How do you find the exact values of ${sin}^{- 1} (\frac{1}{2})$ $sin^-1(1/2)$ ?

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How do you find the exact values of ${sin}^{- 1} (\frac{1}{2})$ $sin^-1(1/2)$ ?

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Answer 1 · 2016-10-15T17:47:23

${sin}^{- 1} (\frac{1}{2}) = \frac{π}{6}$ $sin^-1(1/2)=pi/6$

Explanation:

Find ${sin}^{- 1} (\frac{1}{2})$ $sin^-1(1/2)$

This problem is asking for the ANGLE with a sine of $\frac{1}{2}$ $1/2$ .

The range of ${sin}^{- 1}$ $sin^-1$ or $arcsin$ $arcsin$ is between $\frac{π}{2}$ $pi/2$ and $- \frac{π}{2}$ $-pi/2$ .

If you are finding ${sin}^{- 1}$ $sin^-1$ of a positive value, the answer will be between 0 and $\frac{π}{2}$ $pi/2$ , or the first quadrant in the unit circle. Do NOT use the second quadrant angle with a sine of $\frac{1}{2}$ $1/2$ , because it does not fall within the range of ${sin}^{- 1}$ $sin^-1$ .

Using the unit circle, the angle with a sine of $\frac{1}{2}$ $1/2$ in the first quadrant is $\frac{π}{6}$ $pi/6$ .

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How do you find the exact values of ${sin}^{- 1} (\frac{1}{2})$ $sin^-1(1/2)$ ?

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How do you find the exact values of ${sin}^{- 1} (\frac{1}{2})$ $sin^-1(1/2)$ ?