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I have an example:

$$ \frac{49}{64}\cos^2 \theta + \cos^2 \theta = 1 $$

Then what happens next:

$$ \frac{113}{64}\cos^2 \theta = 1 $$

Where has the other cosine disappeared to? What operation happened here? Any hints please.

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    $\begingroup$ $\frac{49}{64}\cos^2(\theta) + \frac{64}{64}\cos^2(\theta) = 1$. So adding those two fractions gives you the coefficient $\frac{113}{64}$. $\endgroup$ – MathNewbie May 30 '15 at 5:45
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    $\begingroup$ Here's a hint: look at $1 + (49/64)$. Cheers! $\endgroup$ – Robert Lewis May 30 '15 at 5:46
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    $\begingroup$ If you have $\frac{49}{64}$ of a potato, and add a whole potato to that, you have, in total, $1\frac{49}{64}$ of a potato as a mixed number. Converting that mixed number to an improper fraction, you have $\frac{113}{64}$ of a potato. $\endgroup$ – alex.jordan May 30 '15 at 6:03
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    $\begingroup$ I would just like to add a comment here that I appreciate that the answers and comments here were not condescending in the slightest. You can't expect that from other sites on this network, especially stackoverflow. $\endgroup$ – MCT May 30 '15 at 17:34
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This is a simple use of the distributive law (Wikipedia link): $$ac+bc=(a+b)c$$ In this situation, $$\left(\frac{49}{64}\right)\cos^2(\theta)+\left(1\right)\cos^2(\theta)=\left(\frac{49}{64}+1\right)\cos^2(\theta)=\left(\frac{113}{64}\right)\cos^2(\theta)$$

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By distributive law, $$\frac{49}{64}\cos^2 \theta+\cos^2 \theta=\left(\frac{49}{64}+1\right)\cos^2 \theta=\left(\frac{49}{64}+\frac{64}{64}\right)\cos^2 \theta=\left(\frac{49+64}{64}\right)\cos^2 \theta.$$ and the numerator $$49+64=113.$$

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$\dfrac{49}{64}\cos^2 \theta + \cos^2 \theta = \left(\dfrac{49}{64} +1\right) \cos^2 \theta = \dfrac{113}{64}\cos^2 \theta$

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