Question - Finding Tangent of Angle with Known Sine
Solution:
The expression given is:\[ \tan(\sin^{-1}(\frac{\sqrt{2}}{2})) \]Here, we are asked to find the tangent of the angle whose sine is $$\frac{\sqrt{2}}{2}$$.Recognize that $$\sin^{-1}(\frac{\sqrt{2}}{2})$$ corresponds to the angle whose sine is $$\frac{\sqrt{2}}{2}$$. This is an angle in a right triangle where the side opposite the angle has length $$\sqrt{2}$$ and the hypotenuse has length 2. By the Pythagorean theorem, the length of the adjacent side of this right triangle will be:\[ \text{Adjacent side} = \sqrt{\text{Hypotenuse}^2 - \text{Opposite side}^2} \]\[ \text{Adjacent side} = \sqrt{2^2 - (\sqrt{2})^2} \]\[ \text{Adjacent side} = \sqrt{4 - 2} \]\[ \text{Adjacent side} = \sqrt{2} \]Now we have a right triangle where the opposite side is $$\sqrt{2}$$, the adjacent side is also $$\sqrt{2}$$, and the hypotenuse is 2. To find the tangent of the angle, we simply divide the opposite side by the adjacent side:\[ \tan(\sin^{-1}(\frac{\sqrt{2}}{2})) = \frac{\text{Opposite side}}{\text{Adjacent side}} \]\[ \tan(\sin^{-1}(\frac{\sqrt{2}}{2})) = \frac{\sqrt{2}}{\sqrt{2}} \]\[ \tan(\sin^{-1}(\frac{\sqrt{2}}{2})) = 1 \]Therefore, the value of the expression is 1.
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