Tangent of double angle calculate 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 +...
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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]
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