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NCERT Exemplar · Q56

Q.The value of cos⁡248∘−sin⁡212∘\cos^2 48^\circ - \sin^2 12^\circ is
(A) 5+18\dfrac{\sqrt{5} + 1}{8}
(B) 5−18\dfrac{\sqrt{5} - 1}{8}
(C) 5+15\dfrac{\sqrt{5} + 1}{5}
(D) 5+122\dfrac{\sqrt{5} + 1}{2\sqrt{2}}

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The expression cos⁡248∘−sin⁡212∘\cos^2 48^\circ - \sin^2 12^\circ simplifies using the identity cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B), leading to cos⁡60∘cos⁡36∘=12⋅5+14=5+18\cos 60^\circ \cos 36^\circ = \frac12 \cdot \frac{\sqrt{5}+1}{4} = \frac{\sqrt{5}+1}{8}, which matches option (A).

The key insight here is that cos⁡2θ−sin⁡2ϕ\cos^2 \theta - \sin^2 \phi is not a standard double-angle form unless the angles are equal. But we can rewrite it using the product-to-sum identities. Specifically, recall that:

cos⁡2A−sin⁡2B=cos⁡(A+B)cos⁡(A−B)\cos^2 A - \sin^2 B = \cos(A+B)\cos(A-B)

This identity comes from expanding cos⁡(A+B)cos⁡(A−B)\cos(A+B)\cos(A-B) using the cosine addition formulas:

cos⁡(A+B)cos⁡(A−B)=(cos⁡Acos⁡B−sin⁡Asin⁡B)(cos⁡Acos⁡B+sin⁡Asin⁡B)=cos⁡2Acos⁡2B−sin⁡2Asin⁡2B\cos(A+B)\cos(A-B) = (\cos A \cos B - \sin A \sin B)(\cos A \cos B + \sin A \sin B) = \cos^2 A \cos^2 B - \sin^2 A \sin^2 B

But that’s not directly our form. A cleaner derivation:

›Proof

Start with cos⁡2A−sin⁡2B\cos^2 A - \sin^2 B. Write cos⁡2A=1+cos⁡2A2\cos^2 A = \frac{1+\cos 2A}{2} and sin⁡2B=1−cos⁡2B2\sin^2 B = \frac{1-\cos 2B}{2}. Then:

cos⁡2A−sin⁡2B=1+cos⁡2A2−1−cos⁡2B2=cos⁡2A+cos⁡2B2\cos^2 A - \sin^2 B = \frac{1+\cos 2A}{2} - \frac{1-\cos 2B}{2} = \frac{\cos 2A + \cos 2B}{2}

Using the sum-to-product identity cos⁡P+cos⁡Q=2cos⁡P+Q2cos⁡P−Q2\cos P + \cos Q = 2\cos\frac{P+Q}{2}\cos\frac{P-Q}{2}, we get:

cos⁡2A+cos⁡2B2=2cos⁡(A+B)cos⁡(A−B)2=cos⁡(A+B)cos⁡(A−B)\frac{\cos 2A + \cos 2B}{2} = \frac{2\cos(A+B)\cos(A-B)}{2} = \cos(A+B)\cos(A-B)

So the identity is proven. Now apply it with A=48∘A = 48^\circ and B=12∘B = 12^\circ:

  1. Set up the identity:

cos⁡248∘−sin⁡212∘=cos⁡(48∘+12∘)cos⁡(48∘−12∘)\cos^2 48^\circ - \sin^2 12^\circ = \cos(48^\circ + 12^\circ) \cos(48^\circ - 12^\circ)

  1. Simplify the angles: 48∘+12∘=60∘48^\circ + 12^\circ = 60^\circ and 48∘−12∘=36∘48^\circ - 12^\circ = 36^\circ. So the expression becomes:

cos⁡60∘⋅cos⁡36∘\cos 60^\circ \cdot \cos 36^\circ

  1. Evaluate cos⁡60∘\cos 60^\circ:

    This is a standard value: cos⁡60∘=12\cos 60^\circ = \frac12.

  2. Evaluate cos⁡36∘\cos 36^\circ: …

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