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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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 $$49+64=113.$$