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UPSC CDS 2 2022 Math Paper

Question : 58 of 100

Marks: +1, -0

What is

2 \sin^{6}\theta + 2 \cos^{6}\theta - 3 \sin^{4}\theta - 3 \cos^{4}\theta

equal to ?

Solution:

Concept:

The given expression simplifies to a constant value independent of

\theta

Explanation:

We use the identities

\sin^{2}\theta + \cos^{2}\theta = 1

and the expansions for

a^{3}+b^{3}

and

a^{2}+b^{2}

Rewrite the expression:

2(\sin^{6}\theta + \cos^{6}\theta) - 3(\sin^{4}\theta + \cos^{4}\theta)

Let

a = \sin^{2}\theta

and

b = \cos^{2}\theta

. Then

a+b=1

\sin^{6}\theta + \cos^{6}\theta = a^{3}+b^{3} = (a+b)(a^{2}+b^{2}-ab) = 1 \cdot (a^{2}+b^{2}-ab)

\sin^{4}\theta + \cos^{4}\theta = a^{2}+b^{2} = (a+b)^{2} - 2ab = 1 - 2ab

So the expression becomes:

2[ (a^{2}+b^{2}-ab) ] - 3[1-2ab]

But

a^{2}+b^{2} = 1-2ab

, so

a^{2}+b^{2}-ab = 1-2ab - ab = 1-3ab

Thus the expression =

2(1-3ab) - 3(1-2ab) = 2 - 6ab - 3 + 6ab = -1

Alternatively, substitute

\theta = 0^{\circ}

\sin0^{\circ}=0

\cos0^{\circ}=1

, then

2(0+1) - 3(0+1) = 2-3 = -1

The value is always

-1

Answer:

-1

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