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Trigonometric Ratios of Compound Angles

Prove that ...

Question

Prove that
$cos 2 x cos \frac{x}{2} - cos 3 x cos \frac{9 x}{2} = sin 5 x sin \frac{5 x}{2}$ .

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Solution

We know that,
$2 c o s x c o s y = c o s (x + y) + c o s (x - y)$

Now,
LHS = $c o s 2 x c o s \frac{x}{2} = \frac{1}{2} [c o s (2 x + \frac{x}{2}) + c o s (2 x - \frac{x}{2})]$
$= \frac{1}{2} (c o s \frac{5 x}{2} + c o s \frac{3 x}{2})$

And,
$c o s 3 x c o s \frac{9 x}{2} = \frac{1}{2} [c o s (3 x + \frac{9 x}{2}) + c o s (3 x - \frac{9 x}{2})]$
$= \frac{1}{2} (c o s \frac{15 x}{2} + c o s \frac{3 x}{2})$

Therefore,
$c o s 2 x c o s \frac{x}{2} - c o s 3 x c o s \frac{9 x}{2} = \frac{1}{2} (c o s \frac{5 x}{2} + c o s \frac{3 x}{2}) - \frac{1}{2} (c o s \frac{15 x}{2} + c o s \frac{3 x}{2})$

$= \frac{1}{2} (c o s \frac{5 x}{2} - c o s \frac{15 x}{2})$

Now, using $c o s x - c o s y = - 2 s i n \frac{x + y}{2} s i n \frac{x - y}{2}$ , we get,

$= \frac{1}{2} (- 2 s i n \frac{20 x}{4} s i n (\frac{- 10 x}{4}))$

$= s i n 5 x s i n \frac{5 x}{2}$ = RHS

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