Question

Sin^6¢+cos^6¢=1-3/4sin^2¢, prove it.

Answer 1

Sin^6+sin^6 =(sin^2¢)+(cos^2)^3

=(sin^2¢+cos^2¢) ^3sin^2¢*cos^2¢(sin^2+cos^2¢)

=1-3sin^2¢*cos^2¢

=1-3cos^2¢cos^2¢

=1-3sin^2*cos^2¢

=1-3(sin^2¢*cos^2)

=1-3/4(2sin^2*cos^2¢)

=1-3/42sin^22¢rhs.

Answer 2

Given Sin^6¢+cos^6¢ can be written as

= (sin^2¢)^3+(cos^2¢)^3 

= (sin^2¢ + cos^2¢)^3 - 3 sin^2¢ cos^2¢( sin^2¢ + cos^2¢) 

Multiply LHS and RHS by 4.

= 1- 3/4 * 4 sin^2¢ cos^2¢ * 1

= 1 - 3/4 * (2 sin ¢ cos ¢)^2

= 1 - 3/4 * (sin 2¢)^2

= 1 - 3/4 sin^2¢.


Hope this helps!