# Why does $\frac{49}{64}\cos^2 \theta + \cos^2 \theta$ equal $\frac{113}{64}\cos^2 \theta$?

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.

• $\frac{49}{64}\cos^2(\theta) + \frac{64}{64}\cos^2(\theta) = 1$. So adding those two fractions gives you the coefficient $\frac{113}{64}$. – MathNewbie May 30 '15 at 5:45
• Here's a hint: look at $1 + (49/64)$. Cheers! – Robert Lewis May 30 '15 at 5:46
• 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. – alex.jordan May 30 '15 at 6:03
• 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. – MCT May 30 '15 at 17:34

This is a simple use of the distributive law (Wikipedia link): $$ac+bc=(a+b)c$$ In this situation, $$\left(\frac{49}{64}\right)\cos^2(\theta)+\left(1\right)\cos^2(\theta)=\left(\frac{49}{64}+1\right)\cos^2(\theta)=\left(\frac{113}{64}\right)\cos^2(\theta)$$
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 $$49+64=113.$$
$\dfrac{49}{64}\cos^2 \theta + \cos^2 \theta = \left(\dfrac{49}{64} +1\right) \cos^2 \theta = \dfrac{113}{64}\cos^2 \theta$