Right Triangles and Trigonometry Pattern - Trig Relationships

This pattern tests your understanding of cofunction identities (the relationship between sine and cosine of complementary angles) and how trig ratios transfer between similar triangles.

 

The Cofunction Identity: The Big Idea

In a right triangle, the two acute angles always add up to $90°$. If one acute angle is $\theta$, the other is $90° - \theta$. The side that is "opposite" one angle is "adjacent" to the other. This creates the fundamental relationship:

$\sin \theta = \cos(90° - \theta)$ and $\cos \theta = \sin(90° - \theta)$

In plain English: the sine of any angle equals the cosine of its complement.

 

Using the Cofunction Identity Directly

In right triangle $ABC$ with $\angle C = 90°$, $\sin A = \dfrac{12}{13}$. What is $\cos B$?

A) $\dfrac{5}{13}$
B) $\dfrac{5}{12}$
C) $\dfrac{12}{13}$
D) $\dfrac{13}{12}$

Since $\angle A + \angle B = 90°$, angles $A$ and $B$ are complementary.
$\cos B = \sin A = \dfrac{12}{13}$.
The answer is C. It's that simple — no calculation needed. The sine of one acute angle equals the cosine of the other.

Why it works geometrically: $\sin A = \dfrac{\text{opposite } A}{\text{hypotenuse}} = \dfrac{BC}{AB}$. Meanwhile, $\cos B = \dfrac{\text{adjacent to } B}{\text{hypotenuse}} = \dfrac{BC}{AB}$. They're the same fraction because the side opposite $A$ is adjacent to $B$.

 

Solving for an Angle Using $\sin(\alpha) = \cos(\beta)$

When the SAT says $\sin \alpha = \cos \beta$, it means $\alpha + \beta = 90°$. Set up the equation and solve.

If $\sin(3x + 5)° = \cos(2x + 20)°$, what is the value of $x$?

A) $5$
B) $13$
C) $25$
D) $65$

$\sin \alpha = \cos \beta$ means $\alpha + \beta = 90°$:
$(3x + 5) + (2x + 20) = 90$
$5x + 25 = 90$
$5x = 65$
$x = 13$
The answer is B. Check: $\sin(44°) = \cos(46°)$, and $44 + 46 = 90$. ✓

Gotcha: Don't set the expressions equal to each other. The equation is $\alpha + \beta = 90$, not $\alpha = \beta$.

 

Similar Triangles: Same Angles → Same Trig Ratios

If $\triangle PQR \sim \triangle STU$, the corresponding angles are equal. Since trig ratios depend only on the angle (not the size of the triangle), corresponding angles have the same trig values.

$\triangle PQR$ is similar to $\triangle STU$, with $P$ corresponding to $S$ and $Q$ corresponding to $T$. If $\cos S = \dfrac{15}{17}$, what is $\cos P$?

A) $\dfrac{8}{17}$
B) $\dfrac{8}{15}$
C) $\dfrac{15}{17}$
D) $\dfrac{17}{15}$

$P$ corresponds to $S$, so $\angle P = \angle S$. Therefore $\cos P = \cos S = \dfrac{15}{17}$.
The answer is C. Similarity means the angles are equal — and equal angles have equal trig ratios, regardless of side lengths.

$\triangle PQR$ is similar to $\triangle STU$. If $\sin P = \dfrac{5}{13}$, what is $\sin S$?

A) $\dfrac{12}{13}$
B) $\dfrac{5}{12}$
C) $\dfrac{5}{13}$
D) $\dfrac{13}{5}$

$P$ corresponds to $S$: $\sin S = \sin P = \dfrac{5}{13}$. The answer is C.

Key point: It doesn't matter if the similar triangles have different side lengths, different perimeters, or different areas. The trig ratios are determined entirely by the angles.

 

Computing a Trig Ratio from Given Sides

In the right triangle shown, $PQ = 8$, $QR = 15$, and $\angle Q = 90°$. What is $\cos P$?

A) $\dfrac{15}{17}$
B) $\dfrac{8}{17}$
C) $\dfrac{8}{15}$
D) $\dfrac{17}{8}$

First, find the hypotenuse: $PR = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$.
$\cos P = \dfrac{\text{adjacent to } P}{\text{hypotenuse}} = \dfrac{PQ}{PR} = \dfrac{8}{17}$.
The answer is B. Option A ($\dfrac{15}{17}$) gives $\sin P$ (or equivalently $\cos R$). Option C ($\dfrac{8}{15}$) gives $\tan P$ with the wrong denominator.

 

Advanced Cofunction Expressions

Some harder problems combine cofunctions with algebraic manipulation.

What is the value of $(\sin 18°)(\cos 72°) + (\cos 18°)(\sin 72°)$?

A) $1$
B) $0$
C) $\dfrac{1}{2}$
D) $2$

Since $18° + 72° = 90°$: $\cos 72° = \sin 18°$ and $\sin 72° = \cos 18°$.
Substitute: $(\sin 18°)(\sin 18°) + (\cos 18°)(\cos 18°) = \sin^2 18° + \cos^2 18°$.
By the Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$.
The answer is A.

What is the value of $\dfrac{\cos^2 20°}{\sin^2 70°} + (\cos 20°)(\sin 70°)$?

Since $20° + 70° = 90°$: $\sin 70° = \cos 20°$.
$\dfrac{\cos^2 20°}{\cos^2 20°} + (\cos 20°)(\cos 20°) = 1 + \cos^2 20°$.
This simplifies to $1 + \cos^2 20°$ — but actually: $\dfrac{\cos^2 20°}{\sin^2 70°} = \dfrac{\cos^2 20°}{\cos^2 20°} = 1$, and $(\cos 20°)(\sin 70°) = (\cos 20°)(\cos 20°) = \cos^2 20°$.
So the total $= 1 + \cos^2 20° \approx 1 + 0.883 \approx 1.883$.
The answer depends on the specific options.

 

Pythagorean Identity: $\sin^2 \theta + \cos^2 \theta = 1$

This identity is the trig version of the Pythagorean theorem. It shows up when: - You see $\sin^2 + \cos^2$ of the same angle (always equals $1$) - You need to find $\cos$ from $\sin$ (or vice versa): $\cos \theta = \sqrt{1 - \sin^2 \theta}$

 

What to Do on Test Day

• The cofunction identity: $\sin \theta = \cos(90° - \theta)$. In a right triangle, the sine of one acute angle equals the cosine of the other
• When you see $\sin \alpha = \cos \beta$: set $\alpha + \beta = 90°$ and solve. Don't set $\alpha = \beta$
• Similar triangles have equal trig ratios for corresponding angles. The scale factor, perimeter, and area are irrelevant — only the angles matter
• The Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$. Recognize it when you see $(\sin x)(\cos(90° - x)) + (\cos x)(\sin(90° - x))$, which simplifies to $\sin^2 x + \cos^2 x = 1$
• Common complementary pairs: $18° + 72°$, $20° + 70°$, $25° + 65°$, $30° + 60°$, $35° + 55°$, $40° + 50°$. Recognize these to apply the cofunction identity quickly
• When computing a trig ratio from sides: find all three sides first (Pythagorean theorem if needed), then apply SOHCAHTOA carefully for the correct angle
• $\cos$ and $\sin$ of acute angles are always between 0 and 1. If your answer exceeds 1, you've flipped the fraction