How do you find $sin ({sin}^{- 1} (\frac{1}{4}))$ $sin(sin^-1 (1/4))$ ?

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Answer 1 · 2015-04-14T11:37:14

Answer is $\frac{1}{4}$ $1/4$
Recall that $sin [arcsin (x)] = x$ $sin[arcsin(x)] = x$ for $x$ $x$ in the domain of $arcsin (x)$ $arcsin(x)$ .

(The domain of $arcsin (x)$ $arcsin(x)$ is: $- 1 \leq x \leq 1$ $-1 <= x <= 1$ .)

Therefore $sin [arcsin (\frac{1}{4})] = \frac{1}{4} .$ $sin[arcsin(1/4)] = 1/4.$

How do you find sin(sin^-1 (1/4))sin(sin−1(14))sin(sin^-1 (1/4))?