Evaluate $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4})$

Question

My attempt:

Method:- 1

Let $\sin^{-1}\frac{3}{4} = \theta \implies \sin\theta = \frac{3}{4}$ and $\theta \in [0,\frac{π}{2}]$ Therefore $$\cos\theta = \sqrt{1-\sin^2\theta} = \frac{\sqrt 7}{4}$$

Now $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4}) = \tan\frac{\theta}{2} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \sqrt{ \frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \frac{4-\sqrt7}{3}$

Method:-2

Let $\sin^{-1}\frac{3}{4} = \theta \implies \sin\theta = \frac{3}{4} \implies 2\sin\frac{\theta}{2} \cos\frac{\theta}{2} = \frac{3}{4} \implies \sin\frac{\theta}{2} \cos\frac{\theta}{2} = \frac{3}{8}$

Now $\tan\frac{\theta}{2} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}$

Please help me in 2nd method. Thanks in advance.

Hint: $\tan x=\tan2\dfrac x2=\dfrac{2\tan\dfrac x2}{1-\tan^2\dfrac x2}$. This gives you a quadratic equation in $\tan\dfrac x2$. — , Sep 16 '21 at 17:46

score 2 · Answer 1 · answered Sep 16 '21 at 17:38

So you have $\sin\frac{\theta}{2} \cos\frac{\theta}{2} = \sqrt{1-\cos^2\frac{\theta}{2}}\cos\frac{\theta}{2}=\frac{3}{8}$
Solving $(1-x^2)x^2=\frac{9}{64} \implies x^4-x^2+\frac{9}{64}=0 \implies x^2=\frac{1+\sqrt{1-\frac{9}{16}}}{2}=\frac{4+\sqrt{7}}{8} $
$\tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{\cos^2\frac{\theta}{2}}=\frac{3\cdot 8}{8(4+\sqrt{7})}=\frac{3}{4+\sqrt{7}}=\frac{4-\sqrt{7}}{3}$

score 2 · Accepted Answer · answered Sep 16 '21 at 17:50

Your Method 2:

Since $$\sin\theta=2\sin\frac\theta2\cos\frac\theta2\\=2\tan\frac\theta2\cos^2\frac\theta2\\=\frac{2\tan\frac\theta2}{\sec^2\frac\theta2}\\=\frac{2\tan\frac\theta2}{1+\tan^2\frac\theta2}$$ and $$\sin\theta=\frac34,$$ therefore $$\frac{2\tan\frac\theta2}{1+\tan^2\frac\theta2}=\frac34\\3\tan^2\frac\theta2-8\tan\frac\theta2+3=0\\\tan\frac\theta2=\frac{4-\sqrt7}3.$$

Evaluate $\tan(\frac{1}{2}\sin^{-1}\frac{3}{4})$

2 Answers2