Trigonometric functions of specific angles

Trigonometric functions of specific angles

The unit-circle definition lets us find exact trigonometric values at standard angles by locating the coordinates of the corresponding point P geometrically, rather than approximating.

1) Angle of measure 0°. The terminal ray of 0° is the initial ray itself, meeting the unit circle at $P(1,0)$, so $x=1,\ y=0$. Then $\sin 0° = y = 0$, $\cos 0° = x = 1$, $\tan 0° = y/x = 0/1 = 0$. Since $y=0$, both $\text{cosec}\,0°$ (=1/y) and $\cot 0°$ (=x/y) are undefined; $\sec 0° = r/x = 1$.

2) Angle of measure 90° (π/2). The terminal ray meets the unit circle at $P(0,1)$, so $x=0, y=1$. Then $\sin 90° = 1$, $\cos 90° = 0$, and $\tan 90°$ is undefined (division by $\cos 90°=0$). $\text{cosec}\,90° = 1/1 = 1$; $\sec 90°$ is undefined; $\cot 90° = 0/1 = 0$.

(Activity: using the same idea, at 180° the point is $P(-1,0)$ giving $\sin180°=0,\cos180°=-1$, and at 270° the point is $P(0,-1)$ giving $\sin270°=-1,\cos270°=0$.)

3) Angle of measure 360° (2π). Since 360° and 0° are coterminal, all trigonometric functions of 360° equal those of 0°.

4) Angle of measure 120° (2π/3). The terminal ray meets the unit circle at $P(x,y)$ in the second quadrant. Dropping a perpendicular PQ to the x-axis creates a $30°$–$60°$–$90°$ triangle OPQ with hypotenuse $OP=1$, so $OQ = \tfrac12$ and $PQ=\tfrac{\sqrt3}{2}$. As P is in quadrant II, $x=-\tfrac12$ and $y=\tfrac{\sqrt3}{2}$. Hence:

5) Angle of measure 225° (5π/4). The terminal ray meets the unit circle at $P(x,y)$ in the third quadrant. Dropping PQ perpendicular to the x-axis gives a $45°$–$45°$–$90°$ triangle with $OQ = PQ = \tfrac{1}{\sqrt2}$. As P is in quadrant III, $x=-\tfrac1{\sqrt2}$ and $y=-\tfrac1{\sqrt2}$. Hence:

Reference table — 0°, 30°, 45°, 60°, 90°

The remaining four functions at each angle follow immediately from $\tan\theta=\sin\theta/\cos\theta$ and the reciprocal relations.

(Activity: the same 30-60-90 / 45-45-90 triangle method extends this table to 150°, 210°, 330°, −45°, −120°, −3π/4, each identified by its quadrant and reference angle.)

Solved Example 1 — verify sin2θ = 2sinθcosθ at θ = 30°

Step 1. With $\theta = 30°$, $2\theta = 60°$.

Step 2. From the table, $\sin\theta = \sin30° = \tfrac12$, $\cos\theta=\cos30°=\tfrac{\sqrt3}{2}$, and $\sin2\theta = \sin60° = \tfrac{\sqrt3}{2}$.

Step 3. Compute the right side: $2\sin\theta\cos\theta = 2\times\tfrac12\times\tfrac{\sqrt3}{2} = \tfrac{\sqrt3}{2}$.

Step 4. Since $\tfrac{\sqrt3}{2} = \tfrac{\sqrt3}{2}$, LHS = RHS — the relation checks out at θ = 30°.

Solved Example 2 — evaluate four numerical expressions

i) $\cos30°\cos60°+\sin30°\sin60°$. $= \tfrac{\sqrt3}{2}\cdot\tfrac12 + \tfrac12\cdot\tfrac{\sqrt3}{2} = \tfrac{\sqrt3}{4}+\tfrac{\sqrt3}{4} = \dfrac{\sqrt3}{2}$. …