value of an Inverse trigonometry expression

Question

How can i find the value of $\alpha = \arcsin\frac{\sqrt{63}}{8}$ to substitute in the expression , to value of $\sin^2(\frac\alpha4)$ ?

Answer 1

You actually don't need do calculate value of $\arcsin\frac{\sqrt{63}}{8}$.

As $$\sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1 - \cos\alpha}{2}}, \quad \cos\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1 + \cos\alpha}{2}}$$

we get $$ \sin^2\left(\frac{\alpha}{4}\right) = \frac{1 - \cos\left(\frac{\alpha}{2}\right)}{2} = \frac{1 - \sqrt{\frac{1 + \cos\alpha}{2}}}{2} = \frac{1 - \sqrt{\frac{1 + \sqrt{1 - \sin^2\alpha}}{2}}}{2}. $$ Here we substitute $\sin\alpha = \frac{\sqrt{63}}{8}$ and get $$ \sin^2\left(\frac{\alpha}{4}\right) = \frac{1}{8}. $$

Answer 2

Let $\arcsin\dfrac{\sqrt{63}}8=y$

As $-\dfrac\pi2\le\arcsin x\le\dfrac\pi2,0<y<\dfrac\pi2$

$\implies\cos y>0\implies\cos y=+\sqrt{1-\sin^2y}=\dfrac18$

Use $\cos2A=2\cos^2A-1=1-2\sin^2A$

to find $\cos^2\dfrac y2=?$

For obvious reason, $\cos\dfrac y2>0,$

Now $\cos\dfrac y2=1-2\sin^2\dfrac y4\iff\sin^2\dfrac y4=?$

Answer 3

We have, from the formulas for $\sin^2\frac{\theta}2$ and $\cos\frac{\theta}2$ \begin{align*} \sin^2\left(\frac{\alpha}{4}\right)&=\frac{1-\cos\left(\frac{\alpha}{2}\right)}{2}\tag{1}\\[5pt] \cos\left(\frac{\alpha}2\right)&=\sqrt{\frac{1+\cos \alpha}2}\tag{2} \end{align*}

From the Pythagorean identity $\qquad\sin^2\alpha+\cos^2\alpha=1\quad$ we get $$\cos \alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac{63}{64}}=\frac18$$ Plugging this into $(2)$ he have $$\cos\left(\frac{\alpha}2\right)=\sqrt{\frac{1+\frac18}2}=\frac34$$ Finally, from $(1)$ $$\sin^2\left(\frac{\alpha}{4}\right)=\frac{1-\frac34}{2}=\color{blue}{\frac18}$$