Trigonometry Beyond Right Triangles

Trigonometry Beyond Right Triangles

Right-triangle trigonometry only takes you so far. The ACT also asks you to rewrite trig expressions with identities and to solve triangles that have no right angle. Both skills rest on a small set of connected facts.

The identities

Three kinds of trigonometric identities let one ratio be rewritten in terms of another, without needing a calculator or a specific angle.

• Pythagorean identity: $\sin^2\theta + \cos^2\theta = 1$. It connects sine and cosine directly.
• Reciprocal identities: $\csc\theta = \frac{1}{\sin\theta}$, $\sec\theta = \frac{1}{\cos\theta}$, and $\cot\theta = \frac{1}{\tan\theta}$. They invert the three basic ratios.
• Quotient identity: $\tan\theta = \frac{\sin\theta}{\cos\theta}$. It expresses tangent in terms of sine and cosine.

A related identity, $1 + \tan^2\theta = \sec^2\theta$, follows directly from the Pythagorean identity by dividing every term by $\cos^2\theta$. Recognizing that these identities are all connected, instead of memorizing each in isolation, makes simplifying an unfamiliar trig expression far more manageable.

Worked example: find a ratio from another

If $\sin\theta = \frac{5}{13}$ and $\theta$ is in the first quadrant, find $\cos\theta$. Apply the Pythagorean identity:

$\cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$, so $\cos\theta = \frac{12}{13}$, positive in the first quadrant.

Always take the square root last, and always use the quadrant to choose the sign. In the second quadrant, for instance, sine is positive and cosine is negative, so if $\cos\theta = -\frac{1}{2}$ with $\theta$ between $90^\circ$ and $180^\circ$, then $\sin^2\theta = \frac{3}{4}$ and $\sin\theta = +\frac{\sqrt{3}}{2}$.

Worked example: simplify an expression

Simplify $\frac{1 - \cos^2\theta}{\sin\theta \cdot \cos\theta}$.

Recognize. The numerator matches the Pythagorean identity: $1 - \cos^2\theta = \sin^2\theta$. Rewrite it before simplifying, instead of canceling anything directly.

Set up. $\frac{1 - \cos^2\theta}{\sin\theta\cos\theta} = \frac{\sin^2\theta}{\sin\theta\cos\theta}$

Solve. $= \frac{\sin\theta}{\cos\theta} = \tan\theta$

Check. Testing $\theta = 45^\circ$: the original expression gives $\frac{1 - 0.5}{0.7071 \times 0.7071} = \frac{0.5}{0.5} = 1$, and $\tan 45^\circ = 1$, which match.

Best route. Hand simplification using the identity. A calculator can spot-check one specific angle, but it does not simplify the general expression.

Common error. Canceling $\cos^2\theta$ directly against the $\cos\theta$ in the denominator without first rewriting $1 - \cos^2\theta$ as $\sin^2\theta$.

The Law of Sines and the Law of Cosines

These two laws extend right-triangle trigonometry to triangles with no right angle.

The Law of Sines states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$, relating each side to the sine of its opposite angle. It applies when two angles and a side are known, or when two sides and a non-included angle are known.

The Law of Cosines, $c^2 = a^2 + b^2 - 2ab\cos C$, applies when two sides and the included angle are known, or when all three sides are known and an angle is needed.

Choosing the route

Before picking up a calculator, identify which of three routes the triangle actually calls for.

If none of these patterns is visible in the given information, that is a signal to re-read the figure and identify what is actually paired, not to start entering numbers into a calculator.

Worked example: Law of Sines

In triangle $ABC$, angle $A = 55^\circ$, angle $B = 65^\circ$, and side $a = 9$. Find side $b$, rounded to the nearest tenth. Set up the Law of Sines using the known angle-side pair:

$\frac{9}{\sin 55^\circ} = \frac{b}{\sin 65^\circ}$, so $b = \frac{9\sin 65^\circ}{\sin 55^\circ} \approx 10.0$.

Put the unknown side over its own opposite angle. Inverting the ratio, which gives $\frac{9\sin 55^\circ}{\sin 65^\circ}$, is the most common mistake. If only two angles are given, find the third by subtracting from $180^\circ$ first.

Worked example: Law of Cosines

A triangle has sides of length $6$ and $8$, with an included angle of $50^\circ$. Find the length of the third side, rounded to the nearest hundredth.

$c^2 = 6^2 + 8^2 - 2(6)(8)\cos 50^\circ = 100 - 96\cos 50^\circ \approx 38.29$, so $c \approx 6.19$.

Do the multiplication $2ab\cos C$ before subtracting from $a^2 + b^2$. Leaving out the cosine term entirely gives $\sqrt{a^2 + b^2}$, the right-triangle answer, which is a standard wrong choice. To find an angle from three sides, solve for the cosine: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. The largest angle is always opposite the longest side.

The ambiguous case

When the givens are two sides and a non-included angle (SSA), check for the ambiguous case. The sine equation can produce a second possible angle, so the information may describe zero, one, or two triangles. Any second angle must still leave the triangle with a positive third angle: if the first solution is $B$, the other candidate is $180^\circ - B$, and it counts only if it plus the given angle is less than $180^\circ$.

Where this appears as a model

Two paths leaving a shared starting point at an angle to each other make a triangle with no right angle. Two ships leaving the same port, one traveling $30$ miles and the other $40$ miles along paths that form a $70^\circ$ angle, end up about $40.98$ miles apart, found with the Law of Cosines once the triangle is recognized as having no right angle.

Strategy summary

1. For trig expressions, look for $\sin^2 + \cos^2$, a reciprocal, or $\frac{\sin}{\cos}$ before doing anything else.
2. Use the quadrant to choose the sign of a square root.
3. Match the triangle to a route: right angle, Law of Sines (a side paired with its opposite angle), or Law of Cosines (SAS or SSS).
4. Check that your calculator is in degree mode.