Secondary school › Trigonometry
Trigonometric functions
Values, domain, range, signs and the order of trigonometric values. 10 problems with step-by-step solutions.
Solve each problem in your notebook first, then open the solution. In all answers \(k, m \in \mathbb{Z}\).
Problem 1 Values at quadrantal angles
Complete the table.
| \(x\) | \(0^\circ\) | \(90^\circ\) | \(180^\circ\) | \(270^\circ\) | \(360^\circ\) |
|---|---|---|---|---|---|
| \(\sin x\) | |||||
| \(\cos x\) | |||||
| \(\tan x\) | |||||
| \(\cot x\) | |||||
| \(\sec x\) | |||||
| \(\csc x\) |
Show solution
On the unit circle the point for \(0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ\) is \((1,0), (0,1), (-1,0), (0,-1), (1,0)\). The cosine is the \(x\)-coordinate and the sine is the \(y\)-coordinate. “–” means undefined.
| \(x\) | \(0^\circ\) | \(90^\circ\) | \(180^\circ\) | \(270^\circ\) | \(360^\circ\) |
|---|---|---|---|---|---|
| \(\sin x\) | 0 | 1 | 0 | −1 | 0 |
| \(\cos x\) | 1 | 0 | −1 | 0 | 1 |
| \(\tan x\) | 0 | – | 0 | – | 0 |
| \(\cot x\) | – | 0 | – | 0 | – |
| \(\sec x\) | 1 | – | −1 | – | 1 |
| \(\csc x\) | – | 1 | – | −1 | – |
Problem 2 Values at quadrantal angles
Calculate.
a) \(\sin 0^\circ + 2\sin 90^\circ - 3\sin 180^\circ + \sin 270^\circ\)
b) \(\cos 0^\circ - 3\cos 90^\circ + 2\cos 180^\circ - \cos 270^\circ + \cos 360^\circ\)
c) \(\sin 90^\circ - \cos 180^\circ + 3\tan 180^\circ - \cot 90^\circ\)
d) \(2\cos 0^\circ + 5\tan 0^\circ - \cot 270^\circ + 3\sin 270^\circ\)
Show solution
- a) \(0 + 2 - 0 - 1 = 1\)
- b) \(1 - 0 - 2 - 0 + 1 = 0\)
- c) \(1 + 1 + 0 - 0 = 2\)
- d) \(2 + 0 - 0 - 3 = -1\)
Problem 3 Values at multiples of π/2
Calculate.
a) \(\sin\pi + \cos\frac{11\pi}{2} + \sin 15\pi - 3\cos 9\pi + \sin\frac{13\pi}{2}\)
b) \(\sin\frac{3\pi}{2}\cdot\cos 2\pi - \tan 2\pi\cdot\cot\frac{\pi}{2} + \sec\pi - \csc\frac{\pi}{2}\)
c) \(\sec 5\pi + 2\csc\frac{3\pi}{2} - \cot\frac{15\pi}{2}\)
d) \(\sec 12\pi + \csc\frac{7\pi}{2} - \tan 3\pi + \cot\frac{3\pi}{2}\)
Show solution
Remove full turns (\(2\pi\)) first: \(\frac{11\pi}{2} \to \frac{3\pi}{2}\), \(15\pi \to \pi\), \(9\pi \to \pi\), \(\frac{13\pi}{2} \to \frac{\pi}{2}\), \(\frac{15\pi}{2} \to \frac{3\pi}{2}\), \(\frac{7\pi}{2} \to \frac{3\pi}{2}\).
- a) \(0 + 0 + 0 - 3\cdot(-1) + 1 = 4\)
- b) \((-1)\cdot 1 - 0\cdot 0 + (-1) - 1 = -3\)
- c) \(-1 + 2\cdot(-1) - 0 = -3\)
- d) \(1 + (-1) - 0 + 0 = 0\)
Problem 4 Domain and range
Find the domain and the range.
a) \(f(x) = 2\sin x + 5\) b) \(f(x) = 3 - 4\cos x\) c) \(f(x) = \tan 2x - 1\)
d) \(y = \sin 2x\cos 3x + \cos 2x\sin 3x + 2\) e) \(y = 3\cos x + 2\tan 3x\) f) \(y = \sin^2 x - \sin x\)
Show solution
a) Domain \(\mathbb{R}\). From \(-1 \le \sin x \le 1\): \(3 \le 2\sin x + 5 \le 7\). Range \([3, 7]\).
b) Domain \(\mathbb{R}\). \(-4 \le -4\cos x \le 4\), so the range is \([-1, 7]\).
c) \(2x \ne \frac{\pi}{2} + k\pi\), so the domain is \(\mathbb{R}\setminus\left\{\frac{\pi}{4} + \frac{k\pi}{2}\right\}\). Range \(\mathbb{R}\).
d) \(y = \sin 5x + 2\). Domain \(\mathbb{R}\), range \([1, 3]\).
e) The domain is the domain of \(\tan 3x\): \(\mathbb{R}\setminus\left\{\frac{\pi}{6} + \frac{k\pi}{3}\right\}\). The tangent takes all real values, so the range is \(\mathbb{R}\).
f) Domain \(\mathbb{R}\). Put \(t = \sin x \in [-1, 1]\): \(g(t) = t^2 - t\), a parabola with vertex at \(t = \frac12\), \(g\left(\frac12\right) = -\frac14\). Also \(g(-1) = 2\), \(g(1) = 0\). Range \(\left[-\frac14, 2\right]\).
Problem 5 Domain
Find the domain of each function.
a) \(y = 2\sin(3x - 1)\) b) \(y = 5\cos 4x\) c) \(y = 3\tan 2x + 1\) d) \(y = \cot(2x + 1) - 3\)
e) \(y = 2\csc 3x + 1\) f) \(y = \sec\left(3x - \frac{\pi}{4}\right)\) g) \(y = \tan 2x - \cot 3x\)
Show solution
- a), b) Sine and cosine are defined everywhere: \(\mathbb{R}\).
- c) \(2x \ne \frac{\pi}{2} + k\pi \Rightarrow x \ne \frac{\pi}{4} + \frac{k\pi}{2}\).
- d) \(2x + 1 \ne k\pi \Rightarrow x \ne \frac{k\pi - 1}{2}\).
- e) \(\sin 3x \ne 0 \Rightarrow x \ne \frac{k\pi}{3}\).
- f) \(\cos\left(3x - \frac{\pi}{4}\right) \ne 0 \Rightarrow 3x - \frac{\pi}{4} \ne \frac{\pi}{2} + k\pi \Rightarrow x \ne \frac{\pi}{4} + \frac{k\pi}{3}\).
- g) Both must be defined: \(x \ne \frac{\pi}{4} + \frac{k\pi}{2}\) and \(x \ne \frac{k\pi}{3}\), \(k \in \mathbb{Z}\).
Problem 6 Range
Find the range of each function.
a) \(y = 4\sin 3x - 1\) b) \(y = 2\sin(5x + 2) + 3\) c) \(y = 7\cos(x - 1)\) d) \(y = 3\cos 2x + 5\)
e) \(y = \tan 5x + 2\) f) \(y = 3\sec 2x\) g) \(y = 2\csc 5x + 1\) h) \(y = \tan x + \cot 2x\)
Show solution
- a) \([-5, 3]\) b) \([1, 5]\) c) \([-7, 7]\) d) \([2, 8]\) e) \(\mathbb{R}\)
- f) \(|\sec 2x| \ge 1\), so \(|3\sec 2x| \ge 3\): \((-\infty, -3] \cup [3, \infty)\).
- g) \(2\csc 5x \in (-\infty, -2] \cup [2, \infty)\), so \(y \in (-\infty, -1] \cup [3, \infty)\).
- h) With \(t = \tan x\): \(\cot 2x = \frac{1 - t^2}{2t}\), so \(y = t + \frac{1 - t^2}{2t} = \frac{t^2 + 1}{2t}\). Since \(\sin 2x = \frac{2t}{1 + t^2}\), we get \(y = \frac{1}{\sin 2x}\). The range is \((-\infty, -1] \cup [1, \infty)\).
Problem 7 Signs
Find the sign of each ratio.
a) \(\sin 200^\circ\) b) \(\cos 100^\circ\) c) \(\tan 300^\circ\) d) \(\cot 250^\circ\) e) \(\sec 130^\circ\) f) \(\csc 300^\circ\)
g) \(\tan 500^\circ\) h) \(\sin\frac{8\pi}{5}\) i) \(\cos\frac{13\pi}{7}\) j) \(\tan\frac{7\pi}{5}\) k) \(\cot\frac{11\pi}{12}\) l) \(\sec\frac{5\pi}{4}\)
Show solution
Signs by quadrant: I all positive; II only \(\sin\), \(\csc\) positive; III only \(\tan\), \(\cot\) positive; IV only \(\cos\), \(\sec\) positive.
- a) III: − b) II: − c) IV: − d) III: + e) II: − f) IV: −
- g) \(500^\circ \to 140^\circ\), II: − h) \(\frac{8\pi}{5}\) is in IV: − i) \(\frac{13\pi}{7} = 2\pi - \frac{\pi}{7}\), IV: +
- j) \(\frac{7\pi}{5}\) is in III: + k) \(\frac{11\pi}{12}\) is in II: − l) \(\frac{5\pi}{4}\) is in III: −
Problem 8 Ordering values
Write in ascending order.
a) \(a = \sin 70^\circ\), \(b = \cos 140^\circ\), \(c = \tan 50^\circ\)
b) \(a = \sin 40^\circ\), \(b = \cos 40^\circ\), \(c = \tan 40^\circ\), \(d = \sec 40^\circ\)
c) \(x = \sin 50^\circ\), \(y = \cos 80^\circ\), \(z = \tan 160^\circ\)
d) \(a = \sin 140^\circ\), \(b = \cos 140^\circ\), \(c = \tan 140^\circ\), \(d = \sec 140^\circ\)
e) \(m = \sin 100^\circ\), \(n = \tan 220^\circ\), \(r = \cot(-20^\circ)\)
Show solution
a) \(b < 0\); \(0 < a < 1\); \(c = \tan 50^\circ > \tan 45^\circ = 1\). So \(b < a < c\).
b) For \(x \in (0^\circ, 45^\circ)\): \(\sin x < \cos x\). Also \(\tan x = \frac{\sin x}{\cos x} > \sin x\), and \(\tan 40^\circ < 1 < \sec 40^\circ\). Compare \(b\) and \(c\): \(\cos 40^\circ \approx 0.77\), \(\tan 40^\circ \approx 0.84\). So \(a < b < c < d\).
c) \(z < 0\) (II quadrant). \(y = \cos 80^\circ = \sin 10^\circ < \sin 50^\circ = x\). So \(z < y < x\).
d) \(a = \sin 40^\circ > 0\), \(b = -\cos 40^\circ \approx -0.77\), \(c = -\tan 40^\circ \approx -0.84\), \(d = -\sec 40^\circ \approx -1.31\). So \(d < c < b < a\).
e) \(m = \sin 80^\circ \approx 0.98\), \(n = \tan 40^\circ \approx 0.84\), \(r = -\cot 20^\circ < 0\). So \(r < n < m\).
Problem 9 Increasing and decreasing
Write in ascending order.
a) \(\cos 20^\circ\), \(\cos 50^\circ\), \(\cos 80^\circ\) b) \(\tan 20^\circ\), \(\tan 40^\circ\), \(\tan 70^\circ\) c) \(\cot 30^\circ\), \(\cot 50^\circ\), \(\cot 70^\circ\) d) \(\sin 15^\circ\), \(\sin 50^\circ\), \(\sin 80^\circ\)
Show solution
On \((0^\circ, 90^\circ)\) sine and tangent are increasing; cosine and cotangent are decreasing.
- a) \(\cos 80^\circ < \cos 50^\circ < \cos 20^\circ\)
- b) \(\tan 20^\circ < \tan 40^\circ < \tan 70^\circ\)
- c) \(\cot 70^\circ < \cot 50^\circ < \cot 30^\circ\)
- d) \(\sin 15^\circ < \sin 50^\circ < \sin 80^\circ\)
Problem 10 Minimum and maximum
Find the minimum and the maximum value.
a) \(A = 2 - 3\cos x\) b) \(A = -1 + 4\sin x\) c) \(A = 5\cos 4x + 3\)
d) For \(f(x) = 2\cos x - \sin 2x\), find \(f\left(\frac{\pi}{3}\right)\).
Show solution
a) \(\cos x = 1\) gives the minimum \(-1\); \(\cos x = -1\) gives the maximum \(5\).
b) Minimum \(-5\) (\(\sin x = -1\)), maximum \(3\) (\(\sin x = 1\)).
c) Minimum \(-2\), maximum \(8\).
d) \(f\left(\frac{\pi}{3}\right) = 2\cdot\frac12 - \sin\frac{2\pi}{3} = 1 - \frac{\sqrt3}{2}\).