anonymous
  • anonymous
sina-cosatanb/cosa+sinatanb = tan(a+b) Prove the identity
Trigonometry
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
divide both the numerator and denominator by cosa
welshfella
  • welshfella
are you sure the question is correct?
anonymous
  • anonymous
Yeah, the question is correct

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anonymous
  • anonymous
Why divide cosa?
anonymous
  • anonymous
What's the next step after that?
welshfella
  • welshfella
I plugged in 2 values of a and b and didn't get the 2 sides to be equal
anonymous
  • anonymous
Ohh
anonymous
  • anonymous
What about this question? 2cos^23-1
welshfella
  • welshfella
Yes there is a mistake in the first question.
anonymous
  • anonymous
How do you solve 2cos^23-1
welshfella
  • welshfella
is that supposed to be an equation? there is no variable
anonymous
  • anonymous
I think you're meant to simplify it
welshfella
  • welshfella
is the 3 degrees or radians?
anonymous
  • anonymous
Woops I just realised I wrote the question wrong
anonymous
  • anonymous
It's actually meant to be 2cos^23x-1
anonymous
  • anonymous
Sorry
anonymous
  • anonymous
X is the variable
welshfella
  • welshfella
2 cos^2 3x = 1 so cos ^2 3x = 1/2 cos 3x = +/- 1 / sqrt2
Hero
  • Hero
Taking a look at your original question
Hero
  • Hero
The original question says to Prove The Identity, but the given trig equation is not an identity. Double check all the Variables to make sure they are correct.
Hero
  • Hero
Apparently the expression on the left side is equivalent to tan(a-b) not tan (a+b)
Hero
  • Hero
We can prove this identity: (sina-cosatanb )/(cosa+sinatanb) = tan(a - b)
welshfella
  • welshfella
yes one way to do this would be to convert tan b to sin b / cos b and simplify
welshfella
  • welshfella
and finally convert back to a tangent
welshfella
  • welshfella
Hint: use compound angle formula for sin and cos
Hero
  • Hero
@welshfella give @Cococute a chance to respond first
welshfella
  • welshfella
ok
anonymous
  • anonymous
Thank you @welshfella and @Hero
Hero
  • Hero
YW
anonymous
  • anonymous
I changed the tanb to sinb/cosb and got this: sina-cosaxsinb/cosb / cosa+sinaxsinb/cosb
anonymous
  • anonymous
What's the next step?
prizzyjade
  • prizzyjade
\[\frac{ \sin \alpha-\cos \alpha \tan \beta}{\cos \alpha+\sin \\[=\frac{ \frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta }- \tan \beta }{ \cos^2 \alpha -\sin^2 \alpha \tan^2 \beta }\]alpha \tan \beta }= \tan(\alpha+\beta)\]\[\tan (\alpha+\beta)=\frac{ \sin \alpha -\cos \alpha \tan \beta}{ \cos \alpha+ \sin \alpha \tan \beta }\] \[=\frac{ \sin \alpha-\cos \alpha \tan \beta }{ \cos \alpha+ \sin \alpha \tan \beta } \times \frac{ \cos \alpha-\sin \alpha \tan \beta }{ \cos \alpha-\sin \alpha \tan \beta }\] \[=\frac{ \sin \alpha \cos \alpha-\sin ^{2}\alpha \tan \beta- \cos^2 \alpha \tan \beta +\cos \alpha \sin \alpha \tan^2 \beta }{ \cos^2 \alpha-\sin^2 \alpha \tan ^2 \beta}\] \[=\frac{ \sin \alpha \cos \alpha (1+\tan^2 \beta) -\tan \beta (\sin^2 \alpha + \cos^2 \alpha )}{ \cos^2 \alpha - \sin^2 \alpha \tan ^2 \beta}\]\[=\frac{ \sin \alpha \cos \alpha(\sec^2 \beta)- \tan \beta (1) }{ \cos^2 \alpha- \sin^2 \alpha \tan^2 \beta}\]
prizzyjade
  • prizzyjade
\[=\frac{ \frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta }- \tan \beta }{ \cos^2 \alpha - \sin^2 \alpha \tan^2 \beta }\] \[=\frac{\frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta} -\frac{ \sin \beta}{ \cos \beta }}{\cos^2\alpha- \frac{ \sin^ 2 \alpha \sin^2 \beta }{\cos^2 \beta } }\]\[=\frac{ \frac{ \sin \alpha \cos \alpha- \sin \beta \cos \beta }{ \cos^2 \beta } }{ \frac{ \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin ^2 \beta }{\cos^2 \beta } }\]\[=\frac{ \sin \alpha \cos \alpha -\sin \beta \cos \beta }{ \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin^2 \beta }\]\[=\frac{ \frac{ \tan \alpha }{ 1+\tan^2 \alpha } - \frac{ \tan \beta }{ 1+ \tan^\beta } }{ 2-\sin^2 \alpha - \sin^ 2 \beta + \sin^2 \alpha \sin^2 \beta-\sin^2 \alpha \sin^2 \beta}\]
anonymous
  • anonymous
work from left, sub in the addition formula for tan(A+B)

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