How do you find the arcsin⁡(sin⁡(7π6)) ?

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gekraamdbk · Accepted Answer

The answer is:arcsin⁡(sin⁡(7π6))=−π6The range of a function arcsin⁡(x) is, by definition.−π2≤arcsin⁡(x)≤π2It means that we have to find an angle α that lies between −π2 and π2 and whose sin⁡(α) equals to a sin⁡(7π6).From trigonometry we know thatsin⁡(ϕ+π)=−sin⁡(ϕ)for any angle ϕ.This is easy to see if use the definition of a sine as an ordinate of the end of a radius in the unit circle that forms an angle ϕ  with the X-axis (counterclockwise from the X-axis to a radius).We also know that since is an odd function, that is sin⁡(−ϕ)=−sin⁡(ϕ)We will use both properties as follows:sin⁡(7π6)=sin⁡(π6+π)=−sin⁡(π6)=sin⁡(−π6)As we see, the angle α=−π6 first our conditions. It is in the range from −π2 to π2 and its sine equals to sin⁡(7π6). Therefore, it&#039;s a correct answer to a problem.

enguinhispi · Accepted Answer

So:  arcsin=sin−1  sin−1(sin⁡(7π6))  Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant.  sin−1(−sin⁡(π6))  The exact value of sin⁡(π6) is 12.  sin−1(−12)  The exact value of sin−1(−12) is −π6  The result can be shown in multiple forms.  Exact Form:  −π6  Decimal Form:  −0.52359877…

How do you find the \arcsin(\sin(\frac{7\pi}{6})) ?

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