Solve - sin 3A-2sin 5A -sin 7Asin A-2sin 3A -sin 5A-sin 5Asin 3A

sin 3a+2sin 5a +sin 7asin a+2sin 3a +sin 5a =(sin 3a+sin 5a)( sin 5a+sin 7a)(sin a+sin 3a)( sin 3a+sin 5a) ={ 2sin 4acos a }+{ 2sin 6a cos a }{ ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

Solve : sin...

Question

Solve :
$\frac{sin 3 A + 2 sin 5 A + sin 7 A}{sin A + 2 sin 3 A + sin 5 A} = \frac{sin 5 A}{sin 3 A}$

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Solution

$\frac{sin 3 a + 2 sin 5 a + sin 7 a}{sin a + 2 sin 3 a + sin 5 a}$

$= \frac{(sin 3 a + sin 5 a) (sin 5 a + sin 7 a)}{(sin a + sin 3 a) (sin 3 a + sin 5 a)}$

$= \frac{{2 sin 4 a cos a} + {2 sin 6 a cos a}}{{2 sin 2 a cos a} + {2 sin 4 a cos a}}$

$= \frac{sin 4 a + sin 6 a}{sin 2 a + sin 4 a}$

$= \frac{2 sin 5 a cos a}{2 sin 3 a cos a}$

$= \frac{sin 5 a}{sin 3 a}$

$∴ \frac{sin 3 A + 2 sin 5 A + sin 7 A}{sin A + 2 sin 3 A + sin 5 A} = \frac{sin 5 A}{sin 3 A}$

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Similar questions

Prove that sin A - sin 3A + sin 5A - sin 7A/ sin A - sin 3A - sin 5A +sin 7A = cot 2A.

Q. Solve

sin A + sin 3 A + sin 5 A + sin 7 A =

8. Prove that :
(i) $\frac{s i n A + s i n 3 A + s i n 5 A}{c o s A + c o s 3 A + c o s 5 A} = t a n 3 A$
(ii) $(i i) \frac{c o s 3 A + 2 c o s 5 A + c o s 7 A}{c o s A + 2 c o s 3 A + c o s 5 A} = \frac{c o s 5 A}{c o s 3 A}$
(iii) $\frac{c o s 4 A + c o s 3 A + c o s 2 A}{c o s 4 A + s i n 3 A + s i n 2 A} = c o t 3 A$
(iv) $\frac{s i n 3 A + s i n 5 A + s i n 7 A + s i n 9 A}{c o s 3 A + c o s 5 A + c o s 7 A + c o s 9 A} = t a n 6 A$
(v) $\frac{s i n 5 A - s i n 7 A + s i n 8 A - s i n 4 A}{c o s 4 A + c o s 7 A + c o s 7 A - c o s 5 A - c o s 8 A} = c o t 6 A$
(vi) $\frac{s i n 5 A + c o s 2 A - s i n 6 A c o s A}{s i n A s i n 2 A - c o s 2 A c o s 3 A} = t a n A$
(vii) $\frac{s i n 11 A + s i n A + s i n 7 A + s i n 3 A}{c o s 11 A s i n A + c o s 7 A s i n 3 A} = t a n 8 A$
(viii) $\frac{s i n 3 A c o s 4 A - s i n A c o s 2 A}{s i n 4 A s i n A + c o s 6 A c o s A} = t a n 2 A$
(ix) $\frac{s i n A s i n 2 A + s i n 3 A s i n 6 A}{s i n A c o s 2 A + c o s 3 A c o s 6 A} = t a n 5 A$
(x) $\frac{s i n A + 2 s i n 3 A + s i n 5 A}{s i n 3 A + 2 s i n 5 A + s i n 7 A} = \frac{s i n 3 A}{s i n 5 A}$
(xi) $\frac{s i n (θ + ϕ) - 2 s i n θ + s i n (θ + ϕ)}{c o s (θ + ϕ) - 2 c o s θ + c o s (θ + ϕ)}$

sin A + sin 3 A + sin 5 A + sin 7 A

equals

Q. Solve the problem:-

\frac{sin A - sin 3 A + sin 5 A - sin 7 A}{cos A - cos 3 A - cos 5 A + cos 7 A} = 2 cot A

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Solve : sin3A+2sin5A+sin7AsinA+2sin3A+sin5A=sin5Asin3A

sin3a+2sin5a+sin7asina+2sin3a+sin5a=(sin3a+sin5a)(sin5a+sin7a)(sina+sin3a)(sin3a+sin5a)={2sin4acosa}+{2sin6acosa}{2sin2acosa}+{2sin4acosa}=sin4a+sin6asin2a+sin4a=2sin5acosa2sin3acosa=sin5asin3a∴sin3A+2sin5A+sin7AsinA+2sin3A+sin5A=sin5Asin3A

Solve :
$\frac{sin 3 A + 2 sin 5 A + sin 7 A}{sin A + 2 sin 3 A + sin 5 A} = \frac{sin 5 A}{sin 3 A}$