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Question

Mathematics Question on Inverse Trigonometric Functions

The value of sin(2sin10.8)\sin (2\, \sin^{-1}\, 0.8) is equal to

A

sin 1.21.2^{\circ}

B

0.960.96

C

0.480.48

D

sin 1.61.6^{\circ}

Answer

0.960.96

Explanation

Solution

Let E=sin(2sin10.8)E =\sin \left(2 \sin ^{-1} 0.8\right)
Put sin10.8=θ\sin ^{-1} 0.8=\theta
sinθ=0.8\Rightarrow \sin \theta=0.8
cosθ=1sin2θ\therefore \cos \theta =\sqrt{1-\sin ^{2} \theta}
=10.64=0.36=0.6=\sqrt{1-0.64}=\sqrt{0.36}=0.6
E=sin(2θ)=2sinθcosθ\therefore E =\sin (2 \theta)=2 \sin \theta \cos \theta
=2×0.8×0.6=0.96=2 \times 0.8 \times 0.6=0.96