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Product of Trigonometric Ratios in Terms of Their Sum

Prove that si...

Question

Prove that $s i n x + s i n 3 x + s i n 5 x + s i n 7 x = 4 c o s x c o s 2 x s i n 4 x .$

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Solution

We have LHS = $s i n x + s i n 3 x + s i n 5 x + s i n 7 x$

$= (s i n 7 x + s i n x) + (s i n 5 x + s i n 3 x)$

$= 2 s i n (\frac{7 x + x}{2}) c o s (\frac{7 x - x}{2}) + 2 s i n (\frac{5 x + 3 x}{2}) c o s (\frac{5 x - 3 x}{2})$

$[∵ s i n C + s i n D = 2 s i n (\frac{C + D}{2}) c o s (\frac{C + D}{2})]$

$= 2 s i n 4 X c o s 3 X + 2 s i n 4 X c o s x = 2 s i n 4 x (c o s 3 x + c o s x)$

$= 2 s i n 4 x \times 2 c o s (\frac{3 x + x}{2}) c o s (\frac{3 x - x}{2})$

$[∵ c o s C + c o s D = 2 c o s (\frac{C + D}{2}) o c s () \frac{C - D}{2}]$

$= 4 s i n 4 x c o s 2 x s i n 4 x = R H S H e n c e p r o v e d$

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Product of Trigonometric Ratios in Terms of Their Sum

Standard XI Mathematics

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