The value of (1 4 ^ -1 63 8 ) is: - Vidyadip

MCQ

The value of $\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$ is:

A
$\frac{1}{\sqrt2}$
B
$\frac{1}{\sqrt3}$
✓
$\frac{1}{2\sqrt2}$
D
$\frac{1}{3\sqrt3}$

✓

Answer

Correct option: C.

$\frac{1}{2\sqrt2}$

$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
Let, $\sin^{-1}\frac{\sqrt{63}}{8}=\text{x}$
$\sin\text{x}=\frac{\sqrt{63}}{8}$
$\cos\text{x}\sqrt{1-\sin^2\text{x}}$
$\cos\text{x}=\sqrt{1-\frac{63}{64}}$
$\cos\text{x}=\frac{1}{8}$
Consider,
$\sin\Big(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\Big)$
$=\sin\Big(\frac{1}{4}\text{x}\Big)$
$=\sqrt{\frac{1-\cos\frac{\text{x}}{2}}{2}}$ $\Big(\because \sin\text{x}=\frac{1-\cos2\text{x}}{2}\Big)$
$=\sqrt{\frac{1-\sqrt{\frac{1+\cos\text{x}}{2}}}{2}}$ $\Big(\because \cos\text{x}=\frac{1+\cos2\text{x}}{2}\Big)$
$=\sqrt{\frac{1-\sqrt{1-\frac{1}{8}}}{2}}$
$=\sqrt{\frac{1-\frac{3}{4}}{2}}$
$=\sqrt{\frac{1}{8}}$
$=\frac{1}{2\sqrt2}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Inverse Trigonometric Functions→Inverse Trigonometric Functions — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The point at the curve $y=12 x-x^2$ where the slope of the tangent is zero will be:

Let $\mathrm{f}:(-1, \infty) \rightarrow \mathrm{R}$ be defined by $\mathrm{f}(0)=1$ and $f(x)=\frac{1}{x} \log _{e}(1+x), x \neq 0 .$ Then the function $f$

For $x\,\, \in \,R\,,x\, \ne \,0,$ let ${f_0}(x) = \frac{1}{{1 - x}}$ and ${f_{n + 1}}(x) = {f_0}({f_n}(x)),$ $n\, = 0,1,2,....$ Then the value of ${f_{100}}(3) + {f_1}\left( {\frac{2}{3}} \right) + {f_2}\left( {\frac{3}{2}} \right)$ is equal to

Let $f$ be a real-valued differentiable function on $R$ (the set of all real numbers) such that $f(1)=1$. If the $y$ intercept of the tangent at any point $P(x, y)$ on the curve $y=f(x)$ is equal to the cube of the abscissa of $P$, then the value of $f(-3)$ is equal to

The minimum value of function $f(x)=2 \cos x+x$ in interval $\left[0, \frac{\pi}{2}\right]$ :

If $\phi(x)=\frac{1}{\sqrt{x}} \frac{\pi}{4} \int \limits_0^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t, \quad x > 0$ then $\phi^{\prime}\left(\frac{\pi}{4}\right)$ is equal to :

If $a = 2i + 2j - k$ and $|x\,a|\,\, = 1,$ then $ x =$

If R is the largest equivalence relation on a set A and S is any relation on A, then:

The value of objective function is maximum under linear constraints

Let $f : R \to R, f(x) = max.\{|tan^{-1}x|, cot^{-1}x\}.$ Consider the following statements :
$I.$ Function is continuous and derivable $\forall x \in R$
$II.$ Range of function is $\left[ {\frac{\pi }{4},\pi } \right]$
$III.$ $f(x)$ is many one-into. Identify the correct option -