Question

Question #8e717

Answer 1

See the proof below

Explanation:

We need

`cos2x=cos^2x-sin^2x`

`cos^2x+sin^2x=1`

`(a-b)^3=a^3-3a^2b+3ab^2-b^3`

Therefore,

`LHS=cos^3 2x+3cosx`

`=(cos^2x-sin^2x)^3+3(cos^2x-sin^2x)`

`=cos^6x-3cos^4x sin^2x+3cos^2xsin^4x-sin^6x+3cos^2x-3sin^2x`

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

`=4cos^6x-3cos^4x+3sin^4x-4sin^6x+3cos^2x-3sin^2x`

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

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

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

`=4(cos^6x-sin^6x)-3(cos^2x-1-sin^2xcos^2x+sin^2x)+3(sin^2x-1-cos^2xsin^2x+cos^2x)`

`=4(cos^6x-sin^6x)`

`=RHS`

`QED`

Answer 2

`RHS=4(cos^6x-sin^6x)`

`=4((cos^2x)^3-(sin^2x)^3)`

`=4(cos^2x-sin^2x)((cos^2x)^2+(sin^2x)^2+cos^2xsin^2x)`

`=4cos2x((cos^2x+sin^2x)^2-2cos^2xsin^2x+cos^2xsin^2x)`

`=4cos2x(1-cos^2xsin^2x)`

`=cos2x(4-4cos^2xsin^2x)`

`=cos2x(4-(2sinxcosx)^2)`

`=cos2x(3+1-sin^2 2x)`

`=cos2x(3+cos^2 2x)`

`=cos^3 2x+3cos2x=LHS`

Answer 3

See the Proof in the Explanation.

Explanation:

We know that, `cos2x=cos^2x-sin^2x.`

`:. cos^3 2x+3cos2x=cos2x{cos^2 2x+3}.`

`cos^2x+sin^2x=1 rArr (cos^2x+sin^2x)^2=1.` Hence,

`:. cos^3 2x+3cos2x,`

`=cos2x{(cos^2x-sin^2x)^2+3(cos^2x+sin^2x)^2},`

`=cos2x{cos^4x-2cos^2xsin^2x+sin^4x`

`+3cos^4x+6cos^2xsin^2x+3sin^4x},`

`=cos2x{4cos^4x+4cos^2xsin^2x+4sin^4x},`

`=4(cos^2x-sin^2x){(cos^2 x)^2+cos^2xsin^2x+(sin^2 x)^2},`

`=4{(cos^2x)^3-(sin^2x)^3},`

`rArrcos^3 2x+3cos2x=4(cos^6 x- sin^6 x).`

This completes the Proof.