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Sum of Trigonometric Ratios in Terms of Their Product

Solve sin 3...

Question

Solve $sin (3 θ + α) + sin (3 θ - α) + sin (α - θ) - sin (α + θ) = cos α$ .

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Solution

Combining the first two and last two terms, we get
$2 sin 3 θ cos θ + 2 cos α sin (- θ) = cos α$
$2 sin 3 θ - 2 sin θ = 1$
or $2 (3 sin θ - 4 {sin}^{3} θ) - 2 sin θ = 1$
or $8 {sin}^{3} θ - 4 sin θ + 1 = 0$ .
Clearly $sin θ = \frac{1}{2}$ satisfies it
$∴ (2 sin θ - 1) (4 {sin}^{2} θ + 2 sin θ - 1) = 0$
$sin θ = 1 / 2 = sin (π / 6)$
$∴ θ = n π + (- 1)^{n} π / 6$
Also $sin θ = \frac{- 2 \pm \sqrt{4 + 16}}{2} = \frac{\sqrt{5} - 1}{4}, - \frac{\sqrt{5} + 1}{4}$
$sin θ = sin 18^{o} = sin (π / 10)$
$∴ θ = n π + (- 1)^{n} π / 10$
$sin θ = - sin 54^{o} = sin (\frac{- 3 π}{10})$
$∴ θ = n π - (- 1)^{n} \frac{3 π}{10}$ .

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