Prove that cos A 2 A cos 4 A cos 8 A-sin 16 A-16 sin A-

Givne ,LHS= cos A cos 2A cos 4A cos 8A On multiplying and dividing by 2 sin A, we get =1/2 sin A (2 sin A cos A cos 4A cos 8A) =1/2 sin A (sin 2A cos 2 ...

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Standard XI

Mathematics

Trigonometric Ratios of Multiples of an Angle

Prove that co...

Question

Prove that cos A 2A cos 4A cos 8A = $\frac{s i n 16 A}{16 s i n A}$ .

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Solution

Givne , $L H S = c o s A c o s 2 A c o s 4 A c o s 8 A$

On multiplying and dividing by 2 sin A, we get

$= \frac{1}{2 s i n A} (2 s i n A c o s A c o s 4 A c o s 8 A)$

$= \frac{1}{2 s i n A} (s i n 2 A c o s 2 A c o s 4 A c o s 8 A) [∵ s i n 2 x = 2 s i n x c o s x]$

$= \frac{1}{4 s i n A} (2 s i n 2 A c o s 2 A c o s 4 A c o s 8 A) = \frac{1}{4 s i n A} (s i n 4 A c o s 4 A c o s 8 A [∴ 2 s i n 2 x c o s 2 x = s i n 4 x]$

= $\frac{1}{8 s i n A} (2 s i n 4 A c o s 8 A) = \frac{1}{8 s i n A} (s i n 8 A c o s 8 A)$

= $\frac{1}{16 s i n A} (2 s i n 8 A c o s 8 A) = \frac{s i n 16 A}{16 s i n A} = R H S$

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Similar questions

Q. Prove that :

cos a cos 2 a \cdot cos 4 a cos 8 a = \frac{sin 16 a}{16 sin a}

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cosA cos 2 A cos 4 A cos 8 A = \frac{sin 16 A}{16 sinA}

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cos A cos 2 A cos 4 A cos 8 A = \frac{sin 16 A}{16 sin A}

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Prove that cos A 2A cos 4A cos 8A = sin 16A16 sin A.

Prove that cos A 2A cos 4A cos 8A = $\frac{s i n 16 A}{16 s i n A}$ .