1. ## Half-Angle Formulas

Find the value of $\displaystyle cos\;\left(\frac{\alpha}{2}\right)$ if $\displaystyle tan\;\alpha = -0.2917$ and $\displaystyle (90<\alpha <180)$

I'm not exactly sure how to type the degrees sign using latex. And I'm also lost on how to tackle this problem.

2. Originally Posted by TI-84
Find the value of $\displaystyle cos\;\left(\frac{\alpha}{2}\right)$ if $\displaystyle tan\;\alpha = -0.2917$ and $\displaystyle (90<\alpha <180)$

I'm not exactly sure how to type the degrees sign using latex. And I'm also lost on how to tackle this problem.
Double angle formula: $\displaystyle \cos(2A) = 2 \cos^2 A - 1 \Rightarrow \cos^2 A = \frac{\cos (2A) + 1}{2}$.

Let $\displaystyle A = \frac{\alpha}{2}$:

$\displaystyle \cos^2 \left( \frac{\alpha}{2}\right) = \frac{\cos \alpha + 1}{2} \Rightarrow \cos \left( \frac{\alpha}{2}\right) = \pm \sqrt{\frac{\cos \alpha + 1}{2}}$.

You know $\displaystyle \tan\;\alpha = -0.2917$ where $\displaystyle 90<\alpha <180$ ..... from this you can get $\displaystyle \cos \alpha$.

Substitute the value of $\displaystyle \cos \alpha$ into the above formula and use the fact that $\displaystyle 90< \alpha <180 \Rightarrow 45 < \frac{\alpha}{2} < 90$ to justify using the positive root. Capisce?

3. How exactly do I know what the value of $\displaystyle cos\;\alpha$ is?

4. Originally Posted by TI-84
How exactly do I know what the value of $\displaystyle cos\;\alpha$ is?
Substitute $\displaystyle \tan\ \alpha = -0.2917$ into $\displaystyle 1 + \tan^2 \alpha = \sec^2 \alpha$.

Solve for $\displaystyle \sec^2 \alpha$ and hence get $\displaystyle \cos \alpha$. Use the given fact that $\displaystyle 90<\alpha <180$ and hence $\displaystyle \cos \alpha < 0$ to justify keeping the negative value for $\displaystyle \cos \alpha$.

5. Originally Posted by mr fantastic
Substitute $\displaystyle \tan\ \alpha = -0.2917$ into $\displaystyle 1 + \tan^2 \alpha = \sec^2 \alpha$.

Solve for $\displaystyle \sec^2 \alpha$ and hence get $\displaystyle \cos \alpha$. Use the given fact that $\displaystyle 90<\alpha <180$ and hence $\displaystyle \cos \alpha < 0$ to justify keeping the negative value for $\displaystyle \cos \alpha$.
Alright so what I'm getting from what you said is $\displaystyle 1 + (tan\;-.2917)^2 = sec^2\;\alpha$ Which in turn gives $\displaystyle sec^2\;\alpha = 1.00002592\Longrightarrow sec\;\alpha = \sqrt{1.00002592}\Longrightarrow cos\;\alpha = \frac{1}{\sqrt{1.00002592}}$ I'm pretty sure I screwed up. Its some tricky stuff.

6. Originally Posted by TI-84
Alright so what I'm getting from what you said is $\displaystyle 1 + (tan\;-.2917)^2 = sec^2\;\alpha$ Which in turn gives $\displaystyle sec^2\;\alpha = 1.00002592\Longrightarrow sec\;\alpha = \sqrt{1.00002592}\Longrightarrow cos\;\alpha = \frac{1}{\sqrt{1.00002592}}$ I'm pretty sure I screwed up. Its some tricky stuff.
Since $\displaystyle \tan(\alpha)=-0.2917$ you have to plug in only the value (without the tan-function!). Your equation then becomes:

$\displaystyle 1+(-0.2917)^2 = (\sec(\alpha))^2$

Now go on!

7. I got $\displaystyle cos\;\alpha = .9597$ Is that right?

If it is do I follow that up by $\displaystyle \cos \left( \frac{.9597}{2}\right) = \pm \sqrt{\frac{.9597 + 1}{2}}$

8. Originally Posted by TI-84
I got $\displaystyle cos\;\alpha = .9597$ Is that right?

If it is do I follow that up by $\displaystyle \cos \left( \frac{.9597}{2}\right) = \pm \sqrt{\frac{.9597 + 1}{2}}$
Originally Posted by mr fantastic
Double angle formula: $\displaystyle \cos(2A) = 2 \cos^2 A - 1 \Rightarrow \cos^2 A = \frac{\cos (2A) + 1}{2}$.

Let $\displaystyle A = \frac{\alpha}{2}$:

$\displaystyle \cos^2 \left( \frac{\alpha}{2}\right) = \frac{\cos \alpha + 1}{2} \Rightarrow \cos \left( \frac{\alpha}{2}\right) = \pm \sqrt{\frac{\cos \alpha + 1}{2}}$.

You know $\displaystyle \tan\;\alpha = -0.2917$ where $\displaystyle 90<\alpha <180$ ..... from this you can get $\displaystyle \cos \alpha$.

Substitute the value of $\displaystyle \cos \alpha$ into the above formula and use the fact that $\displaystyle 90< \alpha <180 \Rightarrow 45 < \frac{\alpha}{2} < 90$ to justify using the positive root. Capisce?
Originally Posted by mr fantastic
Substitute $\displaystyle \tan \alpha = -0.2917$ into $\displaystyle 1 + \tan^2 \alpha = \sec^2 \alpha$.

Mr F says: So $\displaystyle 1 + (-0.2917)^2 = \sec^2 \alpha$.

Solve for $\displaystyle \sec^2 \alpha$ and hence get $\displaystyle \cos \alpha$.

Mr F says: So $\displaystyle \sec^2 \alpha = 1.0851 \Rightarrow \cos^2 \alpha = 0.9216 \Rightarrow \cos \alpha = \pm 0.9600$.

Use the given fact that $\displaystyle 90<\alpha <180$ and hence $\displaystyle \cos \alpha < 0$ to justify keeping the negative value for $\displaystyle \cos \alpha$.

Mr F says: So $\displaystyle \cos \alpha = - 0.9600$.
Therefore $\displaystyle \cos \left( \frac{\alpha}{2}\right) = \pm \sqrt{\frac{-0.9600 + 1}{2}} = \pm \sqrt{\frac{0.04}{2}} = \pm \sqrt{0.02}$.

But $\displaystyle 45 < \frac{\alpha}{2} < 90$. Therefore $\displaystyle \cos \left( \frac{\alpha}{2}\right) = \sqrt{0.02} = 0.1414$.