The value of $8 {sin}^{3} 40^{o} - 6 sin 40^{o}$ is

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Solution

$\begin{matrix} 8 {sin}^{2} 40^{\circ} - 6 sin 40^{\circ} = 8 (\frac{3 sin 40^{\circ} - sin 120^{\circ}}{4}) - 6 sin 40^{\circ} (sin 3 θ = 3 sin θ - 4 {sin}^{3} θ) = 2 (3 sin 40^{\circ} - sin 120^{\circ}) - 6 sin 40^{\circ} = 6 sin 40^{^{\circ}} - 2 sin 120^{\circ} - 6 sin 40^{\circ} = - 2 sin (180^{\circ} - 60^{\circ}) = - 2 sin 60^{\circ} = - 2 \times \frac{1}{2} = - 1 \end{matrix}$

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\frac{cot 40^{o}}{tan 50^{o}} - \frac{1}{2} (\frac{cos 35^{o}}{sin 55^{o}})

a = 40^{o}

b

cos (40^{o} + θ) - sin (50^{o} - θ) + \frac{{cos}^{2} 40^{o} + {cos}^{2} 50^{o}}{{sin}^{2} 40^{o} + {sin}^{2} 50^{o}}

P Q ∥ R S

∠ R S F = 40^{o}

∠ P Q F = 35^{o}

∠ Q F P = x^{o}

x

O

∠ A E C = 40^{o}

a + b + c

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