Prove that-i cos 3 A-cos 5 A-cos 7 A-cos 15 A-4 cos 4 A cos 5 A cos 6 Aii cos A-cos 3 A-cos 5 A-cos 7 A-4 cos A cos 2 A cos 4 Aiii sin A-sin 2 A-sin 4 A-sin 5 A-4 cos A 2 cos 3 A 2 sin 3 Aiv sin 3 A-sin 2 A-sin A-4 sin A cos A 2 cos 3 A 2v cos 20-cos 100-cos 100-cos 140-140-cos 200-34vi sin- 2 sin 7 - 2-sin 3 - 2 sin 11 - 2-sin 2 -sin 5 -

(i) Consider #160;LHS: #160;cos #160;3A #160;+ #160;cos #160;5A #160;+ #160;cos #160;7A #160;+ #160;cos #160;15A= #160;2cos #160;3A ...

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Byju's Answer

Standard XII

Mathematics

Evaluation of a Determinant

Prove that:i ...

Question

Prove that:
(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A
(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos $\frac{A}{2}$ cos $\frac{3 A}{2}$ sin 3A
(iv) sin 3A + sin 2A − sin A = 4 sin A cos $\frac{A}{2}$ cos $\frac{3 A}{2}$
(v) cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = − $\frac{3}{4}$
(vi) $\sin \frac{θ}{2} \sin \frac{7 θ}{2} + \sin \frac{3 θ}{2} \sin \frac{11 θ}{2} = \sin 2 θ \sin 5 θ .$

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Solution

(i)

$Consider LHS : \cos 3 A + c os 5 A + \cos 7 A + \cos 15 A = 2 \cos (\frac{3 A + 5 A}{2}) \cos (\frac{3 A - 5 A}{2}) + 2 \cos (\frac{7 A + 15 A}{2}) \cos (\frac{7 A - 15 A}{2}) \{∵ \cos A + \cos B = 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2})\} = 2 \cos 4 A \cos (- A) + 2 \cos 11 A \cos (- 4 A)$

$= 2 \cos 4 A \cos A + 2 \cos 11 A \cos 4 A = 2 \cos 4 A \{\cos A + \cos 11 A\} = 2 \cos 4 A \times \{2 \cos (\frac{A + 11 A}{2}) \cos (\frac{A - 11 A}{2})\} = 4 \cos 4 A \cos 6 A \cos (- 5 A) = 4 \cos 4 A \cos 5 A \cos 6 A = RHS Hence, LHS = RHS$

(ii)

$Consider LHS : \cos A + \cos 3 A + \cos 5 A + \cos 7 A = 2 \cos (\frac{A + 3 A}{2}) \cos (\frac{A - 3 A}{2}) + 2 \cos (\frac{5 A + 7 A}{2}) \cos (\frac{5 A - 7 A}{2}) \{∵ \cos A + \cos B = 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2})\} = 2 \cos 2 A \cos (- A) + 2 \cos 6 A \cos (- A)$

$= 2 \cos 2 A \cos A + 2 \cos 6 A \cos A = 2 \cos A (\cos 2 A + \cos 6 A) = 2 \cos A \times 2 \cos (\frac{2 A + 6 A}{2}) \cos (\frac{2 A - 6 A}{2}) = 4 \cos A \cos 4 A \cos (- 2 A) = 4 \cos A \cos 2 A \cos 4 A = RHS Hence, LHS = RHS .$

(iii)

$Consider LHS : \sin A + \sin 2 A + \sin 4 A + \sin 5 A = 2 \sin (\frac{A + 2 A}{2}) \cos (\frac{A - 2 A}{2}) + 2 \sin (\frac{4 A + 5 A}{2}) \cos (\frac{4 A - 5 A}{2}) \{∵ \sin A + \sin B = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})\} = 2 \sin (\frac{3}{2} A) \cos (- \frac{A}{2}) + 2 \sin (\frac{9}{2} A) \cos (- \frac{A}{2})$

$= 2 \sin (\frac{3}{2} A) \cos (\frac{A}{2}) + 2 \sin (\frac{9}{2} A) \cos (\frac{A}{2}) = 2 \cos (\frac{A}{2}) \{\sin \frac{3}{2} A + \sin \frac{9}{2} A\} = 2 \cos (\frac{A}{2}) \times 2 \sin (\frac{\frac{3}{2} A + \frac{9}{2} A}{2}) \cos (\frac{\frac{3}{2} A - \frac{9}{2} A}{2}) = 4 \cos (\frac{A}{2}) \cos 3 A \cos (- \frac{3}{2} A) = 4 \cos \frac{A}{2} \cos (\frac{3}{2} A) \cos 3 A = RHS Hence, LHS = RHS$

(iv)

$Consider LHS : = \sin 3 A + \sin 2 A - \sin A = 2 \sin (\frac{3 A + 2 A}{2}) co s (\frac{3 A - 2 A}{2}) - \sin A \{∵ \sin A + \sin B = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})\} = 2 \sin (\frac{5}{2} A) \cos (\frac{A}{2}) - \sin A$

$= 2 \sin (\frac{5}{2} A) co s (\frac{A}{2}) - 2 \sin \frac{A}{2} \cos \frac{A}{2} = 2 \cos (\frac{A}{2}) \{\sin \frac{5}{2} A - \sin \frac{A}{2}\} = 2 \cos (\frac{A}{2}) \times 2 \sin (\frac{\frac{5}{2} A - \frac{A}{2}}{2}) \cos (\frac{\frac{5}{2} A + \frac{A}{2}}{2}) = 4 \cos (\frac{A}{2}) \sin A \cos (\frac{3}{2} A) = 4 \sin A \cos (\frac{A}{2}) \cos (\frac{3}{2} A) = RHS Hence, LHS = RHS$

(v)

$Consider LHS : \cos 20 ° \cos 100 ° + \cos 100 ° \cos 140 ° - \cos 140 ° \cos 200 ° = \frac{1}{2} (2 \cos 20 ° \cos 100 ° + 2 \cos 100 ° \cos 140 ° - 2 \cos 140 ° \cos 200 °) = \frac{1}{2} [\cos (100 ° + 20 °) \cos (100 ° - 20 °) + \cos (140 ° + 100 °) \cos (140 ° - 100 °) - \cos (200 ° + 140 °) \cos (200 ° - 140 °)] = \frac{1}{2} [\cos 120 ° + \cos 80 ° + \cos 240 ° + \cos 40 ° - \cos 340 ° - \cos 60 °] = \frac{1}{2} [\cos 120 ° + \cos 240 ° - \cos 60 ° + \cos 80 ° + \cos 40 ° - \cos 340 °] = \frac{1}{2} [(- \frac{1}{2} - \frac{1}{2} - \frac{1}{2}) + \cos 80 ° + \cos 40 ° - \cos 340 °] = \frac{1}{2} [- \frac{3}{2} + \{2 \cos (\frac{80 ° + 40 °}{2}) \cos (\frac{80 ° - 40 °}{2}) - \cos (360 ° - 20 °)\}] = \frac{1}{2} [- \frac{3}{2} + \{2 \cos 60 ° \cos 20 ° - \cos 20 °\}] = \frac{1}{2} [- \frac{3}{2} + \cos 20 ° - \cos 20 °] = \frac{1}{2} [- \frac{3}{2}] = - \frac{3}{4} = RHS Hence, LHS = RHS$

(vi)
$LHS = \sin \frac{θ}{2} \sin \frac{7 θ}{2} + \sin \frac{3 θ}{2} \sin \frac{11 θ}{2} = \frac{1}{2} [2 \sin \frac{θ}{2} \sin \frac{7 θ}{2} + 2 \sin \frac{3 θ}{2} \sin \frac{11 θ}{2}] = \frac{1}{2} [\cos (\frac{7 θ}{2} - \frac{θ}{2}) - \cos (\frac{7 θ}{2} + \frac{θ}{2}) + \cos (\frac{11 θ}{2} - \frac{3 θ}{2}) - \cos (\frac{11 θ}{2} + \frac{3 θ}{2})] = \frac{1}{2} [\cos 3 θ - \cos 4 θ + \cos 4 θ - \cos 7 θ] = \frac{1}{2} [\cos 3 θ - \cos 7 θ] = \frac{1}{2} [- 2 \sin (\frac{3 θ + 7 θ}{2}) \sin (\frac{3 θ - 7 θ}{2})] = \frac{1}{2} [- 2 \sin (5 θ) \sin (- 2 θ)] = \sin (5 θ) \sin (2 θ) = RHS$

Hence, LHS = RHS

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Similar questions

Q. Prove that:
(i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A
(ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A
(iii) sin A + sin 2A + sin 4A + sin 5A = 4 cos

\frac{A}{2}

cos

\frac{3 A}{2}

sin 3A
(iv) sin 3A + sin 2A − sin A = 4 sin A cos

\frac{A}{2}

cos

\frac{3 A}{2}

(v) cos 20° cos 100° + cos 100° cos 140° − 140° cos 200° = −

\frac{3}{4}

(vi)

\sin \frac{x}{2} \sin \frac{7 x}{2} + \sin \frac{3 x}{2} \sin \frac{11 x}{2} = \sin 2 x \sin 5 x .

(vii)

\cos x \cos \frac{x}{2} - \cos 3 x \cos \frac{9 x}{2} = \sin 7 x \sin 8 x

Prove that:
(i) $c o s 3 A + c o s S A + c o s 7 A + c o s 15 A 4 c o s 4 A c o s S A c o s 6 A$
(ii) $c o s A + c o s 3 A + c o s 5 A + c o s 7 A = 4 c o s A c o s 2 A c o s 4 A$
(iii) $s i n A + s i n 2 A + s i n 4 A + s i n 5 A 4 c o s \frac{A}{2} c o s \frac{3}{A} s i n 3 A$
(iv) $s i n 3 A + s i n 2 A + s i n A - 4 s i n A c o s \frac{A}{2} c o s \frac{3 A}{2}$
(v) $c o s 20^{\circ} c o s 100^{\circ} + c o s 100^{\circ} c o s 140^{c} o s 200 ° = - \frac{3}{4}$

(vi) $s i n \frac{θ}{2} s i n \frac{7 θ}{2} + s i n \frac{3 θ}{2} s i n \frac{11 θ}{2} = s i n 2 θ s i n 5 θ$
(vii) $s i n \frac{θ}{2} - c o s 3 θ c o s = \frac{9 θ}{2} = s i n 7 θ s i n 8 θ$

\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

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

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