sinfrac{5x}{2}}{2sinfrac{x}{2}}.$$
answered Jan 16 at 6:29
J.G.J.G.
27.5k22843
$endgroup$
The left-hand side is $$Re(1+e^{ix}+e^{2ix})=Refrac{e^{3ix}-1}{e^{ix}-1}=Refrac{2ie^{3ix/2}sinfrac{3x}{2}}{2ie^{ix/2}sinfrac{x}{2}}=Refrac{2e^{ix}sinfrac{3x}{2}}{2sinfrac{x}{2}}\=frac{2cos xsinfrac{3x}{2}}{2sinfrac{x}{2}}=frac{sinfrac{x}{2}+sinfrac{5x}{2}}{2sinfrac{x}{2}}=frac12+frac{sinfrac{5x}{2}}{2sinfrac{x}{2}}.$$
answered Jan 16 at 6:29
J.G.J.G.
27.5k22843
answered Jan 16 at 6:29
J.G.J.G.
27.5k22843
answered Jan 16 at 6:29
J.G.J.G.
27.5k22843
answered Jan 16 at 6:29
J.G.J.G.
27.5k22843
27.5k22843
$begingroup$
I've never seen this math formulas. Please explain using only basic trig identities.
$endgroup$
– McMath
Jan 16 at 14:43
add a comment |
$begingroup$
I've never seen this math formulas. Please explain using only basic trig identities.
$endgroup$
– McMath
Jan 16 at 14:43
$begingroup$
I've never seen this math formulas. Please explain using only basic trig identities.
$endgroup$
– McMath
Jan 16 at 14:43
$begingroup$
I've never seen this math formulas. Please explain using only basic trig identities.
$endgroup$
– McMath
Jan 16 at 14:43
add a comment |
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