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Q.

Given expression: sin6x + cos6x lies between


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a

14 and 1

b

14 and 2

c

0 and 1

d

None of these 

answer is A.

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Detailed Solution

a3+ b3 = ( 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 × 44 sin2xcos2x
= 1 − 34 × (2sinsin xcoscos x)  2
= 1 − 34 (sinsin 2x) 2
Now, we have to find the maximum and minimum value of the expression.
Maximum value of sin6x + cos6x is 1 − 34 × 0 =1
Minimum value of sin6x + cos6x is 1 − 34 × 1 = 14
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 14 and 1. Therefore option (1) is correct.
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