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Q.

Given expression: sin6x + cos6x lies between

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a

\frac{1}{4}

and 1

b

\frac{1}{4}

and 2

c

0 and 1

d

None of these

answer is A.

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Detailed Solution

a^{3} + b^{3}

=

{(a + b)}^{3}

- 3ab (a + b) ..................(1)
First of all we have to convert each term sin6x and cos6x in the form of a cube.
= (sin2x)3 + (cos2x)3 = (sin2x + cos2x)3 − 3sin2xcos2x (sin2 x + cos2 x) from equation (1)
= (1)3 − 3sin2xcos2x × 1 [since sin2 x + cos2 x = 1]
= 1 − 3sin2xcos2x = 1 – 3 ×

\frac{4}{4}

sin2xcos2x
= 1 −

\frac{3}{4}

×

{(2 \sin \sin x \cos \cos x)}^{2}

= 1 −

\frac{3}{4}

{(\sin \sin 2 x)}^{2}

Now, we have to find the maximum and minimum value of the expression.
Maximum value of sin6x + cos6x is 1 −

\frac{3}{4}

× 0 =1
Minimum value of sin6x + cos6x is 1 −

\frac{3}{4}

× 1 =

\frac{1}{4}

Hence, with the help of the formula (1), and (2) we have obtained the value of the given trigonometric expression sin6x + cos6x which lies between

\frac{1}{4}

and 1. Therefore option (1) is correct.

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