Solve the problem- sin A - sin 3A - sin 5A - sin 7Acos A - cos 3A - cos 5A - cos 7A - 2 A

sinC+sinD =2sinC+D2 cosC D2 =cosC+cosD =2cosC+D2 cosC D2 =sinC sinD =2cosC+D2 sinC D2 =cosC+cosD = 2cosC+ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

Solve the pro...

Question

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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Solution

$sin C + sin D$ $= 2 sin \frac{C + D}{2}$ $cos \frac{C - D}{2}$
$= cos C + cos D$ $= 2 cos \frac{C + D}{2}$ $cos \frac{C - D}{2}$
$= sin C - sin D$ $= 2 cos \frac{C + D}{2}$ $sin \frac{C - D}{2}$
$= cos C + cos D$ $= - 2 cos \frac{C + D}{2}$ $cos \frac{C - D}{2}$
$= \frac{2 cos \frac{A + 3 A}{2} . sin \frac{A - 3 A}{2} + 2 cos \frac{5 A + 7 A}{2} . sin \frac{5 A - 7 A}{2}}{2 sin \frac{A + 3 A}{2} . sin \frac{A - 3 A}{2} + 2 sin \frac{5 A + 7 A}{2} . sin \frac{5 A - 7 A}{2}}$
$= \frac{(2 cos 2 A) . (- sin A) + (cos 6 A) . (- sin A)}{(2 sin 2 A) . (sin A) + (sin 6 A) . (sin A)}$
$= \frac{(2 cos 2 A) . [1 + 2 cos 6 A] . sin (- A)}{(2 sin 2 A) . [1 + 2 cos 6 A] . sin (A)}$
$= \frac{cos 2 A}{sin 2 A}$
$cot 2 A$

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Similar questions

\frac{c o s A + c o s 3 A + c o s 5 A + c o s 7 A}{s i n A + s i n 3 A + s i n 5 A + s i n 7 A} =

Q. Prove that

\frac{sinA + s i n 3 A + s i n 5 A + s i n 7 A}{c o s A + cos 3 A + cos 5 A + cos 7 A} = tan 4 A

\frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + cos 5 A + cos 7 A}

is equal to -

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) $\frac{s i n (θ + ϕ) - 2 s i n θ + s i n (θ + ϕ)}{c o s (θ + ϕ) - 2 c o s θ + c o s (θ + ϕ)}$

Q. (i)

\frac{cos 3 θ + 2 cos 5 θ + cos 7 θ}{cos θ + 2 cos 3 θ + cos 5 θ} = cos 2 θ - sin 2 θ tan 3 θ

(ii)

\frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + cos 5 A + cos 7 A} = tan 4 A

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Solve the problem:-sinA−sin3A+sin5A−sin7AcosA−cos3A−cos5A+cos7A=2cotA

Solve the problem:-
$\frac{sin A - sin 3 A + sin 5 A - sin 7 A}{cos A - cos 3 A - cos 5 A + cos 7 A} = 2 cot A$