Mixing
Showing that $f(4x) + 4f(2x) =8cos^4(x)-3$, where $f(x) = cos x$
$begingroup$
It is given that
$$f(x)= sin(x+30^circ) + cos(x+60^circ)$$
A) Show that $$f(x)= cos(x)$$
B) Hence, show that
$$f(4x) + 4f(2x) =8cos^4(x)-3$$
I managed to prove $f(x)$ equals $cos(x)$, but after that I'm stumped.
trigonometry
edited Jan 29 at 20:51
Blue
49.3k870157
asked Jan 29 at 20:31
RhiaRhia
83
$endgroup$
add a comment |
$begingroup$
It is given that
$$f(x)= sin(x+30^circ) + cos(x+60^circ)$$
A) Show that $$f(x)= cos(x)$$
B) Hence, show that
$$f(4x) + 4f(2x) =8cos^4(x)-3$$
I managed to prove $f(x)$ equals $cos(x)$, but after that I'm stumped.
trigonometry
edited Jan 29 at 20:51
Blue
49.3k870157
asked Jan 29 at 20:31
RhiaRhia
83
$endgroup$
$begingroup$
Do you know how to expand $cos 4x$ and $cos 2x$ in terms of $cos x$? Even easier if you know Euler's formula.
$endgroup$
– rogerl
Jan 29 at 20:38
add a comment |
$begingroup$
It is given that
$$f(x)= sin(x+30^circ) + cos(x+60^circ)$$
A) Show that $$f(x)= cos(x)$$
B) Hence, show that
$$f(4x) + 4f(2x) =8cos^4(x)-3$$
I managed to prove $f(x)$ equals $cos(x)$, but after that I'm stumped.
trigonometry
edited Jan 29 at 20:51
Blue
49.3k870157
asked Jan 29 at 20:31
RhiaRhia
83
$endgroup$
It is given that
$$f(x)= sin(x+30^circ) + cos(x+60^circ)$$
A) Show that $$f(x)= cos(x)$$
B) Hence, show that
$$f(4x) + 4f(2x) =8cos^4(x)-3$$
I managed to prove $f(x)$ equals $cos(x)$, but after that I'm stumped.
trigonometry
trigonometry
edited Jan 29 at 20:51
Blue
49.3k870157
asked Jan 29 at 20:31
RhiaRhia
83
edited Jan 29 at 20:51
Blue
49.3k870157
asked Jan 29 at 20:31
RhiaRhia
83
edited Jan 29 at 20:51
Blue
49.3k870157
edited Jan 29 at 20:51
Blue
49.3k870157
edited Jan 29 at 20:51
Blue
49.3k870157
49.3k870157
asked Jan 29 at 20:31
RhiaRhia
83
asked Jan 29 at 20:31
RhiaRhia
83
asked Jan 29 at 20:31
RhiaRhia
83
83
$begingroup$
Do you know how to expand $cos 4x$ and $cos 2x$ in terms of $cos x$? Even easier if you know Euler's formula.
$endgroup$
– rogerl
Jan 29 at 20:38
add a comment |
$begingroup$
Do you know how to expand $cos 4x$ and $cos 2x$ in terms of $cos x$? Even easier if you know Euler's formula.
$endgroup$
– rogerl
Jan 29 at 20:38
$begingroup$
Do you know how to expand $cos 4x$ and $cos 2x$ in terms of $cos x$? Even easier if you know Euler's formula.
$endgroup$
– rogerl
Jan 29 at 20:38
$begingroup$
Do you know how to expand $cos 4x$ and $cos 2x$ in terms of $cos x$? Even easier if you know Euler's formula.
$endgroup$
– rogerl
Jan 29 at 20:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use the property:
$$cos(2x) = cos^2(x)-sin^2(x)$$
Since this will allow you to express $cos(4x)$ =
$$cos^2(2x)-(1-cos^2(2x)) = 2cos^2(2x)-1$$
By substituting again $cos(2x)$ by the aforementioned expression you will be able to express $4cos(2x)+cos(4x)$ in terms of $sin(x)$ and $cos(x)$ raised to some powers. The sinus will cancell out and will obtained the desired result.
answered Jan 29 at 20:44
maxbpmaxbp
1467
$endgroup$
$begingroup$
Thanks so much I got it now
$endgroup$
– Rhia
Jan 29 at 20:57
$begingroup$
You are welcome!
$endgroup$
– maxbp
Jan 29 at 21:04
add a comment |
$begingroup$
$$f(x)=sin(30^{circ}+x)+sin(30^{circ}-x)=2sin30^{circ}cos{x}=cos{x}.$$
Thus, $$f(4x)+4f(2x)=cos4x+4cos2x=8cos^4x-8cos^2x+1+8cos^2x-4=8cos^4x-3.$$
answered Jan 29 at 20:44
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use the property:
$$cos(2x) = cos^2(x)-sin^2(x)$$
Since this will allow you to express $cos(4x)$ =
$$cos^2(2x)-(1-cos^2(2x)) = 2cos^2(2x)-1$$
By substituting again $cos(2x)$ by the aforementioned expression you will be able to express $4cos(2x)+cos(4x)$ in terms of $sin(x)$ and $cos(x)$ raised to some powers. The sinus will cancell out and will obtained the desired result.
answered Jan 29 at 20:44
maxbpmaxbp
1467
$endgroup$
$begingroup$
Thanks so much I got it now
$endgroup$
– Rhia
Jan 29 at 20:57
$begingroup$
You are welcome!
$endgroup$
– maxbp
Jan 29 at 21:04
add a comment |
$begingroup$
You can use the property:
$$cos(2x) = cos^2(x)-sin^2(x)$$
Since this will allow you to express $cos(4x)$ =
$$cos^2(2x)-(1-cos^2(2x)) = 2cos^2(2x)-1$$
By substituting again $cos(2x)$ by the aforementioned expression you will be able to express $4cos(2x)+cos(4x)$ in terms of $sin(x)$ and $cos(x)$ raised to some powers. The sinus will cancell out and will obtained the desired result.
answered Jan 29 at 20:44
maxbpmaxbp
1467
$endgroup$
$begingroup$
Thanks so much I got it now
$endgroup$
– Rhia
Jan 29 at 20:57
$begingroup$
You are welcome!
$endgroup$
– maxbp
Jan 29 at 21:04
add a comment |
$begingroup$
You can use the property:
$$cos(2x) = cos^2(x)-sin^2(x)$$
Since this will allow you to express $cos(4x)$ =
$$cos^2(2x)-(1-cos^2(2x)) = 2cos^2(2x)-1$$
By substituting again $cos(2x)$ by the aforementioned expression you will be able to express $4cos(2x)+cos(4x)$ in terms of $sin(x)$ and $cos(x)$ raised to some powers. The sinus will cancell out and will obtained the desired result.
answered Jan 29 at 20:44
maxbpmaxbp
1467
$endgroup$
You can use the property:
$$cos(2x) = cos^2(x)-sin^2(x)$$
Since this will allow you to express $cos(4x)$ =
$$cos^2(2x)-(1-cos^2(2x)) = 2cos^2(2x)-1$$
By substituting again $cos(2x)$ by the aforementioned expression you will be able to express $4cos(2x)+cos(4x)$ in terms of $sin(x)$ and $cos(x)$ raised to some powers. The sinus will cancell out and will obtained the desired result.
answered Jan 29 at 20:44
maxbpmaxbp
1467
answered Jan 29 at 20:44
maxbpmaxbp
1467
answered Jan 29 at 20:44
maxbpmaxbp
1467
answered Jan 29 at 20:44
maxbpmaxbp
1467
1467
$begingroup$
Thanks so much I got it now
$endgroup$
– Rhia
Jan 29 at 20:57
$begingroup$
You are welcome!
$endgroup$
– maxbp
Jan 29 at 21:04
add a comment |
$begingroup$
Thanks so much I got it now
$endgroup$
– Rhia
Jan 29 at 20:57
$begingroup$
You are welcome!
$endgroup$
– maxbp
Jan 29 at 21:04
$begingroup$
Thanks so much I got it now
$endgroup$
– Rhia
Jan 29 at 20:57
$begingroup$
Thanks so much I got it now
$endgroup$
– Rhia
Jan 29 at 20:57
$begingroup$
You are welcome!
$endgroup$
– maxbp
Jan 29 at 21:04
$begingroup$
You are welcome!
$endgroup$
– maxbp
Jan 29 at 21:04
add a comment |
$begingroup$
$$f(x)=sin(30^{circ}+x)+sin(30^{circ}-x)=2sin30^{circ}cos{x}=cos{x}.$$
Thus, $$f(4x)+4f(2x)=cos4x+4cos2x=8cos^4x-8cos^2x+1+8cos^2x-4=8cos^4x-3.$$
answered Jan 29 at 20:44
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
$$f(x)=sin(30^{circ}+x)+sin(30^{circ}-x)=2sin30^{circ}cos{x}=cos{x}.$$
Thus, $$f(4x)+4f(2x)=cos4x+4cos2x=8cos^4x-8cos^2x+1+8cos^2x-4=8cos^4x-3.$$
answered Jan 29 at 20:44
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
$$f(x)=sin(30^{circ}+x)+sin(30^{circ}-x)=2sin30^{circ}cos{x}=cos{x}.$$
Thus, $$f(4x)+4f(2x)=cos4x+4cos2x=8cos^4x-8cos^2x+1+8cos^2x-4=8cos^4x-3.$$
answered Jan 29 at 20:44
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
$$f(x)=sin(30^{circ}+x)+sin(30^{circ}-x)=2sin30^{circ}cos{x}=cos{x}.$$
Thus, $$f(4x)+4f(2x)=cos4x+4cos2x=8cos^4x-8cos^2x+1+8cos^2x-4=8cos^4x-3.$$
answered Jan 29 at 20:44
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 29 at 20:44
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 29 at 20:44
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 29 at 20:44
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
add a comment |
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$begingroup$
Do you know how to expand $cos 4x$ and $cos 2x$ in terms of $cos x$? Even easier if you know Euler's formula.
$endgroup$
– rogerl
Jan 29 at 20:38