I'm having trouble solving trigonometric equations. For example, let's say I'm solving a problem and I arrive at a trigonometric equation that says, $$\cos\theta = -\frac12 \quad\text{and}\quad \sin\theta = \frac{\sqrt{3}}{2} $$ At this point, I get stuck and I don't have an efficient way to proceed apart from picking up a calculator.

I can figure that the quadrants (from the signs of the ratios) -- but I can't figure out the angles. What is a good way to figure out the angle? Specifically, how do I systematically solve $\sin$, $\cos$, and $\tan$ trigonometric equations? (I can reciprocate the other three into these ratios.)

I don't have trouble figuring out angles between $0^\circ$ to $90^\circ$ (since I have that memorized), but for angles in other quadrants, I get stuck.


3 Answers 3


If you have angle $\theta$ in quadrant $1$, you can find its "corresponding" angle in quadrant $2$ by $(\pi - \theta)$, in quadrant $3$ by $(\pi+\theta)$, and in quadrant $4$ by $(2\pi-\theta)$. For example, $\frac{\pi}{4}$ corresponds to $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, and $\frac{7\pi}{8}$ in quadrants $2$, $3$, and $4$, respectively. (That's how I always think of them at least.)

Also, recall sine functions correspond to the height of the right triangle ($y$-axis), so they are positive in quadrants $1$ and $2$. Cosine functions correspond to base of the right triangle ($x$-axis), so they are positive in quadrants $2$ and $4$. (Tangent functions can be found through sine and cosine functions.)

You can use the following identities (which are derived from the aforementioned facts).

$$\sin\bigg(\frac{\pi}{2}+\theta\bigg) = \cos\theta \quad \sin\bigg(\frac{\pi}{2}-\theta\bigg) = \cos\theta$$

$$\cos\bigg(\frac{\pi}{2}+\theta\bigg) = -\sin\theta \quad \cos\bigg(\frac{\pi}{2}-\theta\bigg) = \sin\theta$$

$$\tan\bigg(\frac{\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{\pi}{2}-\theta\bigg) = \cot\theta$$

$$\sin\bigg(\pi+\theta\bigg) = -\sin\theta \quad \sin\bigg(\pi-\theta\bigg) = \sin\theta$$

$$\cos\bigg(\pi+\theta\bigg) = -\cos\theta \quad \cos\bigg(\pi-\theta\bigg) = -\cos\theta$$

$$\tan\bigg(\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(\pi-\theta\bigg) = -\tan\theta$$

$$\sin\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cos\theta \quad \sin\bigg(\frac{3\pi}{2}-\theta\bigg) = -\cos\theta$$

$$\cos\bigg(\frac{3\pi}{2}+\theta\bigg) = \sin\theta \quad \cos\bigg(\frac{3\pi}{2}-\theta\bigg) = -\sin\theta$$

$$\tan\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{3\pi}{2}-\theta\bigg) = \cot\theta$$

$$\sin\bigg(2\pi+\theta\bigg) = \sin\theta \quad \sin\bigg(2\pi-\theta\bigg) = -\sin\theta$$

$$\cos\bigg(2\pi+\theta\bigg) = \cos\theta \quad \cos\bigg(2\pi-\theta\bigg) = \cos\theta$$

$$\tan\bigg(2\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(2\pi-\theta\bigg) = -\tan\theta$$

I certainly wouldn't recommend memorizing these though since knowing how the unit circle works basically means you know them already.

For example, in an equation you reach $$\cos \theta = -\frac{\sqrt{3}}{2}$$

You already know that $\cos {\frac{\pi}{6}} = \frac{\sqrt{3}}{2}$ and you also know cosine is negative in quadrants $2$ and $3$, so all you need to do is find the corresponding angle for ${\frac{\pi}{6}}$ in those quadrants.

$$\text{Quadrant II} \implies \theta = \pi-{\frac{\pi}{6}} = \frac{5\pi}{6}$$

$$\text{Quadrant III} \implies \theta = \pi+{\frac{\pi}{6}} = \frac{7\pi}{6}$$

This might take a bit of practice, but once you get this whole "corresponding" angle concept, it all becomes simple. Perhaps you can start by trying to visualize this by solving equations with a unit circle. You'll eventually get the hang of it.

  • $\begingroup$ Thank you, that helped. I think you meant $ \cos ( \frac {3\pi}{2} + \theta) $instead of $ \cos ( \frac{\pi}{2} + \theta) $ in one of the identities you've provided (5th from the last row). I tried editing but it failed because edits need to be of 6 characters minimum. $\endgroup$
    – WorldGov
    Nov 25, 2018 at 16:58
  • 1
    $\begingroup$ Oh yes, thanks for pointing out. I'll edit it. $\endgroup$
    – KM101
    Nov 25, 2018 at 17:00
  • $\begingroup$ To anyone else interested in a good way to remember the identities given in the answer, here's what I use: note that the ratios change into their co-ratios only at 90 and 270 degrees. At 180 and 360, the ratio does not change. To figure out the sign, ask yourself whether the original ratio (before changing in case of 90 and 270 degrees) is positive or negative in the appropriate quadrant. Fore example, consider $ \cos ( \frac{\pi}{2} + \theta ) $. Since we have 90 degrees, cos changes to sin. $ 90 + \theta $ falls in the second quadrant, where cos is negative; so we have $ - \sin \theta $. $\endgroup$
    – WorldGov
    Nov 25, 2018 at 17:02

Consider the following equilateral triangle $ABC$, where $D$ is the foot of the altitude $\overline{CD}$ enter image description here Let $[AB]=a$

Thus $$cos\bigl(\angle DAC \bigr)=cos\Bigl(\frac{\pi}{3}\Bigr)=\frac{AC}{AD}=\frac{1}{2}$$

Consider now the trigonometric identities $$cos\bigl(\pi-\theta\bigr)=-cos\bigl(\theta\bigr)\Rightarrow cos\Bigl(\frac{2\pi}{3}\Bigr)=-\frac{1}{2}$$ $$cos\bigl(\theta+2\pi\bigr)=cos(\theta) \Rightarrow cos\Bigl(\frac{2\pi}{3}+2\pi n\Bigr)=-\frac{1}{2}\; \forall n\in \mathbb{Z}$$

Now back to the equilateral triangle, observe that by the Pythagorean theorem $$[CD]=\sqrt{{[AC]^2}-{[AD]^2}}=\sqrt{{a^2}-{\Bigl(\frac{a}{2}\Bigr)^2}}=\frac{\sqrt{3}}{2}*a$$ Hence $$sin\bigl(\angle DAC\bigr)=sin\Bigl(\frac{\pi}{3}\Bigr)=\frac{[DC]}{[AC]}=\frac{\sqrt{3}}{2}$$ Since $sin(\theta)=sin(\theta + 2\pi)$ $$sin\Bigl(\frac{\pi}{3}+2n\pi\Bigr)=\frac{\sqrt{3}}{2} \; \forall n\in \mathbb{Z}$$


Keeping in mind the sine- (red) and cosine (green) functions might help with this problems enter image description here


enter image description here

In the above, $\theta$ is measured anticlockwise from the positive horizontal axis and P is the angle with the nearest horizontal axis. P takes values from 0 to $\frac{\pi}{2}$.

Suppose you want to solve $cos(\theta) = -0.5$,

  1. Solve $cos(P) = +0.5$ using inverse cosine on your calculator. $P =\frac{\pi}{3}$ is called the reference angle.
  2. Ask yourself where $cos(\theta)$ is negative. This is in the second and third quadrants. In the second quadrant $\theta = \pi-P = \frac{2\pi}{3}$ and in the third quadrant $\theta = \pi+P = \frac{4\pi}{3}$.
  3. If you want more solutions then solve $\theta = \pi-P+2k\pi$ and $\theta = \pi+P+2k\pi$ instead, where k is any positive or negative integer.

eg2. Solve $\sin(3\theta+1) = 0.5$ for $0\le \theta \le 2\pi$

  1. Solve $\sin(P) = 0.5$ then $P = \frac{\pi}{3}$

  2. Where is $sin$ positive? In quadrant 1 so put $3\theta+1 = P+2k\pi$ and in quadrant 2 so put $3\theta+1 = \pi-P+2k\pi$.

Now solve for $\theta$ and keep any solutions which fall in the range $0\le \theta \le 2\pi$.


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