Question

I will give 5 points

Prove the identity \frac { \sin ^ { 3 } \theta + \cos ^ { 3 } \theta } { \sin \theta + \cos \theta } = 1 - \sin \theta \cdot \cos \theta​

Answer 1

Use the sum of cubes factoring rule

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

to transform the left hand side into the right hand side.

`\frac { \sin^(3) \theta + \cos^(3) \theta } { \sin \theta + \cos \theta } = 1 - \sin \theta \cdot \cos \theta\\\\\frac { (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cdot \cos\theta + \cos^2 \theta) } { \sin \theta + \cos \theta } = 1 - \sin \theta \cdot \cos \theta\\\\`

`\sin^2 \theta - \sin \theta \cdot \cos\theta + \cos^2 \theta = 1 - \sin \theta \cdot \cos \theta\\\\(\sin^2 \theta + \cos^2\theta)- \sin \theta \cdot \cos\theta = 1 - \sin \theta \cdot \cos \theta\\\\1- \sin \theta \cdot \cos\theta = 1 - \sin \theta \cdot \cos \theta \ \ \checkmark\\\\`

Throughout the entire process, the right hand side stayed the same.

On the last step, I used the pythagorean identity.