Finding value of $\cos \left(\frac{\theta}{2} \right)$ when tangent and quadrant is given

Question

I cannot figure out how to do this problem. It is a problem in my textbook but there is no answer to it. Do you mind helping me?

Within the range of $0 \leq θ \leq 2π$, if $\tan \theta = -3$ and $\sin \theta <0$, find the exact value of $\cos \left(\frac{\theta}{2} \right)$.

Answer 1

Hint: Use Pythagoras' theorem and the following information about trigonometric functions $$\tan\theta=\frac{\text{opposite}}{\text{adjacent}},\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$$ to find $\cos\theta$.

Then use the double angle formula to find $\cos\frac{\theta}{2}$ by considering $$\cos2\theta=2\cos^2\theta-1.$$

Answer 2

Look at all the info separately. Is $\sin \theta < 0$, then $\theta$ must be in the third or fourth quadrant, think of the unit circle drawing. But, on the other side, if $\tan \theta = -3 < 0$, then $\theta$ has to be on the second, or fourth quadrant. The only possibility that satisfies both cases is: $\theta$ is on the fourth quadrant. This tell us that $\frac{3 \pi}{2} \leq \theta < 2 \pi$, so we have $\frac{3 \pi}{4} \leq \frac{\theta}{2} < \pi$, so that means $\frac{\theta}{2}$ is on the second quadrant. So, after we find the absolute value of $\cos \left( \frac{\theta}{2} \right)$, we know that it must be negative. And, apparently, there's no way to find the exact value of that cosine, because $\mathrm{arctan} (-3)$ is not a special angle. Some not too intuitive trigonometric expressions are: $$\cos \left(\frac{\theta}{2} \right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} \qquad \qquad \tan^2\theta = \frac{1 - \cos(2 \theta)}{1 + \cos(2\theta)}$$ Sure, the second formula gives us $\tan^2\left(\frac{\theta}{4}\right) = \dfrac{1 - \cos(\frac{\theta}{2})}{1 + \cos \frac{\theta}{2}}$, and you could solve for $\cos (\frac{\theta}{2})$, but I don't think it would help much, becaus of that $\tan$, which probably can be written in terms of info we don't have, only. Also, Wolfram Alpha gives an horrible decimal expansion as answer, which makes me wonder if there really is way to solve this.

Answer 3

Note first that since $\sin\theta$ is negative, and $\tan\theta$ is negative, it follows that $\cos\theta$ is positive. Thus $\theta$ is in the fourth quadrant. It follows that $\theta/2$ is in the second quadrant, and therefore $\cos(\theta/2)$ is negative.

We use the cosine double-angle identity, which you probably have seen in the form $\cos(2t)=2\cos^2 t-1$. Putting $t=\theta/2$, we get $$\cos \theta=2\cos^2(\theta/2)-1.\tag{1}$$ We calculate $\cos\theta$. The identity $\sec^2\theta=\tan^2\theta+1$ is useful here. From it we get that $\sec^2\theta=10$. It follows that $\cos^2\theta=\frac{1}{10}$. Since $\cos\theta$ is positive, we have $\cos\theta=\frac{1}{\sqrt{10}}$.

Now we use (1). We have $\frac{1}{\sqrt{10}}=2\cos^2(\theta/2)-1$. Thus $$\cos^2(\theta/2)=\frac{1}{2}\left(1+\frac{1}{\sqrt{10}}\right).\tag{2}$$

Now, if we recall that $\cos(\theta/2)$ is negative, we can use (2) to find $\cos(\theta/2)$.

Remark: The identity $1+\tan^2\theta=\sec^2\theta=\frac{1}{\cos^2\theta}$ may be unfamiliar to you. It can be obtained from $\cos^2\theta+\sin^2\theta=1$ by dividing both sides by $\cos^2\theta$.