prove the identity
(sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2
sinX^6= sinx^2 ^3 = (1-cosX^2)^3
= (1-2CosX^2 + cos^4) (1-cosX^2)
then multiply that out
1-2CosX^2 + cos^4 - cosX^2 + 2cos^4 -cos^6
add that on the left to the cos^6, and combine terms..
1-3CosX^2 + 3cos^4
1-3cos^2x (i-cos^2 x)
then use the sin^2 x=1-cos^2 x and you have it.
To prove the identity (sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2, we will follow the steps you provided:
1. Start with sinX^6:
sinX^6= (sinX^2)^3
2. Use the identity sin^2 x = 1 - cos^2 x:
sinX^6= (1 - cosX^2)^3
3. Expand the cube:
sinX^6= (1 - 3cosX^2 + 3cos^4X - cos^6X)
4. Combine like terms:
sinX^6 + cosX^6 = 1 - 3cosX^2 + 3cos^4X - cos^6X + cosX^6
5. Simplify:
sinX^6 + cosX^6 = 1 - 3cosX^2 + 3cos^4X
6. Use the identity sin^2 x=1-cos^2 x:
sinX^6 + cosX^6 = 1 - 3(1-sin^2X)(1-cos^2X)
7. Expand and simplify:
sinX^6 + cosX^6 = 1 - 3(1-sin^2X)(1-cos^2X)
= 1 - 3(1-sin^2X+cos^2X-sin^2Xcos^2X)
= 1 - 3(1-sin^2X+cos^2X-sin^2X(1-sin^2X))
= 1 - 3(1-sin^2X+cos^2X-sin^2X+sin^4X)
= 1 - 3(1-sin^2X-sin^2X+cos^2X+sin^4X)
= 1 - 3(2sin^2X-sin^2X+cos^2X+sin^4X)
= 1 - 3(sin^2X+cos^2X-sin^2X+sin^4X)
= 1 - 3(sin^2X+cos^2X-sin^2X+sin^4X)
= 1 - 3(sin^2X+sin^4X)
= 1 - 3(sin^2X(1+sin^2X))
= 1 - 3(sin^2Xcos^2X)
= 1 - 3(sinX)^2(cosX)^2
Therefore, we have proven the identity (sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2.