Can you find the range

Geometry Level 3

sin ⁡ θ 1 + tan ⁡ 2 θ − cos ⁡ θ 1 + cot ⁡ 2 θ \large \frac {\sin \theta}{\sqrt {1+\tan^2 \theta}} - \frac {\cos \theta}{\sqrt {1+ \cot^2 \theta}}

For − π ≤ θ ≤ π -\pi \leq \theta \leq \pi , find the range of the expression above.

[ − 1 , 1 ] [-1,1] No solution 0 [ 0 , 1 ] [0,1]

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Chinmay Kurade by Chinmay Kurade , 25, India

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1 solution

Chinmay Kurade
Jul 21, 2015

Using basic trigonometric identities, we get: sin ⁡ θ ∣ cos ⁡ θ ∣ − cos ⁡ θ ∣ sin ⁡ θ ∣ \sin \theta |\cos \theta| - \cos\theta |\sin \theta| . Now, taking cases for modulus, this is 0 for − π ≤ θ ≤ − π 2 -\pi \leq \theta \leq \frac {-\pi}{2} & 0 ≤ θ ≤ π 2 0 \leq \theta \leq \frac {\pi}{2} ; but it is 2 sin ⁡ θ cos ⁡ θ = sin ⁡ 2 θ 2\sin \theta \cos\theta =\sin 2\theta for − π 2 < θ < 0 \frac{-\pi}{2}<\theta<0 & similarly − sin ⁡ 2 θ -\sin 2\theta for π 2 < θ < π \frac {\pi}{2}<\theta<\pi . Solving for the range of these 4 cases we get the range of whole function as [ − 1 , 1 ] [-1,1]

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