1. ## Trigonometric identities

Hi,

I'm working with a question that I think mainly involves trig identities, but I'm not sure if I have the right approach and can't figure out the later parts.

If cos(x)=-5/7 where pi < x < 3pi/2, find the values of cos(2x), sin(2x), cos(x/2), sin(x/2)
I'm also not sure how it being in the third quadrant plays a role...

For the first two, I worked out that cos(2x)=1/49 and sin(2x)=4sqrt6/7 with double angle formulas, but I'm not sure if those are correct and I don't know where to go now.

Thanks!

2. Originally Posted by metaname90
Hi,

I'm working with a question that I think mainly involves trig identities, but I'm not sure if I have the right approach and can't figure out the later parts.

I'm also not sure how it being in the third quadrant plays a role...

For the first two, I worked out that cos(2x)=1/49 and sin(2x)=4sqrt6/7 ? with double angle formulas, but I'm not sure if those are correct and I don't know where to go now.

Thanks!
Given that the angle is in the 3rd quadrant, you may express the angle as $x=180^o+A$

Hence, since $Cosx<-\frac{1}{\sqrt{2}}\Rightarrow\ x<225^o\Rightarrow\ 2x<450^o$ or $2x<2{\pi}+\frac{{\pi}}{2}$

$2x=360^o+2A\Rightarrow\ Cos2x=Cos2A=2Cos^2A-1=2\frac{25}{49}-\frac{49}{49}=\frac{1}{49}$

$Cos^2(2x)+Sin^2(2x)=1\Rightarrow\ Sin2x=\sqrt{\frac{49^2-1}{49^2}}$

$\frac{x}{2}$ is in the 2nd quadrant.

$\frac{x}{2}=90^o+\frac{A}{2}$

If you draw a circle and sketch the angles, you will see we can calculate

$Cos\left(90+\frac{A}{2}\right)=-Sin\left(\frac{A}{2}\right)$

and to find this you can use $Sin^2B=\frac{1}{2}(1-Cos2B)\Rightarrow\ Sin^2\left(\frac{A}{2}\right)=\frac{1}{2}(1-CosA)$ etc