Question

Which double-angle or half-angle identity would you use to verify that 1+cos 2α=(2)/(1+tan 2α)

Answer 1

(I will use x instead, if you don't mind)
Your question is like this?

\(\large\color{#000000 }{ \displaystyle 1+\cos(2x) =\frac{2}{1+\tan(2x)} }\)

Answer 2

Sorry

Answer 3

\(\color{#0cbb34}{\text{Originally Posted by}}\) @SolomonZelman
(I will use x instead, if you don't mind)
Your question is like this?

\(\large\color{#000000 }{ \displaystyle 1+\cos(2x) =\frac{2}{1+\tan^2x} }\)
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 4

Are you allowed to play with both sides or you can only use one side?

Answer 5

\(\large\color{#000000 }{ \displaystyle 1+\cos(2x) =\frac{2}{1+\tan^2x} }\)

\(\large\color{#000000 }{ \displaystyle 1+ \cos(2x) =\frac{2}{\sec^2x} }\)

\(\large\color{#000000 }{ \displaystyle 1+ \cos(2x) = 2\cos^2x }\)

\(\large\color{#000000 }{ \displaystyle 1+ \cos(2x) = 2\cos^2x -1+1 }\)

\(\large\color{#000000 }{ \displaystyle 1+ \cos(2x) = \cos^2x+\cos^2x -1+1 }\)

\(\large\color{#000000 }{ \displaystyle 1+ \cos(2x) = \cos^2x -\sin^2x+1 }\)

this is how I would go about solving the problem

Answer 6

you are using
\(\cos(2\theta)=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1 \)

but, you are using it backwards

Answer 7

you should be able to somple the problem without me...

Answer 8

So sorry, I had a customer walk in.