Question #915b6

1 Answer
Jan 17, 2016

Use the identity sin^2theta+cos^2theta=1sin2θ+cos2θ=1 and raise it to powers that would make the equation contain the desired trigonometric functions.

Explanation:

  • First: solving (sin^2theta+cos^2theta)^3=(1)^3(sin2θ+cos2θ)3=(1)3
    (sin^2theta+cos^2theta)(sin^4theta+2sin^2theta*cos^2theta+cos^4theta)=1(sin2θ+cos2θ)(sin4θ+2sin2θcos2θ+cos4θ)=1

(sin^6theta+2sin^4theta*cos^2theta+sin^2theta*cos^4theta)+(cos^2theta*sin^4theta+2sin^2theta*cos^4theta+cos^6theta)=1(sin6θ+2sin4θcos2θ+sin2θcos4θ)+(cos2θsin4θ+2sin2θcos4θ+cos6θ)=1

(sin^6theta+cos^6theta)+[(2sin^4theta*cos^2theta+2sin^2theta*cos^4theta)+(sin^2theta*cos^4theta+cos^2theta*sin^4theta)]=1(sin6θ+cos6θ)+[(2sin4θcos2θ+2sin2θcos4θ)+(sin2θcos4θ+cos2θsin4θ)]=1

(sin^6theta+cos^6theta)+[sin^2theta*cos^2theta*((2sin^2theta+2cos^2theta)+(cos^2theta+sin^2theta))]=1(sin6θ+cos6θ)+[sin2θcos2θ((2sin2θ+2cos2θ)+(cos2θ+sin2θ))]=1

(sin^6theta+cos^6theta)+[sin^2theta*cos^2theta*((2)+(1))]=1(sin6θ+cos6θ)+[sin2θcos2θ((2)+(1))]=1
(sin^6theta+cos^6theta)+3[sin^2theta*cos^2theta]=1(sin6θ+cos6θ)+3[sin2θcos2θ]=1

:.sin^6theta+cos^6theta=1-3[sin^2theta*cos^2theta]

  • Using a the same method starting with (sin^2theta+cos^2theta)^2=(1)^2
    you get:
    sin^4theta+cos^4theta=1-2[sin^2theta*cos^2theta]

  • The left side of the equation =
    2(sin^6theta+cos^6theta)-3(sin^4theta+cos^4theta)+1
    =2(1-3sin^2theta*cos^2theta)-3(1-2sin^2theta*cos^2theta)+1
    =2cancel(-6sin^2theta*cos^2theta)-3+cancel(6sin^2theta*cos^2theta)+1
    =2-3+1 =0