Fundamental Trigonometric Identities

Trigonometric identities are equations that are true for all valid values of the variables. They are the fundamental tools used to simplify and solve trigonometric problems.

Pythagorean Identities: These are derived from the Pythagorean theorem on the unit circle (x2+y2=1x^2+y^2=1, which translates to cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1).

  • 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

Allied Angles

Allied angles are pairs of angles whose sum or difference is a multiple of 90∘90^\circ (π/2\pi/2). The allied angle rules are a powerful tool to express the trigonometric function of any angle in terms of an acute angle (an angle between 0∘0^\circ and 90∘90^\circ).

Rules for Allied Angles:

To find the value of a trigonometric function for an allied angle (e.g., sin⁡(180∘+θ)\sin(180^\circ+\theta)), follow these two steps:

  1. Determine the Sign: Identify the quadrant in which the allied angle lies (assuming θ\theta is acute). Use the ASTC rule (All Students Take Calculus) to determine if the original function is positive or negative in that quadrant. The result will have this sign.

  2. Determine the Function:

    • If the angle is a multiple of 180∘±θ180^\circ \pm \theta (or π±θ\pi \pm \theta), the function remains the same (e.g., sin stays sin).

    • If the angle is an odd multiple of 90∘±θ90^\circ \pm \theta (or π2±θ,3π2±θ\frac{\pi}{2} \pm \theta, \frac{3\pi}{2} \pm \theta), the function changes to its co-function:

      • sin⁡↔cos⁡\sin \leftrightarrow \cos

      • tan⁡↔cot⁡\tan \leftrightarrow \cot

      • sec⁡↔csc⁡\sec \leftrightarrow \csc

Common Allied Angle Formulas:

  • sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ - \theta) = \cos\theta (Q1: sin is +, 90° changes sin to cos)

  • cos⁡(90∘+θ)=−sin⁡θ\cos(90^\circ + \theta) = -\sin\theta (Q2: cos is -, 90° changes cos to sin)

  • tan⁡(180∘−θ)=−tan⁡θ\tan(180^\circ - \theta) = -\tan\theta (Q2: tan is -, 180° does not change tan)

  • sin⁡(180∘+θ)=−sin⁡θ\sin(180^\circ + \theta) = -\sin\theta (Q3: sin is -, 180° does not change sin)

  • cos⁡(270∘−θ)=−sin⁡θ\cos(270^\circ - \theta) = -\sin\theta (Q3: cos is -, 270° changes cos to sin)

  • sec⁡(360∘−θ)=sec⁡θ\sec(360^\circ - \theta) = \sec\theta (Q4: sec is +, 360° does not change sec)

Example 1: Proving a Trigonometric Identity

Question: Prove that (sin⁡θ+csc⁡θ)2+(cos⁡θ+sec⁡θ)2=7+tan⁡2θ+cot⁡2θ(\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 = 7 + \tan^2\theta + \cot^2\theta.

Solution:

We start by expanding the Left Hand Side (LHS):

LHS = (sin⁡2θ+csc⁡2θ+2sin⁡θcsc⁡θ)+(cos⁡2θ+sec⁡2θ+2cos⁡θsec⁡θ)(\sin^2\theta + \csc^2\theta + 2\sin\theta\csc\theta) + (\cos^2\theta + \sec^2\theta + 2\cos\theta\sec\theta)

Since csc⁡θ=1/sin⁡θ\csc\theta = 1/\sin\theta and sec⁡θ=1/cos⁡θ\sec\theta = 1/\cos\theta, the products simplify:

LHS = (sin⁡2θ+csc⁡2θ+2)+(cos⁡2θ+sec⁡2θ+2)(\sin^2\theta + \csc^2\theta + 2) + (\cos^2\theta + \sec^2\theta + 2)

Group the terms: (sin⁡2θ+cos⁡2θ)+csc⁡2θ+sec⁡2θ+4(\sin^2\theta+\cos^2\theta) + \csc^2\theta + \sec^2\theta + 4

Now, use the Pythagorean identities sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, csc⁡2θ=1+cot⁡2θ\csc^2\theta=1+\cot^2\theta, and sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta:

LHS = 1+(1+cot⁡2θ)+(1+tan⁡2θ)+4=7+tan⁡2θ+cot⁡2θ1 + (1+\cot^2\theta) + (1+\tan^2\theta) + 4 = \mathbf{7 + \tan^2\theta + \cot^2\theta} = RHS.

Example 2: Simplifying with Allied Angles

Question: Find the value of sin⁡(420∘)cos⁡(390∘)+cos⁡(−300∘)sin⁡(−330∘)\sin(420^\circ)\cos(390^\circ) + \cos(-300^\circ)\sin(-330^\circ).

Solution:

We simplify each term by finding its coterminal acute angle:

  • sin⁡(420∘)=sin⁡(360∘+60∘)=sin⁡(60∘)=3/2\sin(420^\circ) = \sin(360^\circ+60^\circ) = \sin(60^\circ) = \mathbf{\sqrt{3}/2}.

  • cos⁡(390∘)=cos⁡(360∘+30∘)=cos⁡(30∘)=3/2\cos(390^\circ) = \cos(360^\circ+30^\circ) = \cos(30^\circ) = \mathbf{\sqrt{3}/2}.

  • cos⁡(−300∘)=cos⁡(300∘)=cos⁡(360∘−60∘)=cos⁡(60∘)=1/2\cos(-300^\circ) = \cos(300^\circ) = \cos(360^\circ-60^\circ) = \cos(60^\circ) = \mathbf{1/2}.

  • sin⁡(−330∘)=−sin⁡(330∘)=−sin⁡(360∘−30∘)=−(−sin⁡(30∘))=sin⁡(30∘)=1/2\sin(-330^\circ) = -\sin(330^\circ) = -\sin(360^\circ-30^\circ) = -(-\sin(30^\circ)) = \sin(30^\circ)=\mathbf{1/2}.

The expression becomes (32)(32)+(12)(12)=34+14=1(\frac{\sqrt{3}}{2})(\frac{\sqrt{3}}{2}) + (\frac{1}{2})(\frac{1}{2}) = \frac{3}{4} + \frac{1}{4} = \mathbf{1}.