Htykuut

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,
$$costheta = -frac12 quadtext{and}quad sintheta = frac{sqrt{3}}{2} $$
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.

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 $(2pi-theta)$. For example, $frac{pi}{4}$ corresponds to $frac{3pi}{4}$, $frac{5pi}{4}$, and $frac{7pi}{8}$ 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).

$$sinbigg(frac{pi}{2}+thetabigg) = costheta quad sinbigg(frac{pi}{2}-thetabigg) = costheta$$

$$cosbigg(frac{pi}{2}+thetabigg) = -sintheta quad cosbigg(frac{pi}{2}-thetabigg) = sintheta$$

$$tanbigg(frac{pi}{2}+thetabigg) = -cottheta quad tanbigg(frac{pi}{2}-thetabigg) = cottheta$$

$$sinbigg(pi+thetabigg) = -sintheta quad sinbigg(pi-thetabigg) = sintheta$$

$$cosbigg(pi+thetabigg) = -costheta quad cosbigg(pi-thetabigg) = -costheta$$

$$tanbigg(pi+thetabigg) = tantheta quad tanbigg(pi-thetabigg) = -tantheta$$

$$sinbigg(frac{3pi}{2}+thetabigg) = -costheta quad sinbigg(frac{3pi}{2}-thetabigg) = -costheta$$

$$cosbigg(frac{3pi}{2}+thetabigg) = sintheta quad cosbigg(frac{3pi}{2}-thetabigg) = -sintheta$$

$$tanbigg(frac{3pi}{2}+thetabigg) = -cottheta quad tanbigg(frac{3pi}{2}-thetabigg) = cottheta$$

$$sinbigg(2pi+thetabigg) = sintheta quad sinbigg(2pi-thetabigg) = -sintheta$$

$$cosbigg(2pi+thetabigg) = costheta quad cosbigg(2pi-thetabigg) = costheta$$

$$tanbigg(2pi+thetabigg) = tantheta quad tanbigg(2pi-thetabigg) = -tantheta$$

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.

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

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

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.

Consider the following equilateral triangle $ABC$, where $D$ is the foot of the altitude $overline{CD}$

Let $[AB]=a$

Thus $$cosbigl(angle DAC bigr)=cosBigl(frac{pi}{3}Bigr)=frac{AC}{AD}=frac{1}{2}$$

Consider now the trigonometric identities $$cosbigl(pi-thetabigr)=-cosbigl(thetabigr)Rightarrow cosBigl(frac{2pi}{3}Bigr)=-frac{1}{2}$$
$$cosbigl(theta+2pibigr)=cos(theta) Rightarrow cosBigl(frac{2pi}{3}+2pi nBigr)=-frac{1}{2}; forall nin mathbb{Z}$$

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 $$sinbigl(angle DACbigr)=sinBigl(frac{pi}{3}Bigr)=frac{[DC]}{[AC]}=frac{sqrt{3}}{2}$$
Since $sin(theta)=sin(theta + 2pi)$
$$sinBigl(frac{pi}{3}+2npiBigr)=frac{sqrt{3}}{2} ; forall nin mathbb{Z}$$

$mathbf{PS}$

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

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 $theta = pi-P = frac{2pi}{3}$ and in the third quadrant $theta = pi+P = frac{4pi}{3}$.


3. If you want more solutions then solve $theta = pi-P+2kpi$ and $theta = pi+P+2kpi$ instead, where k is any positive or negative integer.

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

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




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

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