\(\frac{sin\,5θ}{sin\,θ}\) is equal to(a) 16 cos4θ – 12 cos2θ �

1.	\(\frac{sin\,5θ}{sin\,θ}\) is equal to(a) 16 cos4θ – 12 cos2θ – 1 (b) 16 cos4θ – 12 cos2θ + 1 (c) 16 cos4θ + 12 cos2θ – 1 (d) 16 cos4θ + 12 cos2θ + 1
Answer» (b) 16 cos⁴θ – 12 cos²θ + 1 \(\frac{sin\,5θ}{sin\,θ}\) = \(\frac{sin\,(2\theta+3\theta)}{sin\,\theta}\) = \(\frac{1}{sin\,θ}\) {sin 2θ cos3θ + cos 2θ sin 3θ} = \(\frac{1}{sin\,θ}\) \(\big\{\)2 sin θ cos θ (4 cos³ θ - 3 cosθ) + (2 cos² θ -1)(3 sin θ - 4 sin³ θ)\(\big\}\) (∵ cos3A = 4cos³A – 3cosA, sin 3A = 3sinA – 4sin³A) = {2 cos θ (4 cos³θ – 3 cosθ) + (2 cos²θ – 1) (3 – 4 sin²θ)}. (Note the step) = 8 cos⁴θ – 6 cos²θ + 6 cos²θ – 3 – 8 cos²θ sin²θ + 4 sin²θ = 8 cos⁴θ – 3 – 8 cos²θ (1 – cos²θ) + 4 (1 – cos²θ) (∵ sin²θ + cos²θ = 1) = 8 cos⁴θ – 3 – 8 cos²θ + 8 cos⁴θ + 4 – 4 cos²θ = 16 cos⁴θ – 12 cos²θ + 1.

\(\frac{sin\,5θ}{sin\,θ}\) is equal to(a) 16 cos4θ – 12 cos2θ – 1 (b) 16 cos4θ – 12 cos2θ + 1 (c) 16 cos4θ + 12 cos2θ – 1 (d) 16 cos4θ + 12 cos2θ + 1

Answer»

(b) 16 cos⁴θ – 12 cos²θ + 1

\(\frac{sin\,5θ}{sin\,θ}\) = \(\frac{sin\,(2\theta+3\theta)}{sin\,\theta}\)

= \(\frac{1}{sin\,θ}\) {sin 2θ cos3θ + cos 2θ sin 3θ}

= \(\frac{1}{sin\,θ}\) \(\big\{\)2 sin θ cos θ (4 cos³ θ - 3 cosθ) + (2 cos² θ -1)(3 sin θ - 4 sin³ θ)\(\big\}\)

(∵ cos3A = 4cos³A – 3cosA, sin 3A = 3sinA – 4sin³A)

= {2 cos θ (4 cos³θ – 3 cosθ) + (2 cos²θ – 1) (3 – 4 sin²θ)}. (Note the step)

= 8 cos⁴θ – 6 cos²θ + 6 cos²θ – 3 – 8 cos²θ sin²θ + 4 sin²θ

= 8 cos⁴θ – 3 – 8 cos²θ (1 – cos²θ) + 4 (1 – cos²θ) (∵ sin²θ + cos²θ = 1)

= 8 cos⁴θ – 3 – 8 cos²θ + 8 cos⁴θ + 4 – 4 cos²θ

= 16 cos⁴θ – 12 cos²θ + 1.