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Worked solution

Recover cosine from inverse sine

Given that θ=sin⁡−1(x/2)\theta=\sin^{-1}(x/2), where x>0x>0, find in terms of xx (i) cos⁡θ\cos\theta (ii) sin⁡2θ\sin2\theta.

CJC · Discussion · Question 13

Answer

(i) 4−x2/2\sqrt{4-x^2}/2. (ii) x4−x2/2x\sqrt{4-x^2}/2, for 0<x≤20<x\le2.

(i) Use the principal inverse-sine range

Since sin⁡θ=x/2\sin\theta=x/2 and the inverse sine is real, 0<x≤20<x\le2 and 0<θ≤π/20<\theta\le\pi/2. The reference triangle has opposite side xx, hypotenuse 22 and adjacent side 4−x2\sqrt{4-x^2}. Therefore cos⁡θ=4−x2/2\cos\theta=\sqrt{4-x^2}/2. At x=2x=2, the triangle is a limiting picture and the formula still gives zero.

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(ii) Apply the double-angle identity

sin⁡2θ=2sin⁡θcos⁡θ=2(x/2)(4−x2/2)=x4−x2/2\sin2\theta=2\sin\theta\cos\theta=2(x/2)(\sqrt{4-x^2}/2)=x\sqrt{4-x^2}/2.

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Right triangle with angle theta at the left, opposite side x, hypotenuse 2 and adjacent side square root of (4 minus x squared). Valid as a triangle for 0 < x < 2; at x = 2 the formulas extend to theta = pi/2.
Right triangle with angle theta at the left, opposite side x, hypotenuse 2 and adjacent side square root of (4 minus x squared). Valid as a triangle for 0 < x < 2; at x = 2 the formulas extend to theta = pi/2.
Right triangle with angle theta at the left, opposite side x, hypotenuse 2 and adjacent side square root of (4 minus x squared). Valid as a triangle for 0 < x < 2; at x = 2 the formulas extend to theta = pi/2.

Common mistakes

Choose the nonnegative cosine because the principal inverse-sine angle is in the first quadrant here.

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