home > s25 ma2090 - dambroise precalculus > assessment\nhomework #16 - inverse sine, cosine, tangent\nscore…

home > s25 ma2090 - dambroise precalculus > assessment\nhomework #16 - inverse sine, cosine, tangent\nscore: 13/38 answered: 5/16\nquestion 4\nevaluate the following expressions. your answer must be an angle in radians and in the interval -π/2, π/2.\n(a) sin⁻¹(√2/2) =\n(b) sin⁻¹(-√3/2) =\n(c) sin⁻¹(1/2) =\nquestion help: video 1 video 2\nsubmit question

home > s25 ma2090 - dambroise precalculus > assessment\nhomework #16 - inverse sine, cosine, tangent\nscore: 13/38 answered: 5/16\nquestion 4\nevaluate the following expressions. your answer must be an angle in radians and in the interval -π/2, π/2.\n(a) sin⁻¹(√2/2) =\n(b) sin⁻¹(-√3/2) =\n(c) sin⁻¹(1/2) =\nquestion help: video 1 video 2\nsubmit question

Answer

Explanation:

Step1: Recall sine - inverse sine relationship

The inverse - sine function (y = \sin^{-1}(x)) gives an angle (y\in[-\frac{\pi}{2},\frac{\pi}{2}]) such that (\sin(y)=x).

Step2: Solve part (a)

We know that (\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}) and (\frac{\pi}{4}\in[-\frac{\pi}{2},\frac{\pi}{2}]), so (\sin^{-1}(\frac{\sqrt{2}}{2})=\frac{\pi}{4}).

Step3: Solve part (b)

We know that (\sin(-\frac{\pi}{3})=-\frac{\sqrt{3}}{2}) and (-\frac{\pi}{3}\in[-\frac{\pi}{2},\frac{\pi}{2}]), so (\sin^{-1}(-\frac{\sqrt{3}}{2})=-\frac{\pi}{3}).

Step4: Solve part (c)

We know that (\sin(\frac{\pi}{6})=\frac{1}{2}) and (\frac{\pi}{6}\in[-\frac{\pi}{2},\frac{\pi}{2}]), so (\sin^{-1}(\frac{1}{2})=\frac{\pi}{6}).

Answer:

(a) (\frac{\pi}{4}) (b) (-\frac{\pi}{3}) (c) (\frac{\pi}{6})