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

Answer & Explanation

Dakota Cunningham

Beginner2022-01-27Added 9 answers

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

Eleanor Shaffer

Beginner2022-01-28Added 16 answers

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