Honors Pre-Calculus Unit 7 Review: Trig Identities and Equations

An identity is an equation that's true for every input where both sides are defined, like $\sin^2\theta + \cos^2\theta = 1$. You use identities in two different ways. When you verify an identity, you rewrite one side until it matches the other. When you solve an equation, you use identities to get down to one trig function and then find the angles that work.

The fundamental identities do most of the work. These include the reciprocal identities like $\sec\theta = \frac{1}{\cos\theta}$, the quotient identities like $\tan\theta = \frac{\sin\theta}{\cos\theta}$, the three Pythagorean identities, and the even-odd identities $\sin(-\theta) = -\sin\theta$ and $\cos(-\theta) = \cos\theta$. When an equation mixes functions, try converting everything to sine and cosine or making a Pythagorean swap.

Worked example. Solve $2\cos^2 x + \sin x - 1 = 0$ on $[0, 2\pi)$. Replace $\cos^2 x$ with $1 - \sin^2 x$ to get $2 - 2\sin^2 x + \sin x - 1 = 0$, which simplifies to $2\sin^2 x - \sin x - 1 = 0$. Factor it as $(2\sin x + 1)(\sin x - 1) = 0$. If $\sin x = -\frac{1}{2}$, then $x = \frac{7\pi}{6}$ or $\frac{11\pi}{6}$. If $\sin x = 1$, then $x = \frac{\pi}{2}$. That gives three solutions.

The sum and difference identities find trig values of angles you can build from special angles, like $75° = 45° + 30°$ or $15° = 45° - 30°$.

$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$

$\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$

$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}$

Watch the signs. In the sine formula, the sign matches the operation. In the cosine formula, the sign flips, so $\cos(A + B)$ has a minus. In the tangent formula, the numerator matches and the denominator flips.

Worked example. Find the exact value of $\cos 75°$. Write it as $\cos(45° + 30°) = \cos 45°\cos 30° - \sin 45°\sin 30°$. Substituting gives $\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}$, which is about 0.2588. A calculator agrees.

These formulas also prove the cofunction identities. For example, $\sin(90° - \theta) = \sin 90°\cos\theta - \cos 90°\sin\theta = (1)\cos\theta - (0)\sin\theta = \cos\theta$. The same idea works in radians, so $\frac{5\pi}{12}$ splits into $\frac{\pi}{4} + \frac{\pi}{6}$.

Double-angle formulas come from setting $A = B = \theta$ in the sum identities.

$\sin 2\theta = 2\sin\theta\cos\theta$

$\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$

$\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}$

Pick the version of $\cos 2\theta$ that matches what you know or what the problem needs. If an equation already has $\sin\theta$ in it, use $1 - 2\sin^2\theta$ so everything stays in sine.

Solving the cosine forms for the squared term gives the power-reducing formulas, $\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ and $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$. Many textbooks call these reduction formulas, because they replace a square with a first power of cosine. Some older materials use "reduction formula" for the cofunction identities, but the power-reducing meaning is the one this topic is about.

Replace $\theta$ with $\frac{\theta}{2}$ and take a square root to get the half-angle formulas, $\sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}$ and $\cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}$. You choose the sign from the quadrant of $\frac{\theta}{2}$, not $\theta$.

Worked example. Find $\cos 15°$. Use $\theta = 30°$. Since $15°$ is in Quadrant I, take the positive root: $\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2} \approx 0.9659$.

These formulas convert between products of trig functions and sums of them. They aren't new math. You get them by adding or subtracting the sum and difference identities from 7.2.

Product-to-sum formulas turn a product into a sum.

$\sin A\cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)]$

$\cos A\cos B = \frac{1}{2}[\cos(A + B) + \cos(A - B)]$

$\sin A\sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]$

For example, $\cos 75°\cos 15° = \frac{1}{2}[\cos 90° + \cos 60°] = \frac{1}{2}[0 + \frac{1}{2}] = \frac{1}{4}$, even though neither factor is a special angle.

Sum-to-product formulas go the other way, which helps with solving because a product equal to zero can be split into factors.

$\sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A - B}{2}$

$\cos A + \cos B = 2\cos\frac{A + B}{2}\cos\frac{A - B}{2}$

$\cos A - \cos B = -2\sin\frac{A + B}{2}\sin\frac{A - B}{2}$

To solve $\sin 3x + \sin x = 0$, compute $\frac{3x + x}{2} = 2x$ and $\frac{3x - x}{2} = x$. That gives $2\sin 2x\cos x = 0$, so $\sin 2x = 0$ or $\cos x = 0$. The most common slip is forgetting to divide by 2.

For many trig equations, start by isolating a trig function or factoring. Then use a reference angle and the unit circle when they apply, and find every solution in the stated interval.

If you get a value like $\sin x = 3$, there is no solution, because sine and cosine only output values in $[-1, 1]$. An inverse function on a calculator returns one principal value. For $\sin x = k$, use both $x = \theta + 2\pi n$ and $x = \pi - \theta + 2\pi n$. For $\cos x = k$, use $x = \pm\theta + 2\pi n$. The two families can overlap. For $\tan x = k$, use $x = \theta + \pi n$. In each case, $\theta$ is the principal value and $n$ is any integer.

Equations like $2\cos^2 x - 3\cos x + 1 = 0$ are quadratics in disguise. Let $u = \cos x$, factor to $(2u - 1)(u - 1) = 0$, and solve $\cos x = \frac{1}{2}$ and $\cos x = 1$. On $[0, 2\pi)$, that gives $x = 0, \frac{\pi}{3}, \frac{5\pi}{3}$.

Worked example. Solve $\sin 2x = \frac{\sqrt{3}}{2}$ on $[0, 2\pi)$. Since $x$ runs from $0$ to $2\pi$, the angle $2x$ runs from $0$ to $4\pi$, which is two full trips around the circle. In the first trip, $2x = \frac{\pi}{3}$ or $\frac{2\pi}{3}$. In the second, $2x = \frac{7\pi}{3}$ or $\frac{8\pi}{3}$. Divide by 2 to get $x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$. If you only went around once, you'd miss half the answers.

A sinusoidal model has the form $y = A\sin(B(x - C)) + D$ or $y = A\cos(B(x - C)) + D$. To build one from data, find the midline with $D = \frac{\max + \min}{2}$ and the amplitude with $A = \frac{\max - \min}{2}$. The period $T$ comes from the length of one cycle, and $B = \frac{2\pi}{T}$. Cosine is easiest when you know where a maximum happens, so set $C$ to that time.

Worked example. A city averages a minimum of 30°F in January (month 1) and a maximum of 80°F in July (month 7). Assume the temperature follows a sinusoidal model with a 12-month period. Then $D = 55$, $A = 25$ and $B = \frac{2\pi}{12} = \frac{\pi}{6}$. With the maximum in July, the model is $T(t) = 25\cos\left(\frac{\pi}{6}(t - 7)\right) + 55$. To check it, plug in $t = 1$. You get $25\cos(-\pi) + 55 = 30$, which matches. To find when it's 67.5°F, solve $\cos\left(\frac{\pi}{6}(t - 7)\right) = 0.5$. For one annual cycle, $\frac{\pi}{6}(t - 7) = \pm\frac{\pi}{3}$ gives $t = 5$ or $t = 9$. The model predicts 67.5°F in May and September. The times repeat every 12 months.

Simple harmonic motion, like a mass on a spring, uses the same idea. If the mass starts at maximum displacement from equilibrium, a model is $x(t) = A\cos(\omega t)$. Here, the angular frequency is $\omega = \frac{2\pi}{T}$ and the frequency is $f = \frac{1}{T}$ cycles per unit time. Other starting positions can require a phase shift.