# Tangent of double anglecalculate the tangent of double angle sinx+cosx=1, tan2x=? cos2x=?

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Given that: sinx + cosx = 1

We need to find tan 2x

We know that:

tan2x = 2tanx / (1-tan^2 x)

Then, we will determine tanx

Let us divide the equation by cosx

==> sinx + cosx = 1

==> (sinx/cosx) + cosx/cosx = 1/cosx

==> tanx + 1 = 1/cosx

==> tanx = (1/cosx) - 1

==> tanx = (1 - cosx)/cosx

Now we will substitute into tan2x

==> tan2x = 2 (1-cosx)/cosx / [ 1- (1-cosx)^2/cosx)^2]

= 2(1-cosx)*cosx/ (cos^2 x - ( 1- 2cosx + cos^2x)]

= 2(1-cosx)*cosx / (cos^2 x -1 + 2cosx - cos^2 x)

= 2(1-cosx)*cosx/ ( 2cosx -1)

= (2cosx -2cos^2 x)/ (2cosx-1)

sinx+cosx=1

If we'll divide by cos x:

sin x/cos x + 1= 1/cos x

tan x + 1= 1/cos x

tan x= 1/(cos x -1)

We could write the tangent of the double angle as:

tan 2x = tan (x+x)

tan 2x = (tan x + tan x)/[1-(tan x)^2]

tan 2x = 2tan x/[1-(tan x)^2]

tan 2x =2/(cos x -1)/[1-(1/cosa)+1][1+ (1/cosa)-1]

If we'll raise to square, we'll get:

(tan 2x)^2 + 1= 1/(cos 2x)^2

cos 2x = 1/sqrt[(tan 2x)^2 + 1]