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....
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asked Feb 18 '13 at 12:49
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$\begingroup$ Are you familiar with trig formulas like $\cos \alpha \cos \beta +/- \sin \alpha \sin \beta$? Also $2 \cos \alpha \beta = \cos(x-y) + \cos(x+y)$ $\endgroup$ – Alex Feb 18 '13 at 12:55
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Let $a=\cos\alpha,b=\cos\beta$
So, the the given expression $$=\sqrt{(1-ab)^2-(1-a^2)(1-b^2)}=\sqrt{1-2ab+a^2b^2-(1-a^2-b^2+a^2b^2)}=\sqrt{a^2+b^2-2ab}=|a-b|=|\cos\alpha-\cos\beta|$$
answered Feb 18 '13 at 15:08
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$\begingroup$ thanks that was great...thanks a lot.. $\endgroup$ – Sachin Sharmaa Feb 18 '13 at 15:38
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$\begingroup$ @SachinSharmaa, my pleasure. I used $a,b$ to keep the calculation simple. $\endgroup$ – lab bhattacharjee Feb 18 '13 at 15:41
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$$\begin{align} &\phantom{=\;}( 1 - \cos\alpha \cos\beta)^2 - \sin^2\alpha \sin^2\beta \\ &= ( 1 - \cos\alpha \cos\beta)^2 - (\sin\alpha \sin\beta)^2 &(1) \\ &= ( 1 - \cos\alpha \cos\beta - \sin\alpha \sin\beta)( 1 - \cos\alpha \cos\beta + \sin\alpha \sin\beta) &(2) \\ &= \left( 1 - (\cos\alpha \cos\beta + \sin\alpha \sin\beta)\right)\left( 1 - (\cos\alpha \cos\beta - \sin\alpha \sin\beta) \right) &(3)\\ &= \left( 1 - \cos(\alpha-\beta)\right)\left( 1 - \cos(\alpha+\beta) \right) &(4) \\ &= 2 \sin^2\left(\frac{\alpha-\beta}{2}\right) \cdot 2 \sin^2\left(\frac{\alpha+\beta}{2}\right) &(5) \\ &= 4 \sin^2\left(\frac{\alpha-\beta}{2}\right)\sin^2\left(\frac{\alpha+\beta}{2}\right) &(6) \\ &= \left( 2 \sin\left(\frac{\alpha-\beta}{2}\right)\sin\left(\frac{\alpha+\beta}{2}\right) \right)^2 &(7) \\ &= \left( \cos\beta - \cos\alpha \right)^2 &(8) \end{align}$$
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).
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$\begingroup$ @Cortizol: $\sqrt{4}$ isn't $\pm 2$ (it's just $2$), but $\sqrt{x^2} = \pm x$ (whichever sign makes the variable result non-negative); but, of course, I should've used absolute value. Fixed. $\endgroup$ – Blue Feb 18 '13 at 13:56
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$\begingroup$ that was great explanation .... really thankfull to you.. $\endgroup$ – Sachin Sharmaa Feb 18 '13 at 15:38
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$$\begin{align*}(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta&=1+\cos^2{\alpha}\cos^2{\beta}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\ &=1+(1-\sin^2{\alpha)\cos^2{\beta}}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\ &=1+\cos^2{\beta}-\sin^2{\alpha\cos^2{\beta}}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\ &=1+\cos^2{\beta}-\sin^2{\alpha}-2\cos{\alpha}\cos{\beta}\\ &=\cos^2{\beta}+\cos^2{\alpha}-2\cos{\alpha}\cos{\beta}\\ &=(\cos {\alpha}-\cos {\beta})^2\end{align*} $$
So the answer will be $|\cos {\alpha}-\cos {\beta}|$
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$\begingroup$ This isn't good. Look at second and third row, there is mistake: it should be $1+\cos^2 \beta$ instead $2+\cos^2 \beta$ $\endgroup$ – Cortizol Feb 18 '13 at 13:36
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$\begingroup$ So, your expression is equal to $(\cos \alpha - \cos \beta)^2$ which is much better solution if we agree :) $\endgroup$ – Cortizol Feb 18 '13 at 13:41
