# Question on Tangent in Acute quadrant

#### anderson

Hello everyone

Greetings. Need help on this textbook question:

The tangent for an acute angle x, is m. Find cos2x.

cos2x = cos^2 - sin^2
= 1 - (sin^2 /cos^2 )
= 1 - tan^2
= 1 - m^2

Appreciate your help to clarify, thank you.

Best regards

#### Plato

MHF Helper
The tangent for an acute angle x, is m. Find cos2x.
If $\tan(x)=m$ for $x\in I$ then $\cos(x)=\dfrac{1}{\sqrt{1+m^2}}~\&~\sin(x)=\dfrac{m}{\sqrt{1+m^2}}$

anderson

Dear Plato

Best regards

Last edited:

#### Plato

MHF Helper
How did you get the above equation, can you share the working?

#### anderson

Dear Plato

m, tan x = m/1. Triangle Law, cos x = 1/(1+m^2)^1/2

cos 2x = 2 cos^2x - 1
= 2/(m^2 + 1) - 1
= (1 - m^2)/(m^2 + 1)

Best regards

#### Plato

MHF Helper
Dear Plato

m, tan x = m/1. Triangle Law, cos x = 1/(1+m^2)^1/2

cos 2x = 2 cos^2x - 1
= 2/(m^2 + 1) - 1
= (1 - m^2)/(m^2 + 1)
In a right triangle with an acute angle measuring $x$ such that $\tan(x)=\frac{m}{1}$, then the hypotenuse measures $\sqrt{1+m^2}$

From those numbers find $\cos(x)~\&~\sin(x)~!$

#### anderson

Hi Plato

That is the working for the final answer. Look at the question before you comment.

The tangent for an acute angle x, is m. Find cos2x.

Best regards

#### Plato

MHF Helper
Dear Plato

m, tan x = m/1. Triangle Law, cos x = 1/(1+m^2)^1/2

cos 2x = 2 cos^2x - 1
= 2/(m^2 + 1) - 1
= (1 - m^2)/(m^2 + 1)
In a right triangle with an acute angle measuring $x$ such that $\tan(x)=\frac{m}{1}$, then the hypotenuse measures $\sqrt{1+m^2}$

From those numbers find $\cos(x)~\&~\sin(x)~!$

#### anderson

In a right triangle with an acute angle measuring $x$ such that $\tan(x)=\frac{m}{1}$, then the hypotenuse measures $\sqrt{1+m^2}$

From those numbers find $\cos(x)~\&~\sin(x)~!$
m, tan x = m/1. Triangle Law, cos x = 1/(1+m^2)^1/2

cos 2x = 2 cos^2x - 1
= 2/(m^2 + 1) - 1
= (1 - m^2)/(m^2 + 1