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If sin⁡2θ=cos⁡40∘\sin 2\theta = \cos 40^\circ, then the smallest positive value of θ\theta is :

Solution

✅ Correct Option: 1

We use the identity cos⁡x=sin⁡(90∘−x)\cos x = \sin(90^\circ - x).

So cos⁡40∘=sin⁡50∘\cos 40^\circ = \sin 50^\circ.

Thus sin⁡2θ=sin⁡50∘\sin 2\theta = \sin 50^\circ, giving 2θ=50∘2\theta = 50^\circ, hence θ=25∘\theta = 25^\circ.

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