NDA 2016
Exam NDA
Year 2016
Subject Mathematics
Chapter Trigonometric Ratios & Identities NDA 2016 · Mathematics · Trigonometric Ratios & Identities
If x= 70^ ^ · 50^ ^ and y= 60^ ^ · 80^ ^ , then what is x y equal...
If
x = sin 7 0 ∘ ⋅ sin 5 0 ∘ x=\sin 70^{{}^{\circ}} \cdot \sin 50^{{}^{\circ}} x = sin 7 0 ∘ ⋅ sin 5 0 ∘ and
y = cos 6 0 ∘ ⋅ cos 8 0 ∘ y=\cos 60^{{}^{\circ}} \cdot \cos 80^{{}^{\circ}} y = cos 6 0 ∘ ⋅ cos 8 0 ∘ , then what is
x y x y x y equal to?
Options A 1/16 B 1 / 8 1 / 8 1/8 C 1 / 4 1 / 4 1/4 D 1 / 2 1 / 2 1/2
Step-by-step solution
x = sin 7 0 ∘ ⋅ sin 5 0 ∘ and y = cos 6 0 ∘ ⋅ cos 8 0 ∘ ⇒ x y = cos 6 0 ∘ ⋅ sin 7 0 ∘ ⋅ sin 5 0 ∘ ⋅ cos 8 0 ∘ x y = 1 2 ⋅ sin ( 90 − 20 ) ⋅ sin ( 90 − 40 ) ⋅ cos 80 ⇒ x y = 1 2 ⋅ cos 20 ⋅ cos 40 ⋅ cos 80 ( ∵ sin ( 90 − x ) = cos x ) ⇒ x y = 1 2 ⋅ cos 2 0 ∘ ⋅ cos ( 60 − 20 ) ∘ ⋅ cos ( 60 + 20 ) ∘ ⇒ x y = 1 2 [ 1 4 cos 3 ( 2 0 ∘ ) ] = 1 2 × 1 4 × cos 6 0 ∘ = 1 16 . [ ∵ cos θ ⋅ cos ( 60 − θ ) ⋅ cos ( 60 + θ ) = 1 4 cos 3 θ ] \begin{aligned} & x=\sin 70^{{}^{\circ}} \cdot \sin 50^{{}^{\circ}} \text{\text{ and } } y=\cos 60^{{}^{\circ}} \cdot \cos 80^{{}^{\circ}} \\ & \Rightarrow x y=\cos 60^{{}^{\circ}} \cdot \sin 70^{{}^{\circ}} \cdot \sin 50^{{}^{\circ}} \cdot \cos 80^{{}^{\circ}} \\ & x y=\frac{1}{2} \cdot \sin (90-20) \cdot \sin (90-40) \cdot \cos 80 \\ & \Rightarrow x y=\frac{1}{2} \cdot \cos 20 \cdot \cos 40 \cdot \cos 80 \\ & \quad(\because \sin (90-x)=\cos x) \\ & \Rightarrow x y=\frac{1}{2} \cdot \cos 20^{{}^{\circ}} \cdot \cos (60-20)^{{}^{\circ}} \cdot \cos (60+20)^{{}^{\circ}} \\ & \Rightarrow x y=\frac{1}{2}\left[\frac{1}{4} \cos 3\left(20^{{}^{\circ}}\right)\right]=\frac{1}{2} \times \frac{1}{4} \times \cos 60^{{}^{\circ}}=\frac{1}{16} . \\ & \quad\left[\because \cos \theta \cdot \cos (60-\theta) \cdot \cos (60+\theta)=\frac{1}{4} \cos 3 \theta\right]\end{aligned} x = sin 7 0 ∘ ⋅ sin 5 0 ∘ and y = cos 6 0 ∘ ⋅ cos 8 0 ∘ ⇒ x y = cos 6 0 ∘ ⋅ sin 7 0 ∘ ⋅ sin 5 0 ∘ ⋅ cos 8 0 ∘ x y = 2 1 ⋅ sin ( 90 − 20 ) ⋅ sin ( 90 − 40 ) ⋅ cos 80 ⇒ x y = 2 1 ⋅ cos 20 ⋅ cos 40 ⋅ cos 80 ( ∵ sin ( 90 − x ) = cos x ) ⇒ x y = 2 1 ⋅ cos 2 0 ∘ ⋅ cos ( 60 − 20 ) ∘ ⋅ cos ( 60 + 20 ) ∘ ⇒ x y = 2 1 [ 4 1 cos 3 ( 2 0 ∘ ) ] = 2 1 × 4 1 × cos 6 0 ∘ = 16 1 . [ ∵ cos θ ⋅ cos ( 60 − θ ) ⋅ cos ( 60 + θ ) = 4 1 cos 3 θ ]
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