Mixing
How do I prove this trig identity that contains 2 variables?
$begingroup$
$ frac{sin A cos B + cos A sin B } {cos A cos B - sin A sin B} = frac{tan A + tan B}{1 - tan A tan B}$
I've tried multiplying by the reciprocal denominator on the left side to see if I could begin to simplify it to tangent, but I'm still confused. Any help would be appreciated.
trigonometry
edited Jan 29 at 21:26
abc...
3,237739
asked Jan 29 at 21:14
thefirstplanehomethefirstplanehome
41
$endgroup$
add a comment |
$begingroup$
$ frac{sin A cos B + cos A sin B } {cos A cos B - sin A sin B} = frac{tan A + tan B}{1 - tan A tan B}$
I've tried multiplying by the reciprocal denominator on the left side to see if I could begin to simplify it to tangent, but I'm still confused. Any help would be appreciated.
trigonometry
edited Jan 29 at 21:26
abc...
3,237739
asked Jan 29 at 21:14
thefirstplanehomethefirstplanehome
41
$endgroup$
1
$begingroup$
Divide the numerator and denominator on the left by $cos Acos B$, and use the fact that $tan=sin/cos$. Alternatively, replace all the tan's on the right hand side with sin's over cos's and then simplify.
$endgroup$
– Barry Cipra
Jan 29 at 21:20
$begingroup$
Can you use that $tan(A)=frac{sin(A)}{cos(A)}?$ If yes, just do it in the RHS and put everything in terms of sin and cos in the RHS.
$endgroup$
– Phicar
Jan 29 at 21:20
add a comment |
$begingroup$
$ frac{sin A cos B + cos A sin B } {cos A cos B - sin A sin B} = frac{tan A + tan B}{1 - tan A tan B}$
I've tried multiplying by the reciprocal denominator on the left side to see if I could begin to simplify it to tangent, but I'm still confused. Any help would be appreciated.
trigonometry
edited Jan 29 at 21:26
abc...
3,237739
asked Jan 29 at 21:14
thefirstplanehomethefirstplanehome
41
$endgroup$
$ frac{sin A cos B + cos A sin B } {cos A cos B - sin A sin B} = frac{tan A + tan B}{1 - tan A tan B}$
I've tried multiplying by the reciprocal denominator on the left side to see if I could begin to simplify it to tangent, but I'm still confused. Any help would be appreciated.
trigonometry
trigonometry
edited Jan 29 at 21:26
abc...
3,237739
asked Jan 29 at 21:14
thefirstplanehomethefirstplanehome
41
edited Jan 29 at 21:26
abc...
3,237739
asked Jan 29 at 21:14
thefirstplanehomethefirstplanehome
41
edited Jan 29 at 21:26
abc...
3,237739
edited Jan 29 at 21:26
abc...
3,237739
edited Jan 29 at 21:26
abc...
3,237739
3,237739
asked Jan 29 at 21:14
thefirstplanehomethefirstplanehome
41
asked Jan 29 at 21:14
thefirstplanehomethefirstplanehome
41
asked Jan 29 at 21:14
thefirstplanehomethefirstplanehome
41
41
1
$begingroup$
Divide the numerator and denominator on the left by $cos Acos B$, and use the fact that $tan=sin/cos$. Alternatively, replace all the tan's on the right hand side with sin's over cos's and then simplify.
$endgroup$
– Barry Cipra
Jan 29 at 21:20
$begingroup$
Can you use that $tan(A)=frac{sin(A)}{cos(A)}?$ If yes, just do it in the RHS and put everything in terms of sin and cos in the RHS.
$endgroup$
– Phicar
Jan 29 at 21:20
add a comment |
1
$begingroup$
Divide the numerator and denominator on the left by $cos Acos B$, and use the fact that $tan=sin/cos$. Alternatively, replace all the tan's on the right hand side with sin's over cos's and then simplify.
$endgroup$
– Barry Cipra
Jan 29 at 21:20
$begingroup$
Can you use that $tan(A)=frac{sin(A)}{cos(A)}?$ If yes, just do it in the RHS and put everything in terms of sin and cos in the RHS.
$endgroup$
– Phicar
Jan 29 at 21:20
1
1
$begingroup$
Divide the numerator and denominator on the left by $cos Acos B$, and use the fact that $tan=sin/cos$. Alternatively, replace all the tan's on the right hand side with sin's over cos's and then simplify.
$endgroup$
– Barry Cipra
Jan 29 at 21:20
$begingroup$
Divide the numerator and denominator on the left by $cos Acos B$, and use the fact that $tan=sin/cos$. Alternatively, replace all the tan's on the right hand side with sin's over cos's and then simplify.
$endgroup$
– Barry Cipra
Jan 29 at 21:20
$begingroup$
Can you use that $tan(A)=frac{sin(A)}{cos(A)}?$ If yes, just do it in the RHS and put everything in terms of sin and cos in the RHS.
$endgroup$
– Phicar
Jan 29 at 21:20
$begingroup$
Can you use that $tan(A)=frac{sin(A)}{cos(A)}?$ If yes, just do it in the RHS and put everything in terms of sin and cos in the RHS.
$endgroup$
– Phicar
Jan 29 at 21:20
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$LHS=frac{sin(A+B)}{cos(A+B)}=tan (A+B)=RHS $ by the composite angle formula for $tan$.
answered Jan 29 at 21:33
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
$begingroup$
LHS=$frac{sin(A+B)}{cos(A+B)}=tan(A+B)=frac{tan A+tan B}{1-tan A tan B}$=RHS
Compound angle formulae of $sin, cos$ and $tan$ are used.
answered Jan 29 at 21:28
abc...abc...
3,237739
$endgroup$
add a comment |
$begingroup$
On the right hand side, put those tangents in terms of $sin$ and $cos$, and then multiply through to clear the denominators:
$$frac{tan A+tan B}{1-tan Atan B} = frac{frac{sin A}{cos A}+frac{sin B}{cos B}}{1-frac{sin Asin B}{cos Acos B}} = frac{sin Acos Bfrac{cos A}{cos A}+sin Bcos Afrac{cos B}{cos B}}{cos Acos B-sin Asin Bfrac{cos Acos B}{cos Acos B}}=frac{sin Acos B+sin Bcos A}{cos Acos B-sin Asin B}$$
Done.
These are, of course, two ways of expressing $tan(A+B)$.
answered Jan 29 at 21:41
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$LHS=frac{sin(A+B)}{cos(A+B)}=tan (A+B)=RHS $ by the composite angle formula for $tan$.
answered Jan 29 at 21:33
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
$begingroup$
$LHS=frac{sin(A+B)}{cos(A+B)}=tan (A+B)=RHS $ by the composite angle formula for $tan$.
answered Jan 29 at 21:33
Chris CusterChris Custer
14.3k3827
$endgroup$
add a comment |
$begingroup$
$LHS=frac{sin(A+B)}{cos(A+B)}=tan (A+B)=RHS $ by the composite angle formula for $tan$.
answered Jan 29 at 21:33
Chris CusterChris Custer
14.3k3827
$endgroup$
$LHS=frac{sin(A+B)}{cos(A+B)}=tan (A+B)=RHS $ by the composite angle formula for $tan$.
answered Jan 29 at 21:33
Chris CusterChris Custer
14.3k3827
answered Jan 29 at 21:33
Chris CusterChris Custer
14.3k3827
answered Jan 29 at 21:33
Chris CusterChris Custer
14.3k3827
answered Jan 29 at 21:33
Chris CusterChris Custer
14.3k3827
14.3k3827
add a comment |
add a comment |
$begingroup$
LHS=$frac{sin(A+B)}{cos(A+B)}=tan(A+B)=frac{tan A+tan B}{1-tan A tan B}$=RHS
Compound angle formulae of $sin, cos$ and $tan$ are used.
answered Jan 29 at 21:28
abc...abc...
3,237739
$endgroup$
add a comment |
$begingroup$
LHS=$frac{sin(A+B)}{cos(A+B)}=tan(A+B)=frac{tan A+tan B}{1-tan A tan B}$=RHS
Compound angle formulae of $sin, cos$ and $tan$ are used.
answered Jan 29 at 21:28
abc...abc...
3,237739
$endgroup$
add a comment |
$begingroup$
LHS=$frac{sin(A+B)}{cos(A+B)}=tan(A+B)=frac{tan A+tan B}{1-tan A tan B}$=RHS
Compound angle formulae of $sin, cos$ and $tan$ are used.
answered Jan 29 at 21:28
abc...abc...
3,237739
$endgroup$
LHS=$frac{sin(A+B)}{cos(A+B)}=tan(A+B)=frac{tan A+tan B}{1-tan A tan B}$=RHS
Compound angle formulae of $sin, cos$ and $tan$ are used.
answered Jan 29 at 21:28
abc...abc...
3,237739
answered Jan 29 at 21:28
abc...abc...
3,237739
answered Jan 29 at 21:28
abc...abc...
3,237739
answered Jan 29 at 21:28
abc...abc...
3,237739
3,237739
add a comment |
add a comment |
$begingroup$
On the right hand side, put those tangents in terms of $sin$ and $cos$, and then multiply through to clear the denominators:
$$frac{tan A+tan B}{1-tan Atan B} = frac{frac{sin A}{cos A}+frac{sin B}{cos B}}{1-frac{sin Asin B}{cos Acos B}} = frac{sin Acos Bfrac{cos A}{cos A}+sin Bcos Afrac{cos B}{cos B}}{cos Acos B-sin Asin Bfrac{cos Acos B}{cos Acos B}}=frac{sin Acos B+sin Bcos A}{cos Acos B-sin Asin B}$$
Done.
These are, of course, two ways of expressing $tan(A+B)$.
answered Jan 29 at 21:41
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
On the right hand side, put those tangents in terms of $sin$ and $cos$, and then multiply through to clear the denominators:
$$frac{tan A+tan B}{1-tan Atan B} = frac{frac{sin A}{cos A}+frac{sin B}{cos B}}{1-frac{sin Asin B}{cos Acos B}} = frac{sin Acos Bfrac{cos A}{cos A}+sin Bcos Afrac{cos B}{cos B}}{cos Acos B-sin Asin Bfrac{cos Acos B}{cos Acos B}}=frac{sin Acos B+sin Bcos A}{cos Acos B-sin Asin B}$$
Done.
These are, of course, two ways of expressing $tan(A+B)$.
answered Jan 29 at 21:41
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
On the right hand side, put those tangents in terms of $sin$ and $cos$, and then multiply through to clear the denominators:
$$frac{tan A+tan B}{1-tan Atan B} = frac{frac{sin A}{cos A}+frac{sin B}{cos B}}{1-frac{sin Asin B}{cos Acos B}} = frac{sin Acos Bfrac{cos A}{cos A}+sin Bcos Afrac{cos B}{cos B}}{cos Acos B-sin Asin Bfrac{cos Acos B}{cos Acos B}}=frac{sin Acos B+sin Bcos A}{cos Acos B-sin Asin B}$$
Done.
These are, of course, two ways of expressing $tan(A+B)$.
answered Jan 29 at 21:41
jmerryjmerry
16.9k11633
$endgroup$
On the right hand side, put those tangents in terms of $sin$ and $cos$, and then multiply through to clear the denominators:
$$frac{tan A+tan B}{1-tan Atan B} = frac{frac{sin A}{cos A}+frac{sin B}{cos B}}{1-frac{sin Asin B}{cos Acos B}} = frac{sin Acos Bfrac{cos A}{cos A}+sin Bcos Afrac{cos B}{cos B}}{cos Acos B-sin Asin Bfrac{cos Acos B}{cos Acos B}}=frac{sin Acos B+sin Bcos A}{cos Acos B-sin Asin B}$$
Done.
These are, of course, two ways of expressing $tan(A+B)$.
answered Jan 29 at 21:41
jmerryjmerry
16.9k11633
answered Jan 29 at 21:41
jmerryjmerry
16.9k11633
answered Jan 29 at 21:41
jmerryjmerry
16.9k11633
answered Jan 29 at 21:41
jmerryjmerry
16.9k11633
16.9k11633
add a comment |
add a comment |
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1
$begingroup$
Divide the numerator and denominator on the left by $cos Acos B$, and use the fact that $tan=sin/cos$. Alternatively, replace all the tan's on the right hand side with sin's over cos's and then simplify.
$endgroup$
– Barry Cipra
Jan 29 at 21:20
$begingroup$
Can you use that $tan(A)=frac{sin(A)}{cos(A)}?$ If yes, just do it in the RHS and put everything in terms of sin and cos in the RHS.
$endgroup$
– Phicar
Jan 29 at 21:20