If cos(t)=-7/8 with pi<=t<=3pi/2, how do you find the value of cos(t/2)?

Apr 8, 2018

$\cos \left(\frac{t}{2}\right) = - \frac{1}{4}$

Explanation:

$\text{using the "color(blue)"half angle formula}$

•color(white)(x)cos(t/2)=+-sqrt((1+cost)/2)

$\pi \le t \le \frac{3 \pi}{2} \Rightarrow \frac{\pi}{2} \le \frac{t}{2} \le \frac{3 \pi}{4}$

rArrcos(t/2)=-sqrt((1-7/8)/2

$\textcolor{w h i t e}{\Rightarrow \cos \left(\frac{t}{2}\right)} = - \sqrt{\frac{1}{16}} = - \frac{1}{4}$

Apr 8, 2018

$\cos \left(\frac{t}{2}\right) = - \frac{1}{4}$

Explanation:

Identity:

color(red)bb(cos(x/2)=sqrt(1/2(1+cosx)

cos(t/2)=sqrt(1/2(1+(-7/8))

$\cos \left(\frac{t}{2}\right) = \sqrt{\frac{1}{2} \left(\frac{1}{8}\right)}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{\frac{1}{16}} = \pm \frac{1}{4}$

If:

$\pi \le t \le \frac{3 \pi}{2}$

Then:

$\frac{t}{2}$

$\frac{\pi}{2} \le \frac{t}{2} \le \frac{3 \pi}{4}$

This is in the II quadrant.

Where the cosine is negative:

$\cos \left(\frac{t}{2}\right) = - \frac{1}{4}$

Apr 8, 2018

The use of the some trigonometric identities is the key to solving this question.

Explanation:

First let's look at the double angle formula:
$\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$

This cannot be used in its current form since ${\sin}^{2} \left(x\right)$ is present. But, rearranging the pythagorean identity allows for the ${\sin}^{2} \left(x\right)$ to be replaced in terms of $\cos \left(x\right)$ :
${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$
${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$

Substitution of ${\sin}^{2} \left(x\right)$ into double angle formula:
$\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - \left(1 - {\cos}^{2} \left(x\right)\right)$
$\cos \left(2 x\right) = 2 {\cos}^{2} \left(x\right) - 1$

Now, looking at the question, if we let $t = 2 x$ and $\frac{t}{2} = x$ it changes the double angle formula to a form which can be directly used:
$\cos \left(t\right) = 2 {\cos}^{2} \left(\frac{t}{2}\right) - 1$

Finally, substituting in the value of $\cos \left(t\right)$ and rearranging the equation yields an answer:
$- \frac{7}{8} = 2 {\cos}^{2} \left(\frac{t}{2}\right) - 1$
$\frac{1}{8} = 2 {\cos}^{2} \left(\frac{t}{2}\right)$
$\frac{1}{16} = {\cos}^{2} \left(\frac{t}{2}\right)$
$\frac{1}{4} = \cos \left(\frac{t}{2}\right)$

Hence, the value of $\cos \left(\frac{t}{2}\right)$ is $\pm \frac{1}{4}$ .
$\cos \left(\frac{t}{2}\right) = - \frac{1}{4}$