What is the value of ≤ft(2{n}π+5π/6)≤ft(2{n}π-5π/6) where n ∈ Z ?

Correct Answer is : −14-1/4−41

Correct Answer is : −14-1/4−41

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UPSC NDA 2 2022 Mathematics Solved Paper

Question : 54 of 120

Marks: +1, -0

\sin\left(2\text{n}\pi+\frac{5\pi}{6}\right)\sin\left(2\text{n}\pi-\frac{5\pi}{6}\right)

where n ∈ Z ?

Solution:

Concept:

Use the periodicity property:

\sin(2n\pi + \theta) = \sin\theta

for integer

n

Use the property of negative angles:

\sin(-\theta) = -\sin\theta

Use the transformation:

\sin(\pi - \theta) = \sin\theta

Explanation:

Given expression:

\sin\left(2n\pi + \frac{5\pi}{6}\right) \cdot \sin\left(2n\pi - \frac{5\pi}{6}\right)

First, apply periodicity:

\sin(2n\pi + \frac{5\pi}{6}) = \sin\frac{5\pi}{6}

and

\sin(2n\pi - \frac{5\pi}{6}) = \sin\left(-\frac{5\pi}{6}\right)

So expression becomes:

\sin\frac{5\pi}{6} \cdot \sin\left(-\frac{5\pi}{6}\right)

Using negative angle property:

\sin\left(-\frac{5\pi}{6}\right) = -\sin\frac{5\pi}{6}

Thus:

\sin\frac{5\pi}{6} \cdot \left(-\sin\frac{5\pi}{6}\right) = -\sin^2\frac{5\pi}{6}

Now rewrite

\frac{5\pi}{6}

\pi - \frac{\pi}{6}

Then

\sin\frac{5\pi}{6} = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\frac{\pi}{6} = \frac{1}{2}

Therefore

-\sin^2\frac{5\pi}{6} = -\left(\frac{1}{2}\right)^2 = -\frac{1}{4}

Answer:

-\frac{1}{4}

(Option A)

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