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Grade 11 · MCR3U · Functions

MCR3U: Special Angles and the CAST Rule

The CAST rule tells you the sign of each trig ratio in each quadrant, and the special triangles give exact values for 30∘30^\circ, 45∘45^\circ and 60∘60^\circ. Together they give exact ratios for angles all the way to 360∘360^\circ, and every angle that has a given ratio.

How It Works

Standard position. Put the vertex of the angle at the origin, with its initial arm on the positive xx-axis, and rotate counterclockwise to the terminal arm. Take any point (x,y)(x, y) on the terminal arm, a distance r=x2+y2r = \sqrt{x^2 + y^2} from the origin. Then sin⁡θ=yr\sin\theta = \frac{y}{r}, cos⁡θ=xr\cos\theta = \frac{x}{r} and tan⁡θ=yx\tan\theta = \frac{y}{x}.

The CAST rule. Since rr is always positive, the signs of xx and yy decide the signs of the ratios. Label the quadrants starting in quadrant 4 and going counterclockwise: C, A, S, T. In quadrant 1 All three ratios are positive, in quadrant 2 only Sine, in quadrant 3 only Tangent, and in quadrant 4 only Cosine.

The reference angle is the acute angle between the terminal arm and the xx-axis. In quadrant 2 it is 180∘−θ180^\circ - \theta, in quadrant 3 it is θ−180∘\theta - 180^\circ, and in quadrant 4 it is 360∘−θ360^\circ - \theta. A ratio of θ\theta equals the same ratio of its reference angle, with the sign CAST gives.

Special triangles. The 45∘45^\circ-45∘45^\circ-90∘90^\circ triangle has sides 11, 11 and 2\sqrt{2}. The 30∘30^\circ-60∘60^\circ-90∘90^\circ triangle has sides 11, 3\sqrt{3} and 22, with the side 11 across from 30∘30^\circ. Read the ratios off them: sin⁡30∘=12\sin 30^\circ = \frac{1}{2}, cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2}, tan⁡45∘=1\tan 45^\circ = 1 and tan⁡60∘=3\tan 60^\circ = \sqrt{3}.

Solving for the angle. For sin⁡θ=−12\sin\theta = -\frac{1}{2} with 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ, find the reference angle from the positive value, 30∘30^\circ. Then place it in every quadrant where sine is negative, quadrants 3 and 4: θ=180∘+30∘=210∘\theta = 180^\circ + 30^\circ = 210^\circ and θ=360∘−30∘=330∘\theta = 360^\circ - 30^\circ = 330^\circ.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    Find the exact value of cos⁡150∘\cos 150^\circ.
    Show solution
    1. 150∘150^\circ is in quadrant 2, where cosine is negative.
    2. The reference angle is 180∘−150∘=30∘180^\circ - 150^\circ = 30^\circ, and cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2}.

    Answer: cos⁡150∘=−32\cos 150^\circ = -\frac{\sqrt{3}}{2}

  2. Easy

    The point P(−3,4)P(-3, 4) is on the terminal arm of θ\theta. Find sin⁡θ\sin\theta, cos⁡θ\cos\theta and tan⁡θ\tan\theta.
    Show solution
    1. r=(−3)2+42=25=5r = \sqrt{(-3)^2 + 4^2} = \sqrt{25} = 5.
    2. sin⁡θ=45\sin\theta = \frac{4}{5}, cos⁡θ=−35\cos\theta = \frac{-3}{5} and tan⁡θ=4−3\tan\theta = \frac{4}{-3}.
    3. Check: PP is in quadrant 2, where only sine is positive.

    Answer: sin⁡θ=45\sin\theta = \frac{4}{5}, cos⁡θ=−35\cos\theta = -\frac{3}{5}, tan⁡θ=−43\tan\theta = -\frac{4}{3}

  3. Medium

    Find the exact value of sin⁡240∘+tan⁡315∘\sin 240^\circ + \tan 315^\circ.
    Show solution
    1. 240∘240^\circ is in quadrant 3, with reference angle 60∘60^\circ, and sine is negative there: sin⁡240∘=−32\sin 240^\circ = -\frac{\sqrt{3}}{2}.
    2. 315∘315^\circ is in quadrant 4, with reference angle 45∘45^\circ, and tangent is negative there: tan⁡315∘=−1\tan 315^\circ = -1.
    3. Add: −32−1=−3+22-\frac{\sqrt{3}}{2} - 1 = -\frac{\sqrt{3} + 2}{2}.

    Answer: −3+22-\frac{\sqrt{3} + 2}{2}

  4. Medium

    Solve cos⁡θ=−22\cos\theta = -\frac{\sqrt{2}}{2} for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.
    Show solution
    1. From the positive value, cos⁡45∘=22\cos 45^\circ = \frac{\sqrt{2}}{2}, so the reference angle is 45∘45^\circ.
    2. Cosine is negative in quadrants 2 and 3.
    3. Quadrant 2: 180∘−45∘=135∘180^\circ - 45^\circ = 135^\circ. Quadrant 3: 180∘+45∘=225∘180^\circ + 45^\circ = 225^\circ.

    Answer: θ=135∘\theta = 135^\circ or θ=225∘\theta = 225^\circ

  5. Test level

    tan⁡θ=−512\tan\theta = -\frac{5}{12} and cos⁡θ>0\cos\theta > 0. Find the exact values of sin⁡θ\sin\theta and cos⁡θ\cos\theta, and θ\theta to the nearest degree for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.
    Show solution
    1. Tangent is negative in quadrants 2 and 4, and cosine is positive in quadrants 1 and 4, so θ\theta is in quadrant 4.
    2. In quadrant 4, x>0x > 0 and y<0y < 0: take x=12x = 12 and y=−5y = -5, so r=122+(−5)2=13r = \sqrt{12^2 + (-5)^2} = 13.
    3. sin⁡θ=−513\sin\theta = -\frac{5}{13} and cos⁡θ=1213\cos\theta = \frac{12}{13}.
    4. The reference angle is tan⁡−1(512)≈22.6∘\tan^{-1}\left(\frac{5}{12}\right) \approx 22.6^\circ, so θ≈360∘−22.6∘=337.4∘\theta \approx 360^\circ - 22.6^\circ = 337.4^\circ.

    Answer: sin⁡θ=−513\sin\theta = -\frac{5}{13}, cos⁡θ=1213\cos\theta = \frac{12}{13}, θ≈337∘\theta \approx 337^\circ

Learn It on Martian Lab

Martian Lab teaches this in the Trigonometry lesson of its MCR3U Trigonometric Ratios unit. Watch it, then take the Trigonometric Ratios practice tests to see if it stuck.

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Free Trigonometric Ratios practice test, with solutions

More Trigonometric Ratios topics: Sine Law and the Ambiguous Case · Cosine Law and 3D Problems

More MCR3U topics: Inverse Functions · Transformations of Functions · Domain, Range and Function Notation · Simplifying Rational Expressions · Operations with Rational Expressions · Max/Min and Vertex Form · Linear-Quadratic Systems · Quadratic Word Problems · Simplifying Radicals · Exponent Laws and Rational Exponents · Graphing Exponential Functions · Exponential Growth and Decay · Transformations of Sine and Cosine · Sinusoidal Word Problems · Arithmetic Sequences and Series · Geometric Sequences and Series · Pascal's Triangle and the Binomial Theorem · Simple and Compound Interest · Annuities: Future and Present Value

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