Check the solution to "What can we say about acos2θ+bsin2θ?" - Plainmath

Gerald Ritter

Answered question

2022-01-26

What can we say about

a \cos^{2} θ + b \sin^{2} θ

?

Answer & Explanation

Dakota Cunningham

Beginner2022-01-27Added 9 answers

Notice that

\sin^{2} θ + \cos^{2} θ = 1

then your expression is a weighted mean of a and b.

Eleanor Shaffer

Beginner2022-01-28Added 16 answers

Write it as

a \cdot \cos^{2} θ + a \cdot \sin^{2} θ + (b - a) \cdot \sin^{2} θ

which gives you

a + (b - a) \cdot \sin^{2} θ

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