Trig Identities

A2.H.1Pythagorean identity01/04

The Pythagorean identity.

Step01of 04

The single most-used trig identity:

sin^{2} θ + cos^{2} θ = 1

It comes straight from the unit circle: a point at angle $θ$ has coordinates $(cos θ, sin θ)$ on a circle of radius 1, so by Pythagorean, $cos^{2} θ + sin^{2} θ = 1^{2}$ .

A2.H.1 — The Pythagorean identity.

Step 01 of 04

The single most-used trig identity:

sin^{2} θ + cos^{2} θ = 1

It comes straight from the unit circle: a point at angle $θ$ has coordinates $(cos θ, sin θ)$ on a circle of radius 1, so by Pythagorean, $cos^{2} θ + sin^{2} θ = 1^{2}$ .

Step 02 of 04

Two algebraic rearrangements you'll use constantly.

sin^{2} θ = 1 - cos^{2} θ

cos^{2} θ = 1 - sin^{2} θ

Useful when an expression contains both $sin^{2}$ and $cos^{2}$ — substitute one for the other to leave only one function.

Step 03 of 04

Two derived identities for the other ratios.

1 + tan^{2} θ = sec^{2} θ

1 + cot^{2} θ = csc^{2} θ

Derive: divide the original Pythagorean identity by $cos^{2} θ$ for the first, by $sin^{2} θ$ for the second.

$sec θ$	→ $1/ cos θ$
$csc θ$	→ $1/ sin θ$
$cot θ$	→ $cos θ / sin θ$

Step 04 of 04

Worked example. If $sin θ = 3/5$ and $θ$ is in Q2, find $cos θ$ .

cos^{2} θ = 1 - (3/5)^{2} = 16/25 \Rightarrow cos θ = \pm 4/5

In Q2, cosine is NEGATIVE: $cos θ = - 4/5$ .

Key insight

sin^{2} θ + cos^{2} θ = 1

— derive everything else from this one. Two sister identities (1 + tan² = sec², 1 + cot² = csc²) fall out by dividing through. Quadrant determines the SIGN of the answer.

A2.H.2 — Sum and difference identities.

Step 01 of 04

These let you compute trig values of $A + B$ or $A - B$ from values of $A$ and $B$ separately.

$sin (A + B)$	→ $sin A cos B + cos A sin B$
$sin (A - B)$	→ $sin A cos B - cos A sin B$
$cos (A + B)$	→ $cos A cos B - sin A sin B$
$cos (A - B)$	→ $cos A cos B + sin A sin B$

Notice the SIGN PATTERN: sin formulas keep the operation; cos formulas FLIP it. Memorize that pattern, then memorize the building blocks (sin cos and cos sin).

Step 02 of 04

Worked example — find an exact value. Compute $sin 75°$ .

Write $75° = 45° + 30°$ :

sin 75° = sin 45° cos 30° + cos 45° sin 30°

= \frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2}

= \frac{6}{4} + \frac{2}{4} = \frac{6 + 2}{4}

Step 03 of 04

Tangent identities follow from sin/cos divisions.

tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B}

Step 04 of 04

Useful application — proving identities. The sum/difference identities are the building blocks for proving complex trig identities and for simplifying expressions involving angles like $θ + 90°$ or $θ - π$ .

Example: $sin (θ + π) = sin θ cos π + cos θ sin π$ $= - sin θ$ . The "shift by $π$ negates" rule comes right out of the formula.

Key insight

Sum/difference identities turn unfamiliar angles into familiar ones by splitting them. The pattern: sin(A±B) keeps the sign; cos(A±B) flips it. Useful for exact values like sin 75°, cos 15°, etc.

A2.H.3 — Double-angle identities.

Step 01 of 04

Set $A = B = θ$ in the sum identities and you get the double-angle identities. Each lets you express $sin (2 θ)$ or $cos (2 θ)$ in terms of $θ$ alone.

$sin (2 θ)$	→ $2 sin θ cos θ$
$cos (2 θ)$	→ $cos^{2} θ - sin^{2} θ$
(also)	→ $= 2 cos^{2} θ - 1$
(also)	→ $= 1 - 2 sin^{2} θ$

The three forms of $cos (2 θ)$ are equivalent — pick whichever is most useful. (The Pythagorean identity is what lets you swap among them.)

Step 02 of 04

Worked example. Given $sin θ = 3/5$ and $θ$ in Q1, find $sin (2 θ)$ and $cos (2 θ)$ .

From Q1 + Pythagorean: $cos θ = 4/5$ .

sin (2 θ) = 2 (3/5) (4/5) = 24/25

cos (2 θ) = (4/5)^{2} - (3/5)^{2} = 16/25 - 9/25 = 7/25

Step 03 of 04

Tangent double-angle.

tan (2 θ) = \frac{2 tan θ}{1 - tan ^{2} θ}

Same shape as the sum-formula special case $tan (A + B)$ with $A = B = θ$ .

Step 04 of 04

Half-angle bonus. Solving the cos double-angle identities for $θ$ in terms of $2 θ$ :

sin^{2} θ = \frac{1 - cos ( 2 θ )}{2}, cos^{2} θ = \frac{1 + cos ( 2 θ )}{2}

These "power-reduction formulas" let you compute exact half-angle values and are essential for calculus integration.

Key insight

Double-angle = sum identity with both angles equal to

θ

. Cos has THREE equivalent forms — pick the one matching what you know (sin alone, cos alone, or both). Solve them backwards to get power-reduction / half-angle formulas.

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