Evaluate each of the following:\(sin^{-1}\{(sin\frac{17\pi}8)\}\)

1.	Evaluate each of the following:\(sin^{-1}\{(sin\frac{17\pi}8)\}\)
Answer» As we know sin(–θ) is –sin(θ ) ∴ We can write \((sin\frac{-17\pi}8)\) as \(-sin(\frac{17\pi}8)\) Now \(-sin(\frac{17\pi}8)\) = \(-sin(2\pi+\frac{\pi}8)\) As we know sin(2π +θ) = sin(θ ) So \(-sin(2\pi+\frac{\pi}8)\) can be written as \(-sin(\frac{\pi}8)\) And \(-sin(\frac{\pi}8)\) = \(sin(\frac{-\pi}8)\) As sin^–1(sin x) = x Provided x ∈ \([\frac{-\pi}2,\frac{\pi}2]\) ∴ we can write \(sin^{-1}(sin \frac{-\pi}8)=\frac{-\pi}8\)

Answer»

As we know sin(–θ) is –sin(θ )

∴ We can write \((sin\frac{-17\pi}8)\) as \(-sin(\frac{17\pi}8)\)

Now \(-sin(\frac{17\pi}8)\) = \(-sin(2\pi+\frac{\pi}8)\)

As we know sin(2π +θ) = sin(θ )

So \(-sin(2\pi+\frac{\pi}8)\) can be written as \(-sin(\frac{\pi}8)\)

And \(-sin(\frac{\pi}8)\) = \(sin(\frac{-\pi}8)\)

As sin^–1(sin x) = x

Provided x ∈ \([\frac{-\pi}2,\frac{\pi}2]\)

∴ we can write \(sin^{-1}(sin \frac{-\pi}8)=\frac{-\pi}8\)

Discussion