Draw:

• a coordinate plane
• centered circle
• radius labeled 1
• point at (1 , 0)

Label:

• center = origin
• radius = 1

Key Idea

The unit circle is called the “unit” circle because every radius has length 1 unit.

That’s why the hypotenuse of our triangles becomes 1.

Draw:

• unit circle
• radius line at $45^\circ$
• drop a vertical line to create a right triangle

Now you have:

• hypotenuse = 1
• equal legs

Label:

• x-coordinate
• y-coordinate

Start from the standard square.

Step 1 — Standard Square

Draw:

• square with side length 1

Draw diagonal.

• the diagonal splits the square into two congruent triangles
• each angle becomes $45^\circ$

Step 2 — Use Pythagorean Theorem

$x^2+x^2= 1^2$

Then:

$2x^2=1$

Then:

$x=\frac{1}{\sqrt2} =\frac{\sqrt2}{2}$

**we got rid of the radical in the denominator by multiplying by a form of one $(\frac{\sqrt2}{\sqrt2})$**

Final Point

So the coordinate point becomes:

$\left(\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right)$

Therefore:

• cosine = $\frac{\sqrt2}{2}$
• sine = $\frac{\sqrt2}{2}$

We followed the triangle legs up to the point where the radius intersects the unit circle because the point comes FROM the triangle.

Draw equilateral triangle.

Split equilateral triangle down the middle

• equilateral triangles are standard shapes
• all sides equal
• all angles $60^\circ$

When split it creates two 30-60-90 triangles

Label Ratios

If original side length = 2:

Then:

• short leg = 1
• long leg = $\sqrt3$
• hypotenuse = 2

SCALE TO THE UNIT CIRCLE

“But the unit circle requires the hypotenuse (radius) to equal 1.”

So divide everything by 2 (what I do to one side, I must do to the other to keep everything in ratio).

New ratios:

• short leg = $\frac12$
• long leg = $\frac{\sqrt3}{2}$
• hypotenuse = 1

Coordinate Point

At $60^\circ$: $\left(\frac12,\frac{\sqrt3}{2}\right)$

At $30^\circ$: $\left(\frac{\sqrt3}{2},\frac12\right)$

KEY EMPHASIS

“The x- and y-values are not random.”

“They come directly from the triangle ratios.”

Draw:

• full unit circle
• all four quadrants

Label signs:

• QI: (+,+)
• QII: (-,+)
• QIII: (-,-)
• QIV: (+,-)

“The triangle ratios stay the same size, but the coordinate signs change depending on the quadrant.”

This is a VERY important SAT concept.

IMPORTANT To Note

“The triangle gives us the distances.

The unit circle gives us the position.

Together, they give us the coordinate point.”

Practice 1 — Identify Coordinates

Find the coordinate point for:

1. $45^\circ$
2. $30^\circ$
3. $60^\circ$

Practice 2 — Match Coordinates

Match the angle to the coordinate point.

A. $\left(\frac12,\frac{\sqrt3}{2}\right)$

B. $\left(\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right)$

C. $\left(\frac{\sqrt3}{2},\frac12\right)$

Angles:

• $30^\circ$
• $45^\circ$
• $60^\circ$

Practice 3 — Cosine & Sine

Given: $\left(\frac{\sqrt3}{2},\frac12\right)$

Find:

• cosine
• sine

Practice 4 — Quadrant Signs

If the point is in Quadrant II and has reference angle $30^\circ$, determine the coordinate signs.

Practice 5 — Extrapolation

Draw:

• unit circle
• radius at $45^\circ$
• triangle inside

Using the triangle ratios and the unit circle, determine the coordinate point where the radius intersects the circle.

Practice 6 — SAT-Style

Find: $\cos\left(\frac{5\pi}{4}\right)$

This forces:

• radians
• quadrant understanding
• 45-45-90 ratios
• sign changes