(3 ^ -1 (2 5 ) )= - Vidyadip

MCQ

$\sin \left(3 \sin ^{-1}\left(\frac{2}{5}\right)\right)=$

✓
$\frac{118}{125}$
B
$\frac{115}{127}$
C
$\frac{128}{135}$
D
$\frac{110}{118}$

✓

Answer

Correct option: A.

$\frac{118}{125}$

(A) $\sin \left(3 \sin ^{-1}\left(\frac{2}{5}\right)\right)=\sin 3 \theta$,
Where $\theta=\sin ^{-1}\left(\frac{2}{5}\right) \ldots\left[\theta=\sin ^{-1}\left(\frac{2}{5}\right), \sin \theta=\frac{2}{5}\right]$
$=3 \sin \theta-4 \sin ^3 \theta$
$=3\left(\frac{2}{5}\right)-4\left(\frac{2}{5}\right)^3 \ldots\left[\theta=\sin ^{-1}\left(\frac{2}{5}\right), \sin \theta=\frac{2}{5}\right]$
$=\frac{6}{5}-\frac{32}{125}=\frac{118}{125}$

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 "MCQ" from Trigonometric Functions→Trigonometric Functions — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

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

If $A=\left[\begin{array}{cc}x & -2 \\ 3 & 7\end{array}\right]$ and $A^{-1}=\left[\begin{array}{cc}\frac{7}{34} & \frac{1}{17} \\ \frac{-3}{34} & \frac{2}{17}\end{array}\right]$, then the value of $x$ is

$\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\sin|\text{x}|\text{dx}$ is equal to:

If d is the determinant of a square matrix A of order n, then the determinant of its adjoint is

The vectors $\vec{a}=x \hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=\hat{i}+y \hat{j}-z \hat{k}$ are collinear, if

India play two matches each with West indies and Australia. In any match the probability of india getting $0,1$ and $2$ points are $0.45, 0.05$ and $0.50$ respectively. Assuming that the outcomes are indepecdent, the probability of india getting at least 7 points is.

Let $X$ be a random variable which assumes values $\quad x_1, x_2, x_3, x_4$ such that $2 \mathrm{P}\left(\mathrm{X}=x_1\right)=3 \mathrm{P}\left(\mathrm{X}=x_2\right)=\mathrm{P}\left(\mathrm{X}=x_3\right)=5 \mathrm{P}\left(\mathrm{X}=x_4\right)$, Then, the probability distribution of X is

$\cos \theta\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]+\sin \theta\left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta\end{array}\right]=$

The angle between the planes and $\bar{r} \cdot(\bar{i}-2 \bar{j}+3 \bar{k})+4=0$ and

$\bar{r} \cdot(2 \bar{i}+\bar{j}-3 \bar{k})+7=0$ is

If a corner point of the feasible solutions are $(0,10)(2,2)(4,0)(3,2)$ then the point of $\operatorname{minimum} Z=3 x+2 y$ is