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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 

Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.3

{∵ 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.5

 

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)

Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.6

= 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)

Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.7

= 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)

Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.8

= 3cot 3A

LHS = RHS

 

Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.9

Trigonometric Ratios of Multiple and Sub Multiple Angles – Exercise 9.2 – Q.9

= RHS

LHS = RHS


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