Trigonometric functions of negative angles

Trigonometric functions of negative angles

Let P(x, y) be a point on the unit circle with $\angle AOP = \theta$. The angle $-\theta$ (the same rotation but in the opposite direction) has its terminal ray meeting the unit circle at the mirror image of P across the x-axis, i.e. at $Q(x,-y)$. By definition, $\sin\theta = y$ and $\sin(-\theta) = -y$, so

and since $\cos\theta = x$ and $\cos(-\theta) = x$ (the x-coordinate is unchanged by reflection across the x-axis),

All the other relations follow by dividing: $\tan(-\theta) = \dfrac{\sin(-\theta)}{\cos(-\theta)} = \dfrac{-\sin\theta}{\cos\theta} = -\tan\theta$, and likewise $\cot(-\theta) = -\cot\theta$, $\sec(-\theta) = \dfrac{1}{\cos(-\theta)} = \sec\theta$, $\text{cosec}(-\theta) = \dfrac{1}{\sin(-\theta)} = -\text{cosec}\,\theta$.

6) Angle of measure −60° (−π/3). The terminal ray of $-60°$ meets the unit circle at $P(x,y)$ in the fourth quadrant. Dropping PQ perpendicular to the x-axis gives a $30°$–$60°$–$90°$ triangle with $OQ=\tfrac12$, $PQ=\tfrac{\sqrt3}{2}$. Since P is in quadrant IV, $x=\tfrac12$, $y=-\tfrac{\sqrt3}{2}$. Hence:

Note: $-60°$ and $300°$ are coterminal ($300° = -60°+360°$), so their trigonometric values necessarily agree — a useful cross-check.

Solved Example 3 — all trigonometric functions of the angle to (−5, 12)

Step 1. The terminal arm passes through $P(-5,12)$, so $x=-5, y=12$.

Step 2. $r = OP = \sqrt{(-5)^2+12^2} = \sqrt{25+144} = \sqrt{169} = 13$.

Step 3. Read off all six functions directly:

Solved Example 4 — secθ = −3, π < θ < 3π/2, find the rest

Step 1. $\sec\theta=-3 \implies \cos\theta = -\tfrac13$.

Step 2. Using $1+\tan^2\theta=\sec^2\theta$: $\tan^2\theta = 9-1=8$.

Step 3. Since $\pi<\theta<3\pi/2$ (third quadrant, where tanθ is positive), $\tan\theta=2\sqrt2$, hence $\cot\theta=\dfrac{1}{2\sqrt2}$.

Step 4. $\sin\theta=\tan\theta\cos\theta = 2\sqrt2\times\left(-\tfrac13\right) = -\dfrac{2\sqrt2}{3}$, so $\text{cosec}\,\theta = -\dfrac{3}{2\sqrt2}$.

Solved Example 5 — secx = 13/5, x in the fourth quadrant, find the rest

Step 1. $\sec x = \tfrac{13}{5} \implies \cos x = \tfrac5{13}$.

Step 2. $\tan^2x = \sec^2x-1 = \tfrac{169}{25}-1=\tfrac{144}{25}$.

Step 3. In the fourth quadrant tanx is negative, so $\tan x = -\tfrac{12}5$, hence $\cot x=-\tfrac{5}{12}$.

Step 4. $\sin x = \tan x\cos x = -\tfrac{12}5\times\tfrac5{13} = -\tfrac{12}{13}$, so $\text{cosec}\,x = -\tfrac{13}{12}$.

Solved Example 6 — tanA = 4/3, evaluate (2sinA − 3cosA)/(2sinA + 3cosA)

Step 1. Divide numerator and denominator by $\cos A$: $\dfrac{2\sin A - 3\cos A}{2\sin A+3\cos A} = \dfrac{2\tan A - 3}{2\tan A+3}$.

Step 2. Substitute $\tan A = 4/3$: $\dfrac{2(4/3)-3}{2(4/3)+3} = \dfrac{8/3-3}{8/3+3} = \dfrac{-1/3}{17/3} = -\dfrac1{17}$.

Solved Example 7 — secθ = √2, 3π/2 < θ < 2π, evaluate a combined expression

Step 1. $\sec\theta=\sqrt2 \implies \cos\theta = \tfrac1{\sqrt2}$; $\tan^2\theta=\sec^2\theta-1=1$.

Step 2. In the fourth quadrant ($3\pi/2<\theta<2\pi$) tanθ is negative, so $\tan\theta=-1$, hence $\cot\theta=-1$. …