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Neeraj Yadav

· started a discussion

· 1 Months ago

64 quesion (Question:96569.0)

how sin squa.re 2 theta is greater than equal to all values of theta

Question:

If P = sin θ (sin θ + sin 3θ),then

Options:
A) P ≥ 0 only when θ ≥ 0
B)

P ≤ 0 for all real θ

C) P ≥ 0 for all real θ
D) P ≤ 0 only when θ ≤ 0
Solution:

We have

P = sin θ (sin θ + sin3 θ)

= sin θ (2 sin 2 θ. cos θ)


[\(\because sin C + sin D = 2 sin \left ( \cfrac {C+D}{2} \right ).cos\left ( \cfrac {C-D}{2} \right )\)]

⇒ P = sin 2 θ (2 sin θ. cos θ)

= sin 2θ sin 2θ = sin22 θ 

P = sin22 θ ≥ 0

for all real θ

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