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NCERT Exemplar · Q50

Q.If sin⁡θ+cos⁡θ=1\sin\theta + \cos\theta = 1, then the value of sin⁡2θ\sin 2\theta is equal to
(A) 11
(B) 12\dfrac{1}{2}
(C) 00
(D) −1-1

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Squaring the given equation sin⁡θ+cos⁡θ=1\sin\theta + \cos\theta = 1 and using the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 directly gives sin⁡2θ=0\sin 2\theta = 0. The answer is (C).

The key insight here is that sin⁡2θ\sin 2\theta is not a basic function of sin⁡θ\sin\theta or cos⁡θ\cos\theta alone — but it is directly related to the product sin⁡θcos⁡θ\sin\theta \cos\theta through the double-angle identity sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta. So if we can find sin⁡θcos⁡θ\sin\theta\cos\theta, we are done.

How do we get sin⁡θcos⁡θ\sin\theta\cos\theta from sin⁡θ+cos⁡θ\sin\theta + \cos\theta? Square it. Squaring a sum produces a cross-term 2sin⁡θcos⁡θ2\sin\theta\cos\theta, which is exactly sin⁡2θ \sin 2\theta. That’s the entire strategy.

Let’s walk through it carefully.

  1. Start with the given equation

sin⁡θ+cos⁡θ=1\sin\theta + \cos\theta = 1

  1. Square both sides

(sin⁡θ+cos⁡θ)2=12(\sin\theta + \cos\theta)^2 = 1^2

Expanding the left-hand side:

sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=1\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1

  1. Use the Pythagorean identity We know sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 for any θ\theta. Substitute that in:

1+2sin⁡θcos⁡θ=11 + 2\sin\theta\cos\theta = 1

  1. Simplify Subtract 1 from both sides:

2sin⁡θcos⁡θ=02\sin\theta\cos\theta = 0

  1. Apply the double-angle identity Since sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta, we get: sin⁡2θ=0\sin 2\theta = 0 …

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