## Trigonometry (11th Edition) Clone

$$\sin^3\theta+\cos^3\theta+\sin\theta\cos^2\theta+\sin^2\theta\cos\theta=\sin\theta+\cos\theta$$ The details of explanation is in the Work step by step below.
$$\sin^3\theta+\cos^3\theta+\sin\theta\cos^2\theta+\sin^2\theta\cos\theta=\sin\theta+\cos\theta$$ Take a look at the left side. $$X=\sin^3\theta+\cos^3\theta+\sin\theta\cos^2\theta+\sin^2\theta\cos\theta$$ $$X=(\sin^3\theta+\sin\theta+\cos^2\theta)+(\cos^3\theta+\sin^2\theta\cos\theta)$$ $$X=\sin\theta(\sin^2\theta+\cos^2\theta)+\cos\theta(\cos^2\theta+\sin^2\theta)$$ - Now we already see that $\sin^2\theta+\cos^2\theta$ can be written into $1$, as stated in Pythagorean Identities. $$X=\sin\theta\times1+\cos\theta\times1$$ $$X=\sin\theta+\cos\theta$$ So, $$\sin^3\theta+\cos^3\theta+\sin\theta\cos^2\theta+\sin^2\theta\cos\theta=\sin\theta+\cos\theta$$ The equation is verified to be an identity now.