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Wize AP Calculus (BC) Textbook > Review: Functions/Pre-Calculus

Trigonometric Functions

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Trigonometric (Trig) Functions

SOH CAH TOA

Using a right triangle, Trig Functions can be defined by using the hypotenuse, opposite side, and adjacent sides.
sin⁡θ=opphypcsc⁡θ=hypoppcos⁡θ=adjhypsec⁡θ=hypadjtan⁡θ=oppadjcot⁡θ=adjopp\begin{array}{ccc} \sin\theta=\frac{opp}{hyp} & &\csc\theta=\frac{hyp}{opp}\\ \\ \cos\theta=\frac{adj}{hyp} & & \sec\theta=\frac{hyp}{adj}\\ \\ \tan\theta=\frac{opp}{adj} & &\cot\theta=\frac{adj}{opp} \end{array}


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CAST Rule (All Students Take Calculus)

We can determine the sign (positive or negative) for our trig functions based on which quadrant the angle measure falls in.


• Quadrant 1: All trig ratios are positive• Quadrant 2: Only sin ratio is positive• Quadrant 3: Only tan ratio is positive• Quadrant 4: Only cos ratio is positive\begin{array}{l} •\ \text{Quadrant 1: All trig ratios are positive}\\ •\ \text{Quadrant 2: Only sin ratio is positive}\\ •\ \text{Quadrant 3: Only tan ratio is positive}\\ •\ \text{Quadrant 4: Only cos ratio is positive} \end{array}

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Trig Graphs


Special Triangles

So called 30-60-90 Triangles and 45-45-90 Triangles help us find exact values of trig functions.


  1. Draw out the angle given
  2. Identify the relative acute angle (RAA)--angle made with the x-axis
  3. Find the trig ratio of the RAA using special triangles
  4. Make the trig ratio positive or negative based on the CAST rule
Example: tan⁡(π3)=\tan\left(\frac{\pi}{3}\right)=


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Example: Trig Values

Evaluate the following

a) tan⁡7π6\displaystyle \tan\frac{7\pi}{6}
  1. The angle 7π6\frac{7\pi}{6} can be written as π+π6\pi+\frac{\pi}{6}. So, when we draw out the angle, it’s in quadrant 3.
  2. The relative acute angle (RRA) is the angle it makes with the 𝑥-axis, which will be π6\frac{\pi}{6}.
  3. Finding the trig. ratio of the relative acute angle, we get tan⁡(π6)=13\tan\left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}
  4. Since tan⁡\tan is positive in the 3rd quadrant (use CAST rule), our answer is tan⁡(7π6)=13\tan\left(\frac{7\pi}{6}\right)=\frac{1}{\sqrt{3}}


b) sin⁡15π4\displaystyle \sin\frac{15\pi}{4}
  1. The angle 15π4\frac{15\pi}{4} can be written as 3π+3π43\pi+\frac{3\pi}{4}. So, when we draw out the angle, it's in quadrant 4.
  2. The relative acute angle (RRA) is the angle it makes with the 𝑥-axis, which will be π4\frac{\pi}{4}.
  3. Finding the trig. ratio of the relative acute angle, we get sin⁡(π4)=12\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}
  4. Since sin⁡\sin is negative in the 4th quadrant (use CAST rule), out answer is sin⁡(15π4)=−12\sin\left(\frac{15\pi}{4}\right)=-\frac{1}{\sqrt{2}}
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Trig Identities

Basic Identities

  • tan⁡θ=sin⁡θcos⁡θ\displaystyle \tan\theta=\frac{\sin\theta}{\cos\theta}
  • sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
  • tan⁡2θ+1=sec⁡2θ\tan^2\theta+1=\sec^2\theta

Double Angles

  • sin⁡(2θ)=2sin⁡θ cos⁡θ\sin\left(2\theta\right)=2\sin\theta\ \cos\theta
  • cos⁡(2θ)=cos⁡2θ−sin⁡2θ  or  2cos⁡2θ−1  or  1−2sin⁡2θ\cos\left(2\theta\right)=\cos^2\theta-\sin^2\theta\ \ \text{or}\ \ 2\cos^2\theta-1\ \ \text{or}\ \ 1-2\sin^2\theta

Angle Sum & Difference

  • sin⁡(x±y)=sin⁡xcos⁡y±cos⁡xsin⁡y\sin\left(x\pm y\right)=\sin x\cos y\pm\cos x\sin y
  • cos⁡(x±y)=cos⁡xcos⁡y∓sin⁡xsin⁡y\cos\left(x\pm y\right)=\cos x\cos y\mp\sin x\sin y

Even and Odd Properties

  • Odd function: sin⁡(−θ)=−sin⁡θ\sin\left(-\theta\right)=-\sin\theta
  • Even function: cos⁡(−θ)=cos⁡θ\cos\left(-\theta\right)=\cos\theta
Given cos⁡ θ=1213\cos\ \theta=\frac{12}{13} and 3π2<θ<2π\frac{3\pi}{2}<\theta<2\pi , find the exact values of 13sin⁡2θ13\sin2\theta.

Solve the equation 2sin⁡2x=3sin⁡x−12\sin^2x=3\sin x-1 on the interval [0,2π]\left[0,2\pi\right].