James's Math Hub
← Trigonometry

Lesson 2 / 5 · 12 min read

Trig Identities

Pythagorean, reciprocal, sum/difference, and double-angle — the ones you'll keep reusing.

Pythagorean identities

Straight from the unit circle, sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1. Dividing through by cos⁡2\cos^2 or sin⁡2\sin^2 gives two siblings:

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 1+tan⁡2θ=sec⁡2θ1 + \tan^2 \theta = \sec^2 \theta 1+cot⁡2θ=csc⁡2θ1 + \cot^2 \theta = \csc^2 \theta

where sec⁡θ=1/cos⁡θ\sec \theta = 1/\cos \theta, csc⁡θ=1/sin⁡θ\csc \theta = 1/\sin \theta, cot⁡θ=1/tan⁡θ\cot \theta = 1/\tan \theta.

Reciprocal and quotient identities

tan⁡θ=sin⁡θcos⁡θcot⁡θ=cos⁡θsin⁡θ\tan \theta = \dfrac{\sin \theta}{\cos \theta} \qquad \cot \theta = \dfrac{\cos \theta}{\sin \theta} sec⁡θ=1cos⁡θcsc⁡θ=1sin⁡θ\sec \theta = \dfrac{1}{\cos \theta} \qquad \csc \theta = \dfrac{1}{\sin \theta}

Even and odd

sin⁡(−x)=−sin⁡x(odd)\sin(-x) = -\sin x \quad (\text{odd}) cos⁡(−x)=cos⁡x(even)\cos(-x) = \cos x \quad (\text{even}) tan⁡(−x)=−tan⁡x(odd)\tan(-x) = -\tan x \quad (\text{odd})

Sum and difference

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B

Memory tip. Sine: same-different (sin-cos and cos-sin); signs match the ±\pm. Cosine: same-same and different-different (cos-cos minus sin-sin); signs flip.

Double-angle

A special case of the sum identities where A=BA = B:

sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2 \sin \theta \cos \theta cos⁡(2θ)=cos⁡2θ−sin⁡2θ=1−2sin⁡2θ=2cos⁡2θ−1\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = 1 - 2 \sin^2 \theta = 2\cos^2 \theta - 1

The three forms of cos⁡(2θ)\cos(2\theta) are all equivalent via the Pythagorean identity — use whichever is most convenient.

Worked simplification

Simplify sin⁡2x+sin⁡2xcot⁡2x\sin^2 x + \sin^2 x \cot^2 x.

Factor: sin⁡2x(1+cot⁡2x)\sin^2 x (1 + \cot^2 x). Use 1+cot⁡2x=csc⁡2x1 + \cot^2 x = \csc^2 x: sin⁡2x⋅csc⁡2x\sin^2 x \cdot \csc^2 x.

Since csc⁡2x=1sin⁡2x\csc^2 x = \dfrac{1}{\sin^2 x}:

sin⁡2x⋅1sin⁡2x=1\sin^2 x \cdot \dfrac{1}{\sin^2 x} = 1

Key takeaways

  • sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1 is the foundation; the other two Pythagoreans come from dividing it.
  • tan⁡=sin⁡/cos⁡\tan = \sin/\cos. Reciprocal pairs: sin⁡/csc⁡\sin/\csc, cos⁡/sec⁡\cos/\sec, tan⁡/cot⁡\tan/\cot.
  • sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2 \sin \theta \cos \theta; cos⁡(2θ)\cos(2\theta) has three useful forms.