1. Can't start

Given that $sin(A+B) = 2sin(A-B)$, express $tan A$ in terms of $tan B$

2. Dear Punch,

$Sin(A-B)=SinACosB-SinBCosA$

$Sin(A+B)=SinACosB+SinBCosA$

Given the above relations I think you should be able to express TanA interms of TanB.

3. Originally Posted by Punch
Given that $sin(A+B) = 2sin(A-B)$, express $tan A$ in terms of $tan B$
$\sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \mp \cos{\alpha}\sin{\beta}$.

Given that $\sin{(A + B)} = 2\sin{(A - B)}$

$\sin{A}\cos{B} - \cos{A}\sin{B} = 2(\sin{A}\cos{B} + \cos{A}\sin{B})$

$\sin{A}\cos{B} - \cos{A}\sin{B} = 2\sin{A}\cos{B} + 2\cos{A}\sin{B}$

$-3\cos{A}\sin{B} = \sin{A}\cos{B}$

$\frac{-3\sin{B}}{\cos{B}} = \frac{\sin{A}}{\cos{A}}$

$-3\tan{B} = \tan{A}$.

4. Thanks Prove it.. i feel so grateful to you

I have another question,

Given that cos x = p, express sin4x in terms of p

5. Dear Punch,

$Sin4x=2Sin2xCos2x$

$Sin4x=2(2SinxCosx)Cos2x$

$Sin4x=4\sqrt{(1-Cos^2x)}Cosx(2Cos^2x-1)$

By substituting Cosx=p and further simplification will give you the answer.

Hope this helps.

6. Originally Posted by Sudharaka
Dear Punch,

$Sin4x=2Sin2xCos2x$

3. $Sin4x=2(2SinxCosx)Cos2x$

4. $Sin4x=4\sqrt{(1-Cos^2x)}Cosx(2Cos^2x-1)$

By substituting Cosx=p and further simplification will give you the answer.

Hope this helps.

I don't get how you get from the 3 to the 4

7. Originally Posted by Punch
Thanks Prove it.. i feel so grateful to you

I have another question,

Given that cos x = p, express sin4x in terms of p
$\sin{4x} = 2\sin{2x}\cos{2x}$

$= 2\cdot 2\sin{x}\cos{x}(\cos^2{x} - \sin^2{x})$

$= 4\cos{x}\sqrt{1 - \cos^2{x}}[\cos^2{x} - (1 - \cos^2{x})]$

$= 4\cos{x}(2\cos^2{x} - 1)\sqrt{1 - \cos^2{x}}$

$= 4p(2p^2 - 1)\sqrt{1 - p^2}$.

8. Dear Punch,

Do you know that,

$Sin2A=2SinACosA$

$SinA=\sqrt{1-Cos^2A}$

These are the two equations I used. In case you do not know them I think you should refer List of trigonometric identities - Wikipedia, the free encyclopedia When going from 3 to 4 I used the second identity.

Hope this helps.

9. got it! thanks!

10. Hello everyone

If I might make a slight correction:
Originally Posted by Prove It
$\sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \mp \cos{\alpha}\sin{\beta}$...
should, of course, be:
$\sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}$
So the final result will be:
$3\tan B = \tan A$