sin^-1(cos 3π 5) equals

The expression is $\sin^{-1}(\cos\frac{3\pi}{5})$sin−1(cos53π​). To simplify, convert cosine into sine.

Using the identity $\cos\theta = \sin\left(\frac{\pi}{2} - \theta\right)$cosθ=sin(2π​−θ) with $\theta = \frac{3\pi}{5}$θ=53π​:

$$\cos\frac{3\pi}{5} = \sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right)$$cos53π​=sin(2π​−53π​)

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To subtract the fractions, find a common denominator:

$$\frac{\pi}{2} - \frac{3\pi}{5}$$2π​−53π​

$$= \frac{5\pi}{10} - \frac{6\pi}{10}$$=105π​−106π​

$$= \frac{-\pi}{10}$$=10−π​

Therefore:

$$\cos\frac{3\pi}{5} = \sin\left(\frac{-\pi}{10}\right)$$cos53π​=sin(10−π​)

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Substituting back into the original expression:

$$\sin^{-1}(\cos\frac{3\pi}{5})$$sin−1(cos53π​)

$$= \sin^{-1}\left(\sin\frac{-\pi}{10}\right)$$=sin−1(sin10−π​)

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For $\sin^{-1}(\sin x) = x$sin−1(sinx)=x, the value $x$x must be in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$[−2π​,2π​].

Since $-\frac{\pi}{2} < \frac{-\pi}{10} < \frac{\pi}{2}$−2π​<10−π​<2π​, the value is in the valid range.

Therefore:

$$\sin^{-1}\left(\sin\frac{-\pi}{10}\right) = \frac{-\pi}{10}$$sin−1(sin10−π​)=10−π​

The answer is $\frac{-\pi}{10}$10−π​.