Question

extreme values of sin power 8x+cos power 8x is
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Answer 1

Step-by-step explanation:

Sin ⁸ x –Cos ⁸ x = ( 1 – 2 Cos²x) ( 1 – 2 Sin²x Cos²x)

LHS

Sin ⁸ x –Cos ⁸ x

= ( Sin⁴ x ) ² – (cos⁴ x) ²

= ( Sin⁴ x - Cos⁴ x ) ( Sin⁴ x + Cos⁴ x )

= ( Sin² x - Cos² x ) ( Sin² x + Cos² x ) ( Sin⁴ x + Cos⁴ x )

= ( Sin² x - Cos² x )(1) ( Sin⁴ x + Cos⁴ x ) as Sin² x + Cos² x=1

writing Sin² x= 1-Cos² x & adding and subtracting 2 Sin² xCos² x in second factor

=(1 - Cos² x -Cos² x )( Sin⁴ x + Cos⁴ x+ 2 Sin² xCos² x - 2 Sin² xCos² x)

= (1 -2 Cos² x) [( Sin² x + Cos² x )² - 2 Sin² xCos² x]

= (1 -2 Cos² x) [( 1)² - 2 Sin² xCos² x] as Sin² x + Cos² x=1

= ( 1 – 2 Cos²x) ( 1 – 2 Sin²x Cos²x)

Answer 2

Given:

sin^8x + cos^8x.

To Find:

The value of the given question.

Solution:

sin^8x + cos^8x.

= (sin^4x)^2 + (cos^4x)^2.

= (sin^4x + cos^4x)(sin^4x + cos^4x).

= [(sin^2x)^2 + (cos^2x)^2][(sin^2x)^2 + (cos^2x)^2].

= [(sin^2x + cos^2x) (sin^2x + cos^2x)] [(sin^2x + cos^2x) (sin^2x + cos^2x)].

Since, sin^2x + cos^2x = 1.

So,

= 1 * 1 * 1 * 1.

= 1.

Hence, the extreme value of sin^8x + cos^8x is 1.