Trigonometry Mastery | sin · cos

Core Concept

The Right Triangle & SOH-CAH-TOA

\text{For a right triangle with angle } \theta:\] \[\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}}\quad \cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}\quad \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}

Function	Ratio	Memory Cue	Range (0°–90°)
sin θ	Opp / Hyp	Some Officers Have	0 to 1
cos θ	Adj / Hyp	Curly Auburn Hair	1 to 0
tan θ	Opp / Adj	Throughout Over Ages	0 to ∞

SOH · CAH · TOA Sin = Opposite / Hypotenuse · Cos = Adjacent / Hypotenuse · Tan = Opposite / Adjacent

Special Angle Values — Memorize These!

Angle	sin θ	cos θ	tan θ
0°	0	1	0
30°	\(\tfrac{1}{2}\)	\(\tfrac{\sqrt{3}}{2}\)	\(\tfrac{1}{\sqrt{3}}=\tfrac{\sqrt{3}}{3}\)
45°	\(\tfrac{\sqrt{2}}{2}\)	\(\tfrac{\sqrt{2}}{2}\)	1
60°	\(\tfrac{\sqrt{3}}{2}\)	\(\tfrac{1}{2}\)	\(\sqrt{3}\)
90°	1	0	undefined

Pattern for sin: 0, 30°, 45°, 60°, 90° \(\dfrac{\sqrt{0}}{2},\;\dfrac{\sqrt{1}}{2},\;\dfrac{\sqrt{2}}{2},\;\dfrac{\sqrt{3}}{2},\;\dfrac{\sqrt{4}}{2}\) — cos goes in reverse order!

Key Identities to Know

\sin^2\theta + \cos^2\theta = 1 \qquad \tan\theta = \frac{\sin\theta}{\cos\theta}

\sin(180°-\theta)=\sin\theta \qquad \cos(180°-\theta)=-\cos\theta

\cos^2\theta - \sin^2\theta = \cos(2\theta) \qquad 2\sin\theta\cos\theta = \sin(2\theta)

\sin(A+B)=\sin A\cos B + \cos A\sin B

Worked Examples

Example 1 · Basic Ratio

In a right triangle, the side opposite angle θ = 3, adjacent = 4, hypotenuse = 5. Find sin θ, cos θ, and tan θ.

Using SOH-CAH-TOA:
\(\sin\theta = \dfrac{\text{Opp}}{\text{Hyp}} = \dfrac{3}{5}\) · \(\cos\theta = \dfrac{\text{Adj}}{\text{Hyp}} = \dfrac{4}{5}\) · \(\tan\theta = \dfrac{\text{Opp}}{\text{Adj}} = \dfrac{3}{4}\)

Answer: sin θ = 3/5, cos θ = 4/5, tan θ = 3/4

Example 2 · Special Angle

Evaluate: \(\sin 30° + \cos 60°\)

From the special angles table:
\(\sin 30° = \dfrac{1}{2}\) and \(\cos 60° = \dfrac{1}{2}\)
\(\sin 30° + \cos 60° = \dfrac{1}{2} + \dfrac{1}{2} = 1\)

Answer: 1

Example 3 · Pythagorean Identity

If \(\sin\theta = \dfrac{3}{5}\) and θ is in the first quadrant, find cos θ and tan θ.

Use \(\sin^2\theta + \cos^2\theta = 1\):
\(\cos^2\theta = 1 - \left(\dfrac{3}{5}\right)^2 = 1 - \dfrac{9}{25} = \dfrac{16}{25}\)
Since θ is in Q1, \(\cos\theta = \dfrac{4}{5}\) (positive).
\(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{4/5} = \dfrac{3}{4}\)

Answer: cos θ = 4/5, tan θ = 3/4

Example 4 · Supplementary Angle

Find the exact value of \(\cos 150°\).

Use the identity \(\cos(180°-\theta) = -\cos\theta\):
\(\cos 150° = \cos(180°-30°) = -\cos 30° = -\dfrac{\sqrt{3}}{2}\)

Answer: \(-\dfrac{\sqrt{3}}{2}\)

Example 5 · Finding a Side

In a right triangle, angle θ = 60°, and the adjacent side = 4. Find the opposite side.

\(\tan\theta = \dfrac{\text{Opp}}{\text{Adj}}\) so \(\text{Opp} = \text{Adj}\times\tan\theta = 4\times\tan 60° = 4\times\sqrt{3} = 4\sqrt{3}\)

Answer: \(4\sqrt{3} \approx 6.93\)