The equality $sin A + sin 2 A + sin 3 A = 3$ holds for some real value of A

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Solution

Apply the concept of range of \sine function.

Given: $sin A + sin 2 A + sin 3 A = 3$

If holds true only when $sin A, sin 2 A, sin 3 A$ have value of equals to 1 (i.e., maximum value of sine function) But $sin A, sin 2 A, sin 3 A$ never be equal to $1$ simultaneously.

Hence, the statement is false.

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

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

Write the value of $\frac{s i n A + s i n 3 A}{c o s A + c o s 3 A}$

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) $s i n \frac{θ}{2} - c o s 3 θ c o s = \frac{9 θ}{2} = s i n 7 θ s i n 8 θ$