Core Ideas in Plain Terms

Study Sheet

Core Ideas in Plain Terms

The big trig ideas, in everyday language

What the Trig Ratios Really Are

Concept
Sine, cosine, tangent are just fractions
adjacentoppositehypotenuse

In a right triangle, pick one of the non-right angles. Sine is the side across from it over the longest side; cosine is the side next to it over the longest side; tangent is across over next-to. That is all the words "sin, cos, tan" mean — three particular fractions of the sides. "SOH-CAH-TOA" is just a way to remember which fraction is which. As formulas: sin⁡θ=oppositehypotenuse\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}, cos⁡θ=adjacenthypotenuse\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}, tan⁡θ=oppositeadjacent\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}.

Example
A 3-4-5 triangle
345

Take the angle whose opposite side is 33 and adjacent side is 44 (hypotenuse 55). Then sin⁡=35=0.6\sin = \tfrac35 = 0.6, cos⁡=45=0.8\cos = \tfrac45 = 0.8, and tan⁡=34=0.75\tan = \tfrac34 = 0.75. Nothing mysterious — just three ratios of the sides.

The Unit Circle, Plainly

Concept
A circle of radius 1 that stores every angle
(cos θ, sin θ)

Draw a circle of radius 11 centered at the origin and spin a point around it. For an angle θ\theta, the point lands at coordinates (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) — so the cosine is just the xx-coordinate and the sine is the yy-coordinate. That is why sine and cosine never leave the range [−1,1][-1,1]: the point can't go past the circle.

Example
The angle 90°

At 90∘90^\circ the point is straight up at (0,1)(0, 1). So cos⁡90∘=0\cos 90^\circ = 0 (the xx-coordinate) and sin⁡90∘=1\sin 90^\circ = 1 (the yy-coordinate).

Radians and Wave Graphs

Concept
Radians measure angles by arc length

Instead of 360∘360^\circ for a full turn, radians use 2π2\pi — the distance around a radius-11 circle. So 180∘=π180^\circ = \pi, 90∘=π290^\circ = \tfrac\pi2. To convert, multiply degrees by π180\tfrac{\pi}{180}.

Concept
Reading a sine graph
3

For y=asin⁡(bx)y = a\sin(bx), the amplitude ∣a∣|a| is how tall the wave gets, and the period 2πb\tfrac{2\pi}{b} is how wide one full wave is before it repeats. Bigger aa = taller; bigger bb = more scrunched together.

Example
y = 3 sin(2x)

Amplitude 33 (peaks at 33, troughs at −3-3) and period 2π2=π\tfrac{2\pi}{2} = \pi (one full wave every π\pi units).

Going Deeper: Identities and the Law of Sines

Concept
Why sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
distance 1

The unit-circle point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) sits exactly 11 away from the origin, and distance is x2+y2\sqrt{x^2 + y^2}. Squaring that distance gives cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1 — the Pythagorean identity is just the Pythagorean theorem on the unit circle. Divide the whole thing by cos⁡2θ\cos^2\theta and you get the companion identity 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta for free.

Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2
Reminder — The Pythagorean identity:  sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
Concept
The Law of Sines, plainly

In any triangle, asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}: the bigger an angle, the bigger the side across from it, in exact proportion. It is the tool for triangles with no right angle, whenever you know an angle and its opposite side.

Reminder — Law of Sines:  asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
Example
Using the identity

Given sin⁡θ=35\sin\theta = \tfrac{3}{5} with θ\theta in Quadrant I: cos⁡2θ=1−925=1625\cos^2\theta = 1 - \tfrac{9}{25} = \tfrac{16}{25}, so cos⁡θ=45\cos\theta = \tfrac45 and tan⁡θ=3/54/5=34\tan\theta = \dfrac{3/5}{4/5} = \tfrac34 — no triangle drawing needed.

Example
Solving a right triangle (multi-step)

A right triangle has legs 55 and 1212.

Step 1 — Pythagorean theorem: the hypotenuse is 52+122=169=13\sqrt{5^2 + 12^2} = \sqrt{169} = 13.

Step 2 — for the angle θ\theta opposite the side 55: sin⁡θ=513\sin\theta = \tfrac{5}{13}.

Step 3 — and tan⁡θ=512\tan\theta = \tfrac{5}{12}, straight from the ratio definitions.

Problem-Solving Playbook

Concept
Draw, label, choose the tool

Sketch the triangle and label everything you know. Then pick the tool the givens point to: a right angle → SOH-CAH-TOA; an angle paired with its opposite side → Law of Sines; two sides and the included angle → Law of Cosines (c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C).

Reminder — Law of Cosines:  c2=a2+b2−2abcos⁡Cc^{2}=a^{2}+b^{2}-2ab\cos C
Example
Worked: two sides and the included angle

a=5a = 5, b=8b = 8, C=60∘C = 60^\circ. The givens are side–angle–side, so use the Law of Cosines: c2=25+64−2⋅5⋅8⋅cos⁡60∘=89−80⋅12=49c^2 = 25 + 64 - 2 \cdot 5 \cdot 8 \cdot \cos 60^\circ = 89 - 80 \cdot \tfrac12 = 49, so c=7c = 7.