Gerald Ritter

2022-01-26

What can we say about $a{\mathrm{cos}}^{2}\theta +b{\mathrm{sin}}^{2}\theta$?

Dakota Cunningham

Notice that ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$ then your expression is a weighted mean of a and b.

Eleanor Shaffer

Write it as $a\cdot {\mathrm{cos}}^{2}\theta +a\cdot {\mathrm{sin}}^{2}\theta +\left(b-a\right)\cdot {\mathrm{sin}}^{2}\theta$ which gives you $a+\left(b-a\right)\cdot {\mathrm{sin}}^{2}\theta$