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Home » Prove that cos θ cos θ/2 – cos 3θ cos 9θ/2 = sin7θ sin4θ [Hint: Express L.H.S. = ½ [2cos θcos θ/2 – 2cos 3θ cos 9θ / 2]
Prove that cos θ cos θ/2 – cos 3θ cos 9θ/2 = sin7θ sin4θ [Hint: Express L.H.S. = ½ [2cos θcos θ/2 – 2cos 3θ cos 9θ / 2]
CBSE | Class 11 | Maths | NCERT Exemplar | Trigonometric Functions
Prove that cos θ cos θ/2 – cos 3θ cos 9θ/2 = sin7θ sin4θ [Hint: Express L.H.S. = ½ [2cos θcos θ/2 – 2cos 3θ cos 9θ / 2]

Solution:

By applying transformation formula, we have,

2 \cos A \cos B=\cos (A+B)+\cos (A-B)

-2 \sin A \sin B=\cos (A+B)-\cos (A-B)

Now multiply and divide the expression by 2 .

\therefore \text {Left Hand Side}=\frac{1}{2}\left(2 \cos \theta \cos \frac{\theta}{2}-2 \cos 3 \theta \cos \frac{9 \theta}{2}\right)

Now using the transformation formula, we have,

\text { Left Hand Side }=\frac{1}{2}\left(\cos \left(\theta+\frac{\theta}{2}\right)+\cos \left(\theta-\frac{\theta}{2}\right)-\left\{\cos \left(3 \theta+\frac{9 \theta}{2}\right)+\cos \left(3 \theta-\frac{9 \theta}{2}\right)\right\}\right)

\Rightarrow \text { Left Hand Side }=\frac{1}{2}\left(\cos \frac{3 \theta}{2}+\cos \frac{\theta}{2}-\cos \left(\frac{15 \theta}{2}\right)-\cos \left(-\frac{3 \theta}{2}\right)\right)

\Rightarrow \text { Left Hand Side }=\frac{1}{2}\left(\cos \frac{3 \theta}{2}+\cos \frac{\theta}{2}-\cos \frac{15 \theta}{2}-\cos \frac{3 \theta}{2}\right)\{\because \cos (-\mathrm{x})=\cos \mathrm{x}\}

\Rightarrow \text {Left Hand Side}=\frac{1}{2}\left(\cos \frac{\theta}{2}-\cos \frac{15 \theta}{2}\right)

\Rightarrow \text {Left Hand Side}=\frac{1}{2}\left(2 \sin \left(\frac{\frac{\theta}{2}+\frac{15 \theta}{2}}{2}\right) \sin \left(\frac{\frac{15 \theta}{2}-\frac{\theta}{2}}{2}\right)\right)

\Rightarrow \text {Left Hand Side}=\frac{1}{2}\left(2 \sin \left(\frac{8 \theta}{2}\right) \sin \left(\frac{7 \theta}{2}\right)\right)

\therefore \text {Left Hand Side}=\sin 4 \theta \sin \left(\frac{7 \theta}{2}\right)=\text {Right Hand Side}

As a result,

\cos \theta \cos \frac{\theta}{2}-\cos 3 \theta \cos \frac{9 \theta}{2}=\sin 4 \theta \sin \left(\frac{7 \theta}{2}\right)

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