Chapter 9: Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2
Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.1
LHS,
Sin 5θ = sin (3θ + 2θ)
= sin 3θ cos 2θ + cos 3θ. sin 2θ
= (3 sinθ – 4 sin3 θ)(1 – 2 sin2 θ) + (4 cos3θ - 3 cos θ) 2 sin θ cos θ.
= 3 sinθ – 4 sin3θ – 6 sin3θ + 8 sin5θ + (8 cos4θ – 6 cos2θ) sinθ
= 3 sinθ – 10 sin3θ + 8 sin5θ + 8 sinθ – 16 sin3θ + 8 sin5θ – 6 sinθ + 6 sin3θ
= 5 sinθ – 20 sin3θ + 16 sin5θ = RHS
Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.2
Consider the L.H.S of the given equation
4(cos310° + sin320°) = 3(cos 10° + sin 20°)
since sin 30° = cos 60° = 1/2
and sin 60° = cos 30° = √3/2
⟹ sin 3.20° = cos 3.10°
⟹ 3sin20° - 4sin320° = 4 cos310° - 3 cos10°
⟹ 4(cos310° + sin320°) = 3(cos10° + sin 20°)
Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.3
cos3 θ sin3θ + sin3θ cos 3θ = 3/4 sin 4θ
LHS = cos3θ sin3θ + sin3θ cos 3θ
{∵ sin 3θ = 3 sin θ – 4 sin3 θ cos 3θ = 4 cos3 θ – 3 cos θ}
= 1/4 [3(sin 3θ cos θ + sin θ cos 3θ) + cos 3θ sin 3θ – sin 3θ cos 3θ]
= 1/4 [3 sin (3θ + θ) + 0]
= 3/4 sin 4θ
So,
cos3 θ sin3θ + sin3θ cos 3θ = 3/4 sin 4θ
Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.4
We have to prove that
sin 5A = 5cos4 A sin A - 10cos2 A sin3A + sin5 A
L.H.S = sin 5A = sin (3A + 2A)
= sin 3A cos 2A + cos 3A. sin 2A
= (3sin A - 4sin3 A) (2cos2 A -1) + (4cos3 A - 3cos A) 2sin A cos A.
= -3 sin A + 4 sin3 A + 6 sin A cos2A - 8 sin3A cos2A + 8 cos4A sin A - 6 cos2A sin A
= 8cos4A sin A – 8 sin3A cos2A – 3sin A + 4 sin3A
= 5 cos4A sin A - 10 sin3A cos2A – 3 sin A + 3 cos4A sin A + 4 sin3A + 2 sin3A cos2A
= 5 cos4A sin A – 10 sin3A cos2A – 3 sin A (1 -cos4A) + 2 sin3A (2 + cos2A)
= 5cos4A sin A - 10 sin3A cos2A – 3 sin A (1- cos2 A) (1 + cos2A) +2 sin3 A(2 + cos2A)
= 5 cos4A sin A- 10 sin3A cos2A – 3 sin3 A (1 + cos2A) + 2 sin3 A (2 + cos2A)
= 5cos4AsinA-10 sin3Acos2A - sin3 A [3(1 + cos2A) - 2(2 + cos2A)]
= 5 cos4A sin A- 10 sin3A cos2A - sin3A [3 + 3 cos2A - 4 – 2cos2A]
= 5 cos4A sin A - 10 sin3A cos2A - sin3A [cos2A - 1]
= 5 cos4A sin A - 10 sin3A cos2A + sin5A
= RHS
Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.5
tan A × tan(A + 60°) + tan A × tan(A - 60°) + tan(A + 60°) tan(A - 60°)
Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.6
tan A + tan(60° + A) - tan(60° - A) = 3 tan 3A
LHS = tan A + tan (60° + A) - tan (60° - A)
= 3 tan 3A
So,
tan A + tan(60° + A) - tan(60° - A) = 3 tan 3A
Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.7
LHS = cot A + cot(60° + A) - cot(60° - A)
= 3cot 3A
= RHS
LHS = RHS
Hence proved
Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.8
LHS = cotA + cot(60° + A) + cot(120° + A)
= cotA + cot(60° + A) - cot[180° - (120° + A)]
{since - cotθ = cot(180° - θ)}
= cotA + cot(60° + A) - cot(60° - A)
= 3cot 3A
LHS = RHS
Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.9
= RHS
LHS = RHS