Given that tanx=p and sin2x=(2p)/(1+p^2). Without using table or calculator, find the value of tan15.

I let tan15=p, tan30=(2p)/(1-p^2)

1/(√3)= 2p/(1-p^2)

1-p^2= 2√3p

p^2 + 2√3p-1=0

I still can't get the answer..

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- Jun 24th 2013, 03:47 AMTrefoil2727Trigonometric function
Given that tanx=p and sin2x=(2p)/(1+p^2). Without using table or calculator, find the value of tan15.

I let tan15=p, tan30=(2p)/(1-p^2)

1/(√3)= 2p/(1-p^2)

1-p^2= 2√3p

p^2 + 2√3p-1=0

I still can't get the answer.. - Jun 24th 2013, 04:06 AMPlatoRe: Trigonometric function
- Jun 24th 2013, 04:21 AMProve ItRe: Trigonometric function

Now if we have , so

And since that means , which means the solution is . - Jun 25th 2013, 03:27 AMibduttRe: Trigonometric function
Tan x = p,

sin 2x = 2p/(1+p^2) let x = 15, Then we have

sin 30 = 2p /(1+p^2) = ½ that gives

p^2-4p+1=0

Thus p = (4 ±√(16-12))/2 = (4 ±2√3)/2= 2 ±√3

Because 0 < 15 < 90 that is in first quadrant, we have

tan 15 = = 2- √3