haleyelizabeth2017
  • haleyelizabeth2017
Which double-angle or half-angle identity would you use to verify that 1+cos 2α=(2)/(1+tan 2α)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
(I will use x instead, if you don't mind) Your question is like this? \(\large\color{#000000 }{ \displaystyle 1+\cos(2x) =\frac{2}{1+\tan(2x)} }\)
haleyelizabeth2017
  • haleyelizabeth2017
Sorry
haleyelizabeth2017
  • haleyelizabeth2017
\(\color{#0cbb34}{\text{Originally Posted by}}\) @SolomonZelman (I will use x instead, if you don't mind) Your question is like this? \(\large\color{#000000 }{ \displaystyle 1+\cos(2x) =\frac{2}{1+\tan^2x} }\) \(\color{#0cbb34}{\text{End of Quote}}\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

SolomonZelman
  • SolomonZelman
Are you allowed to play with both sides or you can only use one side?
SolomonZelman
  • SolomonZelman
\(\large\color{#000000 }{ \displaystyle 1+\cos(2x) =\frac{2}{1+\tan^2x} }\) \(\large\color{#000000 }{ \displaystyle 1+ \cos(2x) =\frac{2}{\sec^2x} }\) \(\large\color{#000000 }{ \displaystyle 1+ \cos(2x) = 2\cos^2x }\) \(\large\color{#000000 }{ \displaystyle 1+ \cos(2x) = 2\cos^2x -1+1 }\) \(\large\color{#000000 }{ \displaystyle 1+ \cos(2x) = \cos^2x+\cos^2x -1+1 }\) \(\large\color{#000000 }{ \displaystyle 1+ \cos(2x) = \cos^2x -\sin^2x+1 }\) this is how I would go about solving the problem
SolomonZelman
  • SolomonZelman
you are using \(\cos(2\theta)=\cos^2\theta-\sin^2\theta=2\cos^2\theta-1 \) but, you are using it backwards
SolomonZelman
  • SolomonZelman
you should be able to somple the problem without me...
haleyelizabeth2017
  • haleyelizabeth2017
So sorry, I had a customer walk in.

Looking for something else?

Not the answer you are looking for? Search for more explanations.