Trigonometry
The three graphs Tap to enlarge Graphs of y = sin x, y = cos x and y = tan x from 0° to 360°, with the asymptotes of tan x at 90° and 270°
Close Graph Period Range Notes y = sin x y = \sin x y = sin x 360° or 2 π 2\pi 2 π − 1 ≤ y ≤ 1 -1 \leq y \leq 1 − 1 ≤ y ≤ 1 Passes through the origin y = cos x y = \cos x y = cos x 360° or 2 π 2\pi 2 π − 1 ≤ y ≤ 1 -1 \leq y \leq 1 − 1 ≤ y ≤ 1 Starts at y = 1 y = 1 y = 1 y = tan x y = \tan x y = tan x 180° or π \pi π All real values Asymptotes at 90°, 270° and so on
The transformations of Functions apply here:
y = a sin x y = a\sin x y = a sin x a a a y = sin b x y = \sin bx y = sin b x 360 ∘ b \dfrac{360^\circ}{b} b 36 0 ∘ b b b y = sin x + c y = \sin x + c y = sin x + c c c c y = sin ( x + d ) y = \sin(x + d) y = sin ( x + d ) d d d
cos x \cos x cos x − 1 -1 − 1 1 1 1 3 − 2 = 1 3 - 2 = 1 3 − 2 = 1 3 + 2 = 5 3 + 2 = 5 3 + 2 = 5 1 ≤ y ≤ 5 1 \leq y \leq 5 1 ≤ y ≤ 5
Exact values θ \theta θ 0° 30° 45° 60° 90° sin θ \sin\theta sin θ 0 1 2 \dfrac{1}{2} 2 1 2 2 \dfrac{\sqrt{2}}{2} 2 2 3 2 \dfrac{\sqrt{3}}{2} 2 3 1 cos θ \cos\theta cos θ 1 3 2 \dfrac{\sqrt{3}}{2} 2 3 2 2 \dfrac{\sqrt{2}}{2} 2 2 1 2 \dfrac{1}{2} 2 1 0 tan θ \tan\theta tan θ 0 3 3 \dfrac{\sqrt{3}}{3} 3 3 1 3 \sqrt{3} 3 undefined
Related angles have the same size of value with a sign that depends on the quadrant. This diagram shows which ratios are positive in each quadrant:
Tap to enlarge A CAST diagram: all ratios positive from 0° to 90°, sine from 90° to 180°, tangent from 180° to 270°, cosine from 270° to 360°
Close So sin 120 ∘ = sin 60 ∘ = 3 2 \sin 120^\circ = \sin 60^\circ = \dfrac{\sqrt{3}}{2} sin 12 0 ∘ = sin 6 0 ∘ = 2 3 cos 120 ∘ = − cos 60 ∘ = − 1 2 \cos 120^\circ = -\cos 60^\circ = -\dfrac{1}{2} cos 12 0 ∘ = − cos 6 0 ∘ = − 2 1
Inverse notation and principal values sin − 1 x \sin^{-1}x sin − 1 x cos − 1 x \cos^{-1}x cos − 1 x tan − 1 x \tan^{-1}x tan − 1 x principal value , which is what the calculator shows.
Function Principal values sin − 1 x \sin^{-1}x sin − 1 x − 90 ∘ ≤ θ ≤ 90 ∘ -90^\circ \leq \theta \leq 90^\circ − 9 0 ∘ ≤ θ ≤ 9 0 ∘ cos − 1 x \cos^{-1}x cos − 1 x 0 ∘ ≤ θ ≤ 180 ∘ 0^\circ \leq \theta \leq 180^\circ 0 ∘ ≤ θ ≤ 18 0 ∘ tan − 1 x \tan^{-1}x tan − 1 x − 90 ∘ < θ < 90 ∘ -90^\circ < \theta < 90^\circ − 9 0 ∘ < θ < 9 0 ∘
The two identities The second can be rearranged as sin 2 θ ≡ 1 − cos 2 θ \sin^2\theta \equiv 1 - \cos^2\theta sin 2 θ ≡ 1 − cos 2 θ cos 2 θ ≡ 1 − sin 2 θ \cos^2\theta \equiv 1 - \sin^2\theta cos 2 θ ≡ 1 − sin 2 θ
cos 2 θ = 1 − 9 25 = 16 25 \cos^2\theta = 1 - \dfrac{9}{25} = \dfrac{16}{25} cos 2 θ = 1 − 25 9 = 25 16 cos θ = − 4 5 \cos\theta = -\dfrac{4}{5} cos θ = − 5 4 tan θ = − 3 4 \tan\theta = -\dfrac{3}{4} tan θ = − 4 3
LHS = sin θ cos θ + cos θ sin θ = sin 2 θ + cos 2 θ sin θ cos θ = 1 sin θ cos θ = RHS \text{LHS} = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} = \text{RHS} LHS = c o s θ s i n θ + s i n θ c o s θ = s i n θ c o s θ s i n 2 θ + c o s 2 θ = s i n θ c o s θ 1 = RHS
Solving trigonometric equations There are usually several solutions in an interval. Find the principal value, then use the symmetry of the graph:
Equation Second solution from α \alpha α Then sin x = k \sin x = k sin x = k 180 ∘ − α 180^\circ - \alpha 18 0 ∘ − α Add or subtract 360° cos x = k \cos x = k cos x = k 360 ∘ − α 360^\circ - \alpha 36 0 ∘ − α Add or subtract 360° tan x = k \tan x = k tan x = k 180 ∘ + α 180^\circ + \alpha 18 0 ∘ + α Add or subtract 180°
x = 30 ∘ x = 30^\circ x = 3 0 ∘ x = 180 ∘ − 30 ∘ = 150 ∘ x = 180^\circ - 30^\circ = 150^\circ x = 18 0 ∘ − 3 0 ∘ = 15 0 ∘
When the angle is changed Change the interval first, solve for the whole bracket, and only then find x x x
The interval becomes 0 ∘ ≤ 2 x ≤ 360 ∘ 0^\circ \leq 2x \leq 360^\circ 0 ∘ ≤ 2 x ≤ 36 0 ∘ 2 x = 53.13 ∘ 2x = 53.13^\circ 2 x = 53.1 3 ∘ 2 x = 306.87 ∘ 2x = 306.87^\circ 2 x = 306.8 7 ∘ x = 26.6 ∘ x = 26.6^\circ x = 26. 6 ∘ x = 153.4 ∘ x = 153.4^\circ x = 153. 4 ∘
The interval becomes − 30 ∘ ≤ x − 30 ∘ ≤ 330 ∘ -30^\circ \leq x - 30^\circ \leq 330^\circ − 3 0 ∘ ≤ x − 3 0 ∘ ≤ 33 0 ∘ x − 30 ∘ = 45 ∘ x - 30^\circ = 45^\circ x − 3 0 ∘ = 4 5 ∘ 225 ∘ 225^\circ 22 5 ∘ x = 75 ∘ x = 75^\circ x = 7 5 ∘ x = 255 ∘ x = 255^\circ x = 25 5 ∘
Using the identities in equations Use tan θ = sin θ cos θ \tan\theta = \dfrac{\sin\theta}{\cos\theta} tan θ = cos θ sin θ
Divide by cos x \cos x cos x tan x = 4 3 \tan x = \dfrac{4}{3} tan x = 3 4 x = 53.1 ∘ x = 53.1^\circ x = 53. 1 ∘ x = 233.1 ∘ x = 233.1^\circ x = 233. 1 ∘
Use sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1
2 ( 1 − cos 2 x ) + 3 cos x = 0 ⇒ 2 cos 2 x − 3 cos x − 2 = 0 ⇒ ( 2 cos x + 1 ) ( cos x − 2 ) = 0 2(1 - \cos^2 x) + 3\cos x = 0 \quad\Rightarrow\quad 2\cos^2 x - 3\cos x - 2 = 0 \quad\Rightarrow\quad (2\cos x + 1)(\cos x - 2) = 0 2 ( 1 − cos 2 x ) + 3 cos x = 0 ⇒ 2 cos 2 x − 3 cos x − 2 = 0 ⇒ ( 2 cos x + 1 ) ( cos x − 2 ) = 0
cos x = 2 \cos x = 2 cos x = 2 cos x = − 1 2 \cos x = -\dfrac{1}{2} cos x = − 2 1 x = 120 ∘ x = 120^\circ x = 12 0 ∘ x = 240 ∘ x = 240^\circ x = 24 0 ∘
