Question

***Medal AND Fan Rewarded***

Solve for cosθ in sin^2θ+cos^2θ=1

Answer 1

subtract \(\sin^2(\theta)\) from both sides

Answer 2

This is an identity.

Answer 3

can you tell me the answer i should have gotten? @satellite73

Answer 4

\[\sin^2θ+\cos^2θ=1\]\[\cos^2θ=1-\sin^2θ\]\[\cos\theta=\sqrt{1-\sin^2\theta}\]

Answer 5

dont be confused by the fact that its "cos \(\theta\)" and "sin\(\theta\)"
if it really throws you off that much just look at it with variables instead:\[sin^2\theta+\cos^2\theta=1\]\[~~~\color{blue}{a}^2~~+~~~\color{blue}{b}^2~~=1\]\[~~~~~~~~~~~~~~~~b^2~~=1~-~a^2\]\[~~~~~~~~~~~~~~~~b~~~=\sqrt{~1~-~a^2~}\]now just replace \(a\) and \(b\) with \(\sin\theta\) and \(\cos\theta\):\[b=\sqrt{1-a^2}\]\[cos\theta=\sqrt{1-\sin^2\theta}\]

Answer 6

Thanks for your response!

Answer 7

hope it helped :)