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Inverse Trigonometric function example 02

Inverse Trigonometric function example 02

To find the value of \arcsin\left(\frac{1}{2}\right), we need to determine the angle whose sine is \frac{1}{2}.Let’s call this angle \theta. So, we have:

    \[\sin\theta = \frac{1}{2}\]

We know that for 0 \leq \theta \leq \frac{\pi}{2}, the sine function is positive, and it corresponds to the first quadrant. Therefore:

    \[\theta = \arcsin\left(\frac{1}{2}\right)\]

Using the Pythagorean identity \sin^2\theta + \cos^2\theta = 1, we can find the cosine of \theta:

    \[\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{1}{4} = \frac{3}{4}\]

    \[\cos\theta = \pm\sqrt{\frac{3}{4}} = \pm\frac{\sqrt{3}}{2}\]

Since \theta is in the first quadrant, \cos\theta is positive:

    \[\cos\theta = \frac{\sqrt{3}}{2}\]

Therefore, the exact value of \arcsin\left(\frac{1}{2}\right) is:

    \[\arcsin\left(\frac{1}{2}\right) = \theta = \frac{\pi}{6}\]

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