Question

tan x=1/2a, pi <= x <= 3pi/2, determine an exact value for cos(2x)

Answer 1

is it this \(tanx=\dfrac{1}{2a}\) or this \(tanx=\dfrac{1}{2}a\)

Answer 2

The first, sorry I should have added brackets.

Answer 3

\(\large\tt \begin{align} \color{black}{tan x=\dfrac{1}{2a}\\~\\
tan^2x=\dfrac{1}{4a^2}\\~\\
tan^2x+1=\dfrac{1}{4a^2}+1\\~\\
sec^2x=\dfrac{1+4a^2}{4a^2}\\~\\
cos^2x=\dfrac{4a^2}{1+4a^2}----- \color{red}{1}\\~\\
1-sin^2x=\dfrac{4a^2}{1+4a^2}\\~\\

-sin^2x=\dfrac{4a^2}{1+4a^2}-1----- \color{red}{2}\\~\\

\text{by adding 1 and 2}\\~\\
cos^2x-sin^2x=\dfrac{4a^2}{1+4a^2}+\dfrac{4a^2}{1+4a^2}-1\\~\\
cos(2x)=\dfrac{4a^2}{1+4a^2}+\dfrac{4a^2}{1+4a^2}-1\\~\\
=\dfrac{8a^2-1-4a^2}{1+4a^2}\\~\\
=\dfrac{4a^2-1}{4a^2+1}\\~\\

}\end{align}\)

Answer 4

Ahh okay, I see what you did. I was going about it the wrong way. Thank you kindly. It's much appreciated.

Answer 5

yw