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### Find $\tan2\theta$

Posted: Wed Dec 19, 2012 9:27 pm
Suppose $\theta$ is an angle between $0$ and $\frac{\pi}{2}$ such that $\sin5\theta=\sin^5\theta$. The number $\tan2\theta$ can be uniquely written as $a\sqrt{b}$, where $a$ and $b$ are positive integers, and $b$ is not divisible by the square of a prime. What is the value of $a+b$ ?

### Re: Find $\tan2\theta$

Posted: Thu Dec 20, 2012 12:49 pm
Hint:

### Re: Find $\tan2\theta$

Posted: Thu Dec 20, 2012 12:56 pm
$sin5\theta= 5 sin\theta- 20 sin^3 \theta+ 16 sin^5 \theta$

$sin^5\theta= 5 sin\theta- 20 sin^3 \theta+ 16 sin^5 \theta$
$15sin^5\theta - 20 sin^3 \theta + 5 sin\theta = 0$
$3sin^4\theta - 4 sin^2 \theta + 1 = 0$
$(sin^2\theta-1)(3sin^2\theta-1)=0$

We will get $sin\theta=1, -1 , \frac{1}{√3}, -\frac{1}{√3}$
As $0<\theta<\frac{\pi}{2}$, $0<sin\theta<1$.

So $sin\theta = \frac{1}{√3}$
$sin^2\theta = \frac{1}{3}$
$cosec^2\theta = 3$
$cot^2\theta+1 = 3$
$cot^2\theta = 2$
$tan\theta = \frac{1}{√2}$

$tan 2\theta= \frac{2tan\theta}{1-tan^2\theta}$
$tan 2\theta= \frac{2*\frac{1}{√2}}{1-\frac{1}{2}}$
$tan 2\theta= 2√2$

$2+2=4$

### Re: Find $\tan2\theta$

Posted: Thu Dec 20, 2012 1:17 pm
HOW $sin5\theta= 5 sin\theta- 20 sin^3 \theta+ 16 sin^5 \theta$ ?

$sin5\theta$
$=sin(2\theta +3\theta)$
$= sin2\theta cos3\theta +cos2\theta sin3\theta$

$= 2sin\theta cos\theta(4cos^3\theta - 3cos\theta) + (1-2sin^2\theta)(3sin \theta-4sin^3\theta)$

$=2sin\theta cos^2\theta(4cos^2\theta-3) + sin\theta (1-2sin^2\theta)(3-4sin^2\theta)$

$=2 sin\theta (1- sin^2\theta)[4(1-sin^2\theta)-3] + 3sin \theta-10sin^3\theta + 8sin^5\theta$

$=(2sin\theta -2sin^3\theta)(1-4sin^2\theta) + 3sin \theta -10sin^3\theta + 8sin^5\theta$

$=2sin\theta - 2sin^3\theta -8sin^3\theta +8sin^5\theta+3sin \theta -10sin^3\theta + 8sin^5 \theta$

$= 5sin\theta- 20sin^3\theta +16 sin^5\theta$

### Re: Find $\tan2\theta$

Posted: Sun Mar 11, 2018 12:53 pm
0+0
Take theetha 0