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Math II

Trigonometric functions: practice problems

In Math II angles are measured in radians and extended to general angles. The unit circle is the main tool for values, equations and inequalities.

Basic, Standard: Grade 11 · Term 2 / Advanced: Grade 12 · Exam prep

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Math problem generator

Level

Radians and general angles

180∘=π180^\circ = \pi radians, so 30∘=π630^\circ = \frac{\pi}{6}, 45∘=π445^\circ = \frac{\pi}{4}, 60∘=π360^\circ = \frac{\pi}{3} and 90∘=π290^\circ = \frac{\pi}{2}.

Sectors

A sector of radius rr and angle θ\theta (radians) has arc length l=rθl = r\theta and area S=12r2θ=12lrS = \frac{1}{2}r^2\theta = \frac{1}{2}lr.

If the terminal side of θ\theta meets the unit circle at (x, y)(x,\ y), then cos⁡θ=x\cos\theta = x, sin⁡θ=y\sin\theta = y and tan⁡θ=yx\tan\theta = \frac{y}{x}. Angles differing by 2nπ2n\pi have the same values.

Identities
sin⁡2θ+cos⁡2θ=1,tan⁡θ=sin⁡θcos⁡θ,1+tan⁡2θ=1cos⁡2θ\sin^2\theta + \cos^2\theta = 1,\qquad \tan\theta = \frac{\sin\theta}{\cos\theta},\qquad 1 + \tan^2\theta = \frac{1}{\cos^2\theta}

Periods and translations of graphs

The graph of y=asin⁡(bθ−c)+dy = a\sin(b\theta - c) + d (a>0a > 0, b>0b > 0) is the graph of y=asin⁡bθy = a\sin b\theta shifted by cb\frac{c}{b} in the θ\theta-direction and by dd vertically.

  • Period: 2πb\frac{2\pi}{b} (πb\frac{\pi}{b} for tan⁡\tan)
  • Maximum a+da + d, minimum −a+d-a + d
Watch out

sin⁡(2θ−π3)=sin⁡2(θ−π6)\sin\left(2\theta - \frac{\pi}{3}\right) = \sin 2\left(\theta - \frac{\pi}{6}\right), so the shift is π6\frac{\pi}{6}, not π3\frac{\pi}{3}. Factor out the coefficient of θ\theta first.

Trigonometric equations and inequalities

To solve sin⁡θ=12\sin\theta = \frac{1}{2} on 0≤θ<2π0 \leq \theta < 2\pi, find the points of the unit circle with yy-coordinate 12\frac{1}{2}: θ=π6, 5π6\theta = \frac{\pi}{6},\ \frac{5\pi}{6}. For sin⁡θ>12\sin\theta > \frac{1}{2}, take the arc between them: π6<θ<5π6\frac{\pi}{6} < \theta < \frac{5\pi}{6}.

For sin⁡(θ+π3)=12\sin\left(\theta + \frac{\pi}{3}\right) = \frac{1}{2} or cos⁡2θ=12\cos 2\theta = \frac{1}{2}, substitute t=θ+π3t = \theta + \frac{\pi}{3} or t=2θt = 2\theta and find the range of tt first.

For quadratics such as 2cos⁡2θ+3sin⁡θ−3=02\cos^2\theta + 3\sin\theta - 3 = 0, use cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta to get an equation in sin⁡θ\sin\theta, remembering −1≤sin⁡θ≤1-1 \leq \sin\theta \leq 1.