# write sin2x and cos2x w.r.t. tangent? calculate 2^1/2 sin2x+3^1/2 cos2x, tanx=6^1/2/3

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### 1 Answer

You need to write sin 2x and cos 2x in terms of tanx such that:

`sin 2x = (2 tan x)/(1 + tan^2 x); cos 2x = (1-tan^2 x)/(1 + tan^2 x)`

Plugging `tan x = sqrt6/3` in the formulas above yields:

`sin 2x = (2(sqrt6)/3)/(1 + 6/9)=gt sin 2x = (2(sqrt6)/3)/(1 + 2/3)`

`` `sin 2x = (2(sqrt6)/3)/(5/3) =gt sin 2x = 2(sqrt6)/5`

`cos 2x = (1 - 2/3)/(1 + 2/3) =gt cos 2x = (1/3)/(5/3) =gt cos 2x = 1/5`

Evaluating the value of expression yields:

`sqrt2*sin 2x + sqrt3 cos 2x = (sqrt2)*2(sqrt6)/5 + (sqrt3)*(1/5)`

Notice that the fractions have like denominators such that:

`sqrt2*sin 2x + sqrt3 cos 2x = (4sqrt3+sqrt3)/5`

`sqrt2*sin 2x + sqrt3 cos 2x = 5sqrt3/5`

`sqrt2*sin 2x + sqrt3 cos 2x = sqrt3`

**Hence, evaluating the value of expression yields `sqrt2*sin 2x + sqrt3 cos 2x = sqrt3.` **