To find
\[sin(2 sin^{-1}x)\]
substitute \[\theta = sin^{-1}x\]
then\[sin(2 sin^{-1}x)=sin(2 \theta)=2 sin \theta cos \theta =2x \sqrt{1-x^2}\]
(
\[cos \theta = \sqrt{1-sin^2 \theta}=\sqrt{1-x^2}\]
) Simplifying Expressions With Trigonometric Functions of Inverse Trigonometric Functions
We can simplify expressions with sines and cosines of inverse trigonometric functions using substitutions.
To find
(
To find
(