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NCERT Exemplar · Q6

Q.Prove that cos⁡θ cos⁡θ2−cos⁡3θ cos⁡9θ2=sin⁡7θ2 sin⁡4θ\cos\theta\,\cos\dfrac{\theta}{2} - \cos 3\theta\,\cos\dfrac{9\theta}{2} = \sin\dfrac{7\theta}{2}\,\sin 4\theta.

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Convert each product on the left to a sum using 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B=\cos(A+B)+\cos(A-B), simplify, then convert the resulting difference of cosines to a product using cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C-\cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2} to reach the right-hand side exactly.

Step 1 — Convert each product to a sum.

Using 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B=\cos(A+B)+\cos(A-B):

For cos⁡θcos⁡θ2\cos\theta\cos\dfrac{\theta}{2} (with A=θA=\theta, B=θ2B=\dfrac{\theta}{2}):

cos⁡θcos⁡θ2=12[cos⁡3θ2+cos⁡θ2]\cos\theta\cos\frac{\theta}{2}=\frac{1}{2}\left[\cos\frac{3\theta}{2}+\cos\frac{\theta}{2}\right]

For cos⁡3θcos⁡9θ2\cos3\theta\cos\dfrac{9\theta}{2} (with A=3θA=3\theta, B=9θ2B=\dfrac{9\theta}{2}), and using cos⁡(−x)=cos⁡x\cos(-x)=\cos x:

cos⁡3θcos⁡9θ2=12[cos⁡15θ2+cos⁡(−3θ2)]=12[cos⁡15θ2+cos⁡3θ2]\cos3\theta\cos\frac{9\theta}{2}=\frac{1}{2}\left[\cos\frac{15\theta}{2}+\cos\left(-\frac{3\theta}{2}\right)\right]=\frac{1}{2}\left[\cos\frac{15\theta}{2}+\cos\frac{3\theta}{2}\right]

Step 2 — Subtract and simplify.

LHS=12[cos⁡3θ2+cos⁡θ2]−12[cos⁡15θ2+cos⁡3θ2]\text{LHS}=\frac{1}{2}\left[\cos\frac{3\theta}{2}+\cos\frac{\theta}{2}\right]-\frac{1}{2}\left[\cos\frac{15\theta}{2}+\cos\frac{3\theta}{2}\right]

The cos⁡3θ2\cos\dfrac{3\theta}{2} terms cancel:

LHS=12[cos⁡θ2−cos⁡15θ2]\text{LHS}=\frac{1}{2}\left[\cos\frac{\theta}{2}-\cos\frac{15\theta}{2}\right]

Step 3 — Convert this difference to a product.

Using cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2} with C=θ2C=\dfrac{\theta}{2}, D=15θ2D=\dfrac{15\theta}{2}: …

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