MathIsimple
Lesson 4-1

Trigonometric Identities – Derivation & Applications

Build proof fluency and expression mastery using Pythagorean, reciprocal, quotient, sum/difference, and double-angle identities with practical applications.

Learning Objectives

  • Prove core trigonometric identities from first principles and geometric reasoning.
  • Transform expressions to equivalent forms using identity toolkits.
  • Apply sum/difference and double-angle formulas to simplify and evaluate.
  • Diagnose common pitfalls and design consistent proof strategies.
  • Model physical and signal phenomena using trigonometric identities.

Identity Toolkit

Pythagorean

sin2θ+cos2θ=1 \sin^2\theta+\cos^2\theta=1

Reciprocal

secθ=1cosθ,  cscθ=1sinθ,  cotθ=1tanθ \sec\theta=\tfrac{1}{\cos\theta},\; \csc\theta=\tfrac{1}{\sin\theta},\; \cot\theta=\tfrac{1}{\tan\theta}

Quotient

tanθ=sinθcosθ,  cotθ=cosθsinθ \tan\theta=\tfrac{\sin\theta}{\cos\theta},\; \cot\theta=\tfrac{\cos\theta}{\sin\theta}

Co-Function

sin(π2x)=cosx,  tan(π2x)=cotx \sin(\tfrac{\pi}{2}-x)=\cos x,\; \tan(\tfrac{\pi}{2}-x)=\cot x

Sum/Diff (sin)

sin(A±B)=sinAcosB±cosAsinB \sin(A\pm B)=\sin A\cos B \pm \cos A\sin B

Sum/Diff (cos)

cos(A±B)=cosAcosB\-sinAsinB \cos(A\pm B)=\cos A\cos B \- \sin A\sin B

Sum/Diff (tan)

tan(A±B)=tanA±tanB1\-tanAtanB \tan(A\pm B)=\tfrac{\tan A \pm \tan B}{1 \- \tan A\tan B}

Double-Angle

sin2x=2sinxcosx,  cos2x=cos2xsin2x=12sin2x=2cos2x1 \sin 2x=2\sin x\cos x,\; \cos 2x=\cos^2 x-\sin^2 x=1-2\sin^2 x=2\cos^2 x-1

Key Derivations

Pythagorean Identity

On the unit circle, a point has coordinates (cosθ,sinθ)(\cos\theta,\sin\theta) so the radius condition yields

cos2θ+sin2θ=1\cos^2\theta+\sin^2\theta=1

Sum/Difference (via rotation)

Rotations add: using matrix R(α)R(\alpha) and R(β)R(\beta), we get

sin(A+B)=sinAcosB+cosAsinB\sin(A+B)=\sin A\cos B+\cos A\sin B
cos(A+B)=cosAcosBsinAsinB\cos(A+B)=\cos A\cos B-\sin A\sin B

Double-Angle

Set A=B=xA=B=x in sum identities:

sin2x=2sinxcosx\sin 2x=2\sin x\cos x
cos2x=cos2xsin2x=12sin2x=2cos2x1\cos 2x=\cos^2 x-\sin^2 x=1-2\sin^2 x=2\cos^2 x-1

Product-to-Sum (outline)

From sum/difference, derive cosAcosB=12[cos(A+B)+cos(AB)]\cos A\cos B=\tfrac{1}{2}[\cos(A+B)+\cos(A-B)] etc.

cosAcosB=12[cos(A+B)+cos(AB)]\cos A\cos B=\tfrac{1}{2}[\cos(A+B)+\cos(A-B)]
sinAsinB=12[cos(AB)cos(A+B)]\sin A\sin B=\tfrac{1}{2}[\cos(A-B)-\cos(A+B)]

Worked Examples

Example 1: Simplify using identities

E=sin2x1+cos2xcotxE=\tfrac{\sin 2x}{1+\cos 2x}\cdot \cot x.

Show Solution

Use sin2x=2sinxcosx\sin 2x=2\sin x\cos x, 1+cos2x=2cos2x1+\cos 2x=2\cos^2 x:

E=2sinxcosx2cos2xcosxsinx=1E=\tfrac{2\sin x\cos x}{2\cos^2 x}\cdot \tfrac{\cos x}{\sin x}=1

Example 2: Prove identity

Prove sin4x+cos4x=12sin2xcos2x\sin^4 x+\cos^4 x=1-2\sin^2 x\cos^2 x.

Show Proof

(sin2x+cos2x)2=sin4x+2sin2xcos2x+cos4x=1(\sin^2 x+\cos^2 x)^2=\sin^4 x+2\sin^2 x\cos^2 x+\cos^4 x=1

Rearrange: sin4x+cos4x=12sin2xcos2x\sin^4 x+\cos^4 x=1-2\sin^2 x\cos^2 x.

Example 3: Evaluate expression

Compute cos75\cos 75^\circ using sum/difference.

Show Solution

75=45+3075^\circ=45^\circ+30^\circ:

cos75=cos45cos30sin45sin30=22322212=624\cos 75^\circ=\cos 45^\circ\cos 30^\circ-\sin 45^\circ\sin 30^\circ=\tfrac{\sqrt{2}}{2}\cdot\tfrac{\sqrt{3}}{2}-\tfrac{\sqrt{2}}{2}\cdot\tfrac{1}{2}=\tfrac{\sqrt{6}-\sqrt{2}}{4}

Example 4: Solve for angle

Given sin2x=32\sin 2x=\tfrac{\sqrt{3}}{2}, find all solutions in [0,2π)[0,2\pi).

Show Solution

2x=π3+2kπ,  2x=2π3+2kπ2x=\tfrac{\pi}{3}+2k\pi,\; 2x=\tfrac{2\pi}{3}+2k\pix=π6+kπ,  x=π3+kπx=\tfrac{\pi}{6}+k\pi,\; x=\tfrac{\pi}{3}+k\pi.

Strategy & Common Pitfalls

Strategy

  • Convert everything to sinx,cosx\sin x, \cos x when stuck.
  • Use conjugates to handle expressions like 1±sinx1\pm \sin x or 1±cosx1\pm \cos x.
  • Prefer identities that reduce powers: e.g., use cos2x\cos 2x to eliminate sin2x\sin^2 x or cos2x\cos^2 x.
  • Track domains to avoid dividing by zero.

Pitfalls

  • Misusing sin(A+B)\sin(A+B) as sinA+sinB\sin A+\sin B (incorrect).
  • Dropping absolute values when inverting functions or taking square roots.
  • Ignoring angle units (degrees vs. radians).
  • Forgetting extraneous solutions when multiplying by factors like cosx\cos x.

Practice Problems

Problem 1

Prove sin^2 x = (1 - cos 2x)/2 and cos^2 x = (1 + cos 2x)/2.

Show Solution

Use cos 2x = cos^2 x - sin^2 x and cos^2 x + sin^2 x = 1; solve the 2×2 system.

Problem 2

Simplify (1 - cos x)/(sin x) - (sin x)/(1 + cos x).

Show Solution

Use conjugate on the first term or common denominator: result is 0.

Problem 3

Show tan(A) + tan(B) = (sin(A + B))/(cos A cos B) when A + B ≠ (π/2) + kπ.

Show Solution

Use tan = sin/cos and the sum formula for sin(A+B).

Problem 4

Prove 1 + tan^2 x = sec^2 x.

Show Solution

Divide sin^2 + cos^2 = 1 by cos^2.

Problem 5

Given sin α = 3/5, cos β = 4/5 with α, β in first quadrant, compute cos(α + β).

Show Solution

Find cos α = 4/5, sin β = 3/5, then apply cos sum formula.

Problem 6

Derive sin 3x and cos 3x using angle addition repeatedly.

Show Solution

Use sin(2x + x), cos(2x + x); express via sin x, cos x.

Extended Identities Library

Identity Case 1: Conjugate Simplification

Multiply by the conjugate to remove 1−cos x in numerator.

1cosxsinx\frac{1-\cos x}{\sin x}
=(1cosx)(1+cosx)sinx(1+cosx)=\frac{(1-\cos x)(1+\cos x)}{\sin x(1+\cos x)}
=sin2xsinx(1+cosx)=\frac{\sin^2 x}{\sin x(1+\cos x)}
=sinx1+cosx=\frac{\sin x}{1+\cos x}

Identity Case 2: Quotient to Tangent

Use conjugate on denominator and Pythagorean identity.

sinxcosx+1\frac{\sin x}{\cos x+1}
=sinx(cosx1)(cosx+1)(cosx1)=\frac{\sin x(\cos x-1)}{(\cos x+1)(\cos x-1)}
=sinx(1cosx)1cos2x=\frac{-\sin x(1-\cos x)}{1-\cos^2 x}
=1cosxsinx=-\frac{1-\cos x}{\sin x}

Identity Case 3: Product-to-Sum

Apply product-to-sum identity.

2sinAcosB2\sin A\cos B
=sin(A+B)+sin(AB)=\sin(A+B)+\sin(A-B)

Identity Case 4: Sum-to-Product

Use sum-to-product to factor cosine difference.

cosAcosB\cos A-\cos B
=2sin(A+B2)sin(AB2)=-2\sin\left(\tfrac{A+B}{2}\right)\sin\left(\tfrac{A-B}{2}\right)

Identity Case 5: Half-Angle (Sine)

Half-angle reduces powers.

sin2x\sin^2 x
=1cos2x2=\tfrac{1-\cos 2x}{2}

Identity Case 6: Half-Angle (Cosine)

Use power-reduction for integrals/summations.

cos2x\cos^2 x
=1+cos2x2=\tfrac{1+\cos 2x}{2}

Identity Case 7: Double-Angle to Tangent

Divide by cos^2 x and use tan = sin/cos.

cos2x\cos 2x
=1tan2x1+tan2x=\frac{1-\tan^2 x}{1+\tan^2 x}

Identity Case 8: Reciprocal Pair

Divide sin^2+cos^2=1 by cos^2.

1+tan2x1+\tan^2 x
=sec2x=\sec^2 x

Identity Case 9: Canonical Form via Sum

Apply sum-to-product to numerator and denominator.

sinx+sinycosx+cosy\frac{\sin x+\sin y}{\cos x+\cos y}
=2sinx+y2cosxy22cosx+y2cosxy2=\frac{2\sin\tfrac{x+y}{2}\cos\tfrac{x-y}{2}}{2\cos\tfrac{x+y}{2}\cos\tfrac{x-y}{2}}
=tanx+y2=\tan\tfrac{x+y}{2}

Identity Case 10: Rationalizing with t-substitution

Universal t-substitution for rational trig forms.

tanx2=t\tan\frac{x}{2}=t
sinx=2t1+t2,  cosx=1t21+t2\Rightarrow \sin x=\frac{2t}{1+t^2},\; \cos x=\frac{1-t^2}{1+t^2}

Identity Case 11: Angle Addition (Cosine)

From rotation matrices or unit-circle geometry.

cos(α+β)\cos(\alpha+\beta)
=cosαcosβsinαsinβ=\cos\alpha\cos\beta-\sin\alpha\sin\beta

Identity Case 12: Auxiliary-Angle Method

Rewrite linear combo as phase-shifted sinusoid.

asinx+bcosxa\sin x+b\cos x
=Rsin(x+ϕ),  R=a2+b2=R\sin(x+\phi),\; R=\sqrt{a^2+b^2}

Applications & Modeling

Signal Phasors

Sum of sinusoids with same frequency can be expressed as a single sinusoid using sum identities.

asin(ωt)+bcos(ωt)=Rsin(ωt+ϕ),  R=a2+b2a\sin(\omega t)+b\cos(\omega t)=R\sin(\omega t+\phi),\; R=\sqrt{a^2+b^2}

Useful in AC circuits and oscillations.

Harmonic Identities

Double-angle helps reduce integrands or derive envelope relationships in physics.

sin2x=1cos2x2,  cos2x=1+cos2x2\sin^2 x=\tfrac{1-\cos 2x}{2},\; \cos^2 x=\tfrac{1+\cos 2x}{2}