Question
Been trying this forever so it says: Simplify 1+sin2x+cos2x/1+sin2x-cos2x
ive tried this problem so many times I need help with it...
ive tried this problem so many times I need help with it...
Answers
I believe you mean
(1+sin2x + cos2x)/(1+sin2x - cos2x)
= (sin^2 x + cos^2 x + 2sinxcosx + cos^2 x - sin^2x)/(sin^2 x + cos^2 x + 2sinxcosx - cos^2x + sin^2x)
= (2cos^2 x + 2sinxcosx)/(2sin^2 x + 2sinxcosx)
= 2cosx(cosx+sinx)/((2sinx(sinx+cos))
= 2cosx/2sinx
= cot x
(1+sin2x + cos2x)/(1+sin2x - cos2x)
= (sin^2 x + cos^2 x + 2sinxcosx + cos^2 x - sin^2x)/(sin^2 x + cos^2 x + 2sinxcosx - cos^2x + sin^2x)
= (2cos^2 x + 2sinxcosx)/(2sin^2 x + 2sinxcosx)
= 2cosx(cosx+sinx)/((2sinx(sinx+cos))
= 2cosx/2sinx
= cot x
Related Questions
sinx = 4/5 and x terminates in Quadrant II
Find sin2x and cos2x
How to get the answers, which...
Solve the equation of the interval (0, 2pi)
cosx=sinx
I squared both sides to get :cos²x=sin²x...
Solve identity,
(1-sin2x)/cos2x = cos2x/(1+sin2x)
I tried starting from the right side,
RS:
=(...
solve for (<_ = less than or equal to / pie = pie sign / -pie = negative pie)
3 sin²x = cos²x ; 0...