Wize High School Grade 11 Math Textbook > Trigonometry

Solving Simple Trigonometric Equations
A simple trigonometric equation is in the form:
f(θ)=kf(\theta)=kf(θ)=k

where kkk is a constant and f(θ)f(\theta)f(θ) is a trigonometric function. 

Step 1. 
Identify what quadrant(s) the angle lies in. 

Step 2.
Identify the reference angle. 

Step 3. 
Solve for the solutions in the appropriate quadrants. 

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Example

Solve for θ\thetaθ over the domain 0°≤θ<360°0\degree\leq{\theta}<360\degree0°≤θ<360°.

cos⁡θ=12\cos{\theta}=\displaystyle\frac{1}{2}cosθ=21​

Step 1. 
Identify what quadrant(s) the angle lies in. 

Since cos⁡θ>0\cos{\theta}>0cosθ>0, then cos⁡θ\cos{\theta}cosθ is in quadrants 1 and 4 by the CAST rule. 

Step 2.
Identify the reference angle. 

cos⁡θ=AdjHyp=12\cos{\theta}=\dfrac{Adj}{Hyp}=\dfrac{1}{2}cosθ=HypAdj​=21​, and the side-lengths of 1 and 2 appear in the 30°- 60°- 90°30\degree \text{- }60\degree\text{- }90\degree30°- 60°- 90° triangle:

The reference angle is 60°60\degree60°.

Step 3. 
Solve for the solutions in the appropriate quadrants. 

Quadrant 1 (the principal angle is the same as our reference angle): 

θ=60°\theta=60\degreeθ=60°

Quadrant 4 (almost a full circle -- the reference angle completes the circle): 

θ=360°−60°=300°\theta = 360\degree - 60\degree = 300\degreeθ=360°−60°=300°

Therefore, the solutions are θ=60°, 300°\boxed{\theta=60\degree, \ 300\degree }θ=60°, 300°​

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Example: Solving Simple Trigonometric Equations

Solve tan⁡θ=−0.1051\tan{\theta}=-0.1051tanθ=−0.1051  for θ\thetaθ over the domain 0°≤θ<360°0\degree\leq{}\theta<360\degree0°≤θ<360°.

Step 1. 

Since tan⁡θ\tan{\theta}tanθ is negative, θ\thetaθ is in quadrants 2 and 4 by the CAST rule.

Step 2.

We can't use a special triangle since we don't know anything that relates to −0.1051-0.1051−0.1051.
Instead, solve for the reference angle β\betaβ using the inverse:

tan⁡β=−0.1051β=tan⁡−1(−0.1051)β=−6°\begin{aligned}
\tan{\beta}&=-0.1051 \\[0.5em]
\beta&=\tan^{-1}(-0.1051) \\[0.5em]
\beta &= -6\degree

\end{aligned}tanβββ​=−0.1051=tan−1(−0.1051)=−6°​

So the reference angle must be 6°6\degree6°.
(The negative sign just means it is measured clockwise from the positive x-axis, but it doesn't change how we complete the question).

Step 3. 

Quadrant 2 -- almost a semi-circle; the reference angle completes the half-circle: 

θ=180°−6°=174°\theta = 180\degree - 6\degree = \boxed{174\degree}θ=180°−6°=174°​

Quadrant 4 -- almost a full circle; the reference completes the circle:

θ=360°−6°=354°\theta=360\degree - 6\degree = \boxed{354\degree}θ=360°−6°=354°​

Practice: Solving Simple Trigonometric Equations
Solve cos⁡θ=0\cos{\theta}=0cosθ=0 over the domain 0°≤θ<360°0\degree\leq{}\theta<360\degree0°≤θ<360°.

Practice: Solving Simple Trigonometric Equations
Solve for all possible values of θ\thetaθ between 0°0\degree0° and 360°360\degree360° such that sin⁡θ=0.342\sin\theta=0.342sinθ=0.342.

Smaller angle

\theta

in degrees:

Larger angle

\theta

in degrees:

I don't know

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Example: Trig Ratios for Angles Between 0° and 360°
Consider the point P(3,−4)P(3,-4)P(3,−4) on the circle of radius 5 defined by the equation x2+y2=25x^2 + y^2=25x2+y2=25.

If θ\thetaθ is the principal angle of the terminal arm to the point PPP, find tan⁡θ\tan\thetatanθ and cos⁡θ\cos\thetacosθ.

See video solution for "shorter method" for finding tan⁡θ\tan\thetatanθ and cos⁡θ\cos\thetacosθ.

Here's the full written solution:

1. We can graph the point P(3,−4)P(3,-4)P(3,−4) and draw the terminal arm.

In the image, x=3, y=−4x=3,\ y=-4x=3, y=−4.

2. We first find the reference angle β\betaβ using tan⁡\tantan to relate xxx and yyy. [Keep all side-lengths positive at this step]

tan⁡β=∣y∣∣x∣=43  ⟹  β=tan⁡−1(43)≈53.13°\begin{aligned}
\tan \beta&=\dfrac{|y|}{|x|}=\dfrac{4}{3}\\[1em]

\implies \beta &= \tan^{-1}\left( \dfrac{4}{3} \right) \approx \boxed{53.13\degree}
\end{aligned}tanβ⟹β​=∣x∣∣y∣​=34​=tan−1(34​)≈53.13°​​

3. The principal angle is almost a full circle (360°)(360\degree)(360°), and the reference angle completes the rest of the circle:

θ=360°−53.13°=306.87°\theta=360\degree-53.13\degree = \boxed{306.87\degree}θ=360°−53.13°=306.87°​

4. By the Pythagorean Theorem:

r2=x2+y2r2=32+42r=32+42=5\begin{aligned}
r^2 &= x^2 + y^2\\
r^2 &= 3^2 + 4^2\\
r&=\sqrt{3^2+4^2}=5\\
\end{aligned}r2r2r​=x2+y2=32+42=32+42​=5​

5. Trig ratios of θ\thetaθ are the same as trig ratios of β\betaβ, except possibly for the sign (+/-).
By the CAST rule, since we are in quadrant 4 (C), only Cosine is positive.

Therefore, tan⁡θ=−tan⁡β=−43\boxed{\tan\theta=\bct-\tan\beta=-\dfrac{4}{3}}tanθ=−tanβ=−34​​ and cos⁡θ=cos⁡β=xr=35\boxed{\cos\theta=\cos\beta=\dfrac{x}{r}=\dfrac{3}{5}}cosθ=cosβ=rx​=53​​

Practice: Trig Ratios for Angles Between 0° and 360°
Suppose tan⁡θ=54\tan{\theta}=\displaystyle\frac{5}{4}tanθ=45​ and sin⁡θ>0\sin\theta>0sinθ>0 for some angle θ\thetaθ between 0°0\degree0° and 360°360\degree360°.

Determine the other two primary trigonometric ratios. [Make sure your answers are exact values]

\cos{\theta}

\sin{\theta}

I don't know