If we know ... $$P = (1+ \cos x)(1+ \cos y)(1+ \cos z) = (1- \cos x)(1- \cos y)(1- \cos z)$$ ... then prove ... $$P = |\sin x\sin y \sin z|$$


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  • $\begingroup$ What is $\P$? what did you try? $\endgroup$ – Iuli Nov 24 '16 at 7:37

Compute $P^2$ by multiplying the two different expressions. Use difference of squares and a basic trig identity.



$$P^2 = P \cdot P = (1+\cos x)(1+\cos y)(1+\cos z) \cdot (1-\cos x)(1-\cos y)(1-\cos z)$$ $$=(1+\cos x)(1-\cos x)\cdot (1+\cos y)(1-\cos y)\cdot(1+\cos z)(1-\cos z)$$ $$=(1-\cos^2x)(1-\cos^2y)(1-\cos^2z)$$

Then, applying the identity $\sin^2x+\cos^2x = 1$ yields,

$$P^2 = \sin^2x \cdot \sin^2y \cdot \sin^2z = (\sin x \cdot \sin y \cdot \sin z)^2.$$

Applying the square root to both sides, $$|P| = |\sin x \cdot \sin y \cdot \sin z|.$$

Since $1+\cos x \geq0$ for all $x$, one can show that $P$ is non-negative and thus $|P| = P$ so we finally have, $$P = |\sin x \cdot \sin y \cdot \sin z|.$$


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