Simplification of $\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$ Simplify the expression $$\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$$ 
I have done this way : $(1-\cos\alpha \cos\beta)^2 = 1-2\cos\alpha \cos\beta +\cos^2\alpha \cos^2\beta$ 
Please guide further....
 A: Let $a=\cos\alpha,b=\cos\beta$
So, the the given expression $$=\sqrt{(1-ab)^2-(1-a^2)(1-b^2)}=\sqrt{1-2ab+a^2b^2-(1-a^2-b^2+a^2b^2)}=\sqrt{a^2+b^2-2ab}=|a-b|=|\cos\alpha-\cos\beta|$$
A: $$\begin{align}
&\phantom{=\;}( 1 - \cos\alpha \cos\beta)^2 - \sin^2\alpha \sin^2\beta \\
&= ( 1 - \cos\alpha \cos\beta)^2 - (\sin\alpha \sin\beta)^2 &(1) \\
&= ( 1 - \cos\alpha \cos\beta - \sin\alpha \sin\beta)( 1 - \cos\alpha \cos\beta + \sin\alpha \sin\beta) &(2) \\
&= \left( 1 - (\cos\alpha \cos\beta + \sin\alpha \sin\beta)\right)\left( 1 - (\cos\alpha \cos\beta - \sin\alpha \sin\beta) \right) &(3)\\
&= \left( 1 - \cos(\alpha-\beta)\right)\left( 1 - \cos(\alpha+\beta) \right) &(4) \\
&= 2 \sin^2\left(\frac{\alpha-\beta}{2}\right) \cdot 2 \sin^2\left(\frac{\alpha+\beta}{2}\right) &(5) \\
&= 4 \sin^2\left(\frac{\alpha-\beta}{2}\right)\sin^2\left(\frac{\alpha+\beta}{2}\right) &(6) \\
&= \left( 2 \sin\left(\frac{\alpha-\beta}{2}\right)\sin\left(\frac{\alpha+\beta}{2}\right) \right)^2 &(7) \\
&= \left( \cos\beta - \cos\alpha \right)^2 &(8)
\end{align}$$ 
so that
$$
\sqrt{( 1 - \cos\alpha \cos\beta)^2 - \sin^2\alpha \sin^2\beta} = \left| \cos\beta - \cos\alpha \right|$$
Steps:


*

*Regroup

*Difference of squares: $(x-y)^2 = (x-y)(x+y)$

*Regroup

*Angle Addition formulas for cosine

*Half-angle formulas for sine

*Simplification

*Regroup

*Product-to-Sum (well, -Difference here) "Prosthaphaeresis" formula  


See Wikipedia's "List of Trigonometric Identities" page for various formulas (especially the one for Step 8, which isn't so well known but comes in handy).
A: $$\begin{align*}(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta&=1+\cos^2{\alpha}\cos^2{\beta}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\
&=1+(1-\sin^2{\alpha)\cos^2{\beta}}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\
&=1+\cos^2{\beta}-\sin^2{\alpha\cos^2{\beta}}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\
&=1+\cos^2{\beta}-\sin^2{\alpha}-2\cos{\alpha}\cos{\beta}\\
&=\cos^2{\beta}+\cos^2{\alpha}-2\cos{\alpha}\cos{\beta}\\
&=(\cos {\alpha}-\cos {\beta})^2\end{align*}
$$
So the answer will be $|\cos {\alpha}-\cos {\beta}|$
