Mathematics

Mathematics

Problem

If sin⁡θ=12\sin\theta=\dfrac{1}{2} and 0<θ<π0<\theta<\pi, then the value of cos⁡(θ−π2)\cos\left(\theta-\dfrac{\pi}{2}\right) is a) −12-\dfrac{1}{2} b) 00 c) 12\dfrac{1}{2} d) 32\dfrac{\sqrt{3}}{2}
Intuitive explanation

The key is recognizing that cos⁡(θ−π2)\cos\left(\theta-\dfrac{\pi}{2}\right) can be simplified using a cofunction identity rather than finding θ\theta explicitly. Since cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B, or more directly, recall that cos⁡(x−π2)=sin⁡x\cos\left(x - \dfrac{\pi}{2}\right) = \sin x. Once you apply this identity, your answer depends only on the given value of sin⁡θ\sin\theta, not on which specific angle θ\theta is.

Intuitive explanation

The key is recognizing that cos⁡(θ−π2)\cos\left(\theta-\dfrac{\pi}{2}\right) can be simplified using a cofunction identity rather than finding θ\theta explicitly. Since cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin B, or more directly, recall that cos⁡(x−π2)=sin⁡x\cos\left(x - \dfrac{\pi}{2}\right) = \sin x. Once you apply this identity, your answer depends only on the given value of sin⁡θ\sin\theta, not on which specific angle θ\theta is.

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