# What is cos^2x if tanx=4?

Asked on by ingw33

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have tan x = 4.

Now we have to find (cos x)^2.

tan x = sin x / cos x

(tan x)^2 = (sin x)^2 / (cos x)^2

Also (sin x)^2 + (cos x)^2 = 1

=> (sin x)^2 = 1 - (cos x)^2

So (tan x)^2 = (sin x)^2 / (cos x)^2

=> (tan x)^2 = [1 - (cos x)^2] / (cos x)^2 = 4^2 = 16

=> [1 - (cos x)^2] = 16*(cos x)^2

=> 1 = 17*(cos x)^2

=> (cos x)^2 = 1/17.

Therefore (cos x)^2 = 1/17.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll start from the fundamental formula of trigonometry:

(sin x)^2 + (cos x)^2 = 1

We'll divide by (cos x)^2 both sides:

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

But the ratio (sin x)^2/ (cos x)^2 = (tan x)^2

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

From enunciation, we know that tan x = 4.

We'll square raise both sides:

(tan x)^2 = 16

16 + 1 = 1/(cos x)^2

1/(cos x)^2 = 17

(cos x)^2 = 1/17

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