# Thread: using the half angle formula correctly

1. ## using the half angle formula correctly

Here is the problem- If cos(t) = -9/11 where pi<t<3pi/2, find the values of the following trigonometric functions. the answer has to be a fraction or an arithmetic expression, exact answers.
for:
cos(2t)
sin(2t)
cos(t/2)
sin(t/2)
Here is what I did. Given that cos(t)= -9/11
(opp)^2 + (-9)^2 + (11)^2
(opp)^2 =202
opp = -sqrt(202)

I also thought since cos(2t) = 2cos^2(x)-1 I could do this
-9/11 = 2cos^2(x)-1
2cos^2(x)=2/11
cos^2(x)=1/11
cos(x)=squrt(1/11) and the answer in the back of the book tells me I am wrong. If you could get me on track I would really appreciate it. I will reread the chapter and the section on this and maybe it will come to me.
Keith Stevens

2. Hello, Keith!

If $\cos\theta = -\frac{9}{11}$, where $\pi < \theta < \frac{3\pi}{2}$,
find the values of the following trigonometric functions.

$(1)\;\cos2\theta\quad(2)\;\sin2\theta\quad(3)\;\co s\left(\frac{\theta}{2}\right)\quad (4)\;\sin\left(\frac{\theta}{2}\right)$

Here is what I did.
Given that: $\cos\theta = -\frac{9}{11}$
. . $(opp)^2 + (adj)^2\:=\:(hyp)^2$
. . $(opp)^2 + (-9)^2 \:=\:(11)^2$
. . $(opp)^2 \:=\:202$ . . . no

You had: . $(opp)^2 + 81 \;=\;121\quad\Rightarrow\quad (opp)^2\,=\,40\quad\Rightarrow\quad opp \,= \,\pm2\sqrt{10}$

Since $\theta$ is in Quadrant 3: $opp = -2\sqrt{20}$

So we have: . $\begin{array}{ccc}\sin\theta \:= \\ \\ \cos\theta \:=\end{array}\!\!
\begin{array}{ccc}-\frac{2\sqrt{10}}{11}\\ \\ -\frac{9}{11}\end{array}$

$(1)\;\cos2\theta\;=\;2\cos^2\theta-1 \;=\;2\left(-\frac{9}{11}\right)^2 - 1\;=\;\frac{162}{121} - 1\;=\;\frac{41}{121}$

$(2)\;\sin2\theta\;=\;2\sin\theta\cos\theta \;=\;2\left(-\frac{2\sqrt{10}}{11}\right)\left(-\frac{9}{11}\right)\;=\;\frac{36\sqrt{10}}{121}$

$(3)\;\cos\left(\frac{\theta}{2}\right)\;=\;\pm\sqr t{\frac{1 + \cos\theta}{2}} \;=\;\pm\sqrt{\frac{1 + \left(-\frac{9}{11}\right)}{2}} \;=\;\boxed{-\sqrt{\frac{1}{11}}}$ *

$(4)\;\sin\left(\frac{\theta}{2}\right)\;=\;\pm\sqr t{\frac{1 - \cos\theta}{2}} \;=\;\pm\sqrt{\frac{1 - \left(-\frac{9}{11}\right)}{2}}\;=\;\boxed{\sqrt{\frac{10 }{11}}}$ *

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

*
Since $\cos\theta = -\frac{9}{11}$, then $\theta \,\approx\,215^o$

Hence: $\frac{\theta}{2} \,\approx\,107.5^o$ is in Quadrant 2.