show that the following equalities are valid for every number x

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You need to evaluate the left side using the pythagorean identity, `sin^2 x + cos^2 x = 1` , such that:

`(sin^2 x + cos^2 x)^3 = sin^6 x + cos^6 x + 3sin^2 x*cos^2 x*(sin^2 x + cos^2 x)`

Since `sin^2 x + cos^2 x = 1` yields:

`1^3 = sin^6 x + cos^6 x + 3sin^2 x*cos^2 x*1`

`sin^6 x + cos^6 x = 1 - 3sin^2 x*cos^2 x`

Since `2 sin x*cos x = sin 2x` yields:

`sin^6 x + cos^6 x = 1 - 3*(sin 2x)^2/4`

Hence, evaluating the left side yields:

`sin^6 x + cos^6 x = 1 - (3/4)*(sin^2 2x)`

You need to evaluate the expression to the right side, that is `5/8 + 3/8*cos 4x`

Using the double angle identity for the factor `cos 4x` yields:

`cos 4x = cos 2*(2x) = 1 - 2sin^2 2x`

Replacing `1 - 2sin^2 2x` for `cos 4x` yields:

`5/8 + 3/8*cos 4x = 5/8 + 3/8*(1 - 2sin^2 2x)`

`5/8 + 3/8*cos 4x = 5/8 + 3/8 - 3/4(sin^2 2x)`

`5/8 + 3/8*cos 4x = 8/8 - 3/4(sin^2 2x)`

`5/8 + 3/8*cos 4x = 1 - (3/4)(sin^2 2x)`

Replacing `1 - (3/4)*(sin^2 2x)` for the left side expression, `sin^6 x + cos^6 x` ,  and `1 - (3/4)(sin^2 2x)` for the right side expression `5/8 + 3/8*cos 4x` , yields:

`1 - (3/4)(sin^2 2x) = 1 - (3/4)(sin^2 2x)`

Hence, comparing the results of evaluations of expressions both sides yields that they coincide, thus, the statement `sin^6 x + cos^6 x = 5/8 + 3/8*cos 4x` holds for every number x.

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