Honors Algebra II Unit 11 Review: Trigonometric Functions and Identities

An angle in standard position has its vertex at the origin and its initial side on the positive x-axis. Counterclockwise rotation is positive and clockwise is negative.

A radian is the central angle whose arc length equals the radius. A full circle is $360°$, or $2\pi$ radians, so $180° = \pi$. To convert degrees to radians, multiply by $\frac{\pi}{180}$. To go the other way, multiply by $\frac{180}{\pi}$. One radian is about $57.3°$.

Coterminal angles share a terminal side and differ by multiples of $2\pi$ (or $360°$). For an angle whose terminal side is not on an axis, the reference angle is the acute angle between that side and the x-axis. Some conventions assign reference angles to axis angles too.

For an angle $\theta$ from $0$ to $2\pi$ radians, arc length is $s = r\theta$ and sector area is $A = \frac{1}{2}r^2\theta$. Use the magnitude of a negative angle when finding an ordinary length or area.

Worked example. Find the arc length cut off by a $150°$ central angle in a circle of radius 8. First convert, $150 \cdot \frac{\pi}{180} = \frac{5\pi}{6}$. Then $s = 8 \cdot \frac{5\pi}{6} = \frac{20\pi}{3} \approx 20.94$ units. The sector area is $\frac{1}{2}(64)\cdot\frac{5\pi}{6} = \frac{80\pi}{3} \approx 83.78$ square units.

The unit circle has radius 1 and center $(0, 0)$. Rotate by $\theta$ from $(1, 0)$, and the point where you land is $(\cos\theta, \sin\theta)$. For an acute angle, this matches right-triangle trig: the hypotenuse is 1, so $\cos\theta = \frac{x}{1}$ and $\sin\theta = \frac{y}{1}$. In other quadrants, the coordinates can be negative.

The other four functions come from these two. $\tan\theta = \frac{\sin\theta}{\cos\theta}$, $\sec\theta = \frac{1}{\cos\theta}$, $\csc\theta = \frac{1}{\sin\theta}$ and $\cot\theta = \frac{\cos\theta}{\sin\theta}$. Any of them is undefined when its denominator is 0, so $\tan\frac{\pi}{2}$ does not exist.

Signs depend on the quadrant. All six are positive in Quadrant I, only sine and cosecant in II, only tangent and cotangent in III, and only cosine and secant in IV (the ASTC pattern).

To find an exact value, take the reference angle's value from the table and attach the sign for the quadrant.

Worked example. Find $\cos\frac{4\pi}{3}$, $\sin\frac{4\pi}{3}$ and $\tan\frac{4\pi}{3}$. The angle is $\pi + \frac{\pi}{3}$, so it sits in Quadrant III with reference angle $\frac{\pi}{3}$. Both coordinates are negative, which gives $\cos\frac{4\pi}{3} = -\frac{1}{2}$ and $\sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$. Then $\tan\frac{4\pi}{3} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$, which is positive, as ASTC predicts.

Unwrap the unit circle onto an x-axis and you get the graphs. $y = \sin x$ starts at $(0, 0)$, peaks at $\left(\frac{\pi}{2}, 1\right)$ and repeats every $2\pi$. $y = \cos x$ has the same shape but starts at its maximum, $(0, 1)$. $y = \tan x$ repeats every $\pi$, crosses zero at multiples of $\pi$, and has vertical asymptotes wherever $\cos x = 0$, at odd multiples of $\frac{\pi}{2}$.

The transformed form is $y = A\sin(B(x - C)) + D$ (the same form works for cosine). When $A$ and $B$ are nonzero, the amplitude is $|A|$, the period is $\frac{2\pi}{|B|}$, the phase shift is $C$, and the midline is $y = D$. For tangent, the period is $\frac{\pi}{|B|}$, but there is no amplitude. A negative $A$ reflects the graph over its midline. If $A$ or $B$ is zero, these functions are constant and have no least positive period.

To graph a transformed sine curve with nonzero $A$ and $B$, split one period into four equal steps starting at $x = C$. If $A$ and $B$ have the same sign, the five points follow midline, maximum, midline, minimum, midline. If their signs differ, the minimum comes before the maximum.

Worked example. Graph $y = 3\sin\left(2\left(x - \frac{\pi}{4}\right)\right) + 1$. The amplitude is 3 and the midline is $y = 1$, so the maximum is 4 and the minimum is $-2$. The period is $\frac{2\pi}{2} = \pi$, so each step is $\frac{\pi}{4}$. Starting at the shift $x = \frac{\pi}{4}$, plot $\left(\frac{\pi}{4}, 1\right)$, $\left(\frac{\pi}{2}, 4\right)$, $\left(\frac{3\pi}{4}, 1\right)$, $(\pi, -2)$ and $\left(\frac{5\pi}{4}, 1\right)$.

An identity is an equation that is true for every value where both sides are defined. That is different from a trig equation like $\sin\theta = \frac{1}{2}$, which is only true for certain angles.

The main Pythagorean identity, $\sin^2\theta + \cos^2\theta = 1$, is the unit circle equation $x^2 + y^2 = 1$. Dividing it by $\cos^2\theta$ gives $\tan^2\theta + 1 = \sec^2\theta$, and dividing by $\sin^2\theta$ gives $1 + \cot^2\theta = \csc^2\theta$. You also use the reciprocal and quotient identities, the even-odd identities $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$, and the sum formulas such as $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$.

To prove an identity, start with the messier side and rewrite it until it matches the other side. Converting to sine and cosine, combining fractions and factoring usually get you there.

Worked example. Prove $\sec\theta - \cos\theta = \sin\theta\tan\theta$. Rewrite the left side as $\frac{1}{\cos\theta} - \cos\theta$. Use a common denominator to get $\frac{1 - \cos^2\theta}{\cos\theta}$. By the Pythagorean identity, that is $\frac{\sin^2\theta}{\cos\theta} = \sin\theta \cdot \frac{\sin\theta}{\cos\theta} = \sin\theta\tan\theta$, which matches the right side.