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....
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....
Let $a=\cos\alpha,b=\cos\beta$
So, the the given expression $$=\sqrt{(1-ab)^2-(1-a^2)(1-b^2)}=\sqrt{1-2ab+a^2b^2-(1-a^2-b^2+a^2b^2)}=\sqrt{a^2+b^2-2ab}=|a-b|=|\cos\alpha-\cos\beta|$$
$$\begin{align} &\phantom{=\;}( 1 - \cos\alpha \cos\beta)^2 - \sin^2\alpha \sin^2\beta \\ &= ( 1 - \cos\alpha \cos\beta)^2 - (\sin\alpha \sin\beta)^2 &(1) \\ &= ( 1 - \cos\alpha \cos\beta - \sin\alpha \sin\beta)( 1 - \cos\alpha \cos\beta + \sin\alpha \sin\beta) &(2) \\ &= \left( 1 - (\cos\alpha \cos\beta + \sin\alpha \sin\beta)\right)\left( 1 - (\cos\alpha \cos\beta - \sin\alpha \sin\beta) \right) &(3)\\ &= \left( 1 - \cos(\alpha-\beta)\right)\left( 1 - \cos(\alpha+\beta) \right) &(4) \\ &= 2 \sin^2\left(\frac{\alpha-\beta}{2}\right) \cdot 2 \sin^2\left(\frac{\alpha+\beta}{2}\right) &(5) \\ &= 4 \sin^2\left(\frac{\alpha-\beta}{2}\right)\sin^2\left(\frac{\alpha+\beta}{2}\right) &(6) \\ &= \left( 2 \sin\left(\frac{\alpha-\beta}{2}\right)\sin\left(\frac{\alpha+\beta}{2}\right) \right)^2 &(7) \\ &= \left( \cos\beta - \cos\alpha \right)^2 &(8) \end{align}$$
so that
$$ \sqrt{( 1 - \cos\alpha \cos\beta)^2 - \sin^2\alpha \sin^2\beta} = \left| \cos\beta - \cos\alpha \right|$$
Steps:
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).
$$\begin{align*}(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta&=1+\cos^2{\alpha}\cos^2{\beta}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\ &=1+(1-\sin^2{\alpha)\cos^2{\beta}}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\ &=1+\cos^2{\beta}-\sin^2{\alpha\cos^2{\beta}}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\ &=1+\cos^2{\beta}-\sin^2{\alpha}-2\cos{\alpha}\cos{\beta}\\ &=\cos^2{\beta}+\cos^2{\alpha}-2\cos{\alpha}\cos{\beta}\\ &=(\cos {\alpha}-\cos {\beta})^2\end{align*} $$
So the answer will be $|\cos {\alpha}-\cos {\beta}|$