Proof of Trigonometric Identity Using Sum-to-Product Formulas

Trigonometric Identity Proof Problem

  • Problem Statement:
    • Prove the trigonometric identity:

sin⁡(θ+15∘)+sin⁡(θ−15∘)cos⁡(θ+15∘)+cos⁡(θ−15∘)=tan⁡(θ)\frac{\sin(\theta + 15^\circ) + \sin(\theta - 15^\circ)}{\cos(\theta + 15^\circ) + \cos(\theta - 15^\circ)} = \tan(\theta)

Step-by-step trigonometric proof derivation

Fundamental Trigonometric Identities

  • Sum-to-Product Identity for Sine:

    • sin⁡(A)+sin⁡(B)=2sin⁡(A+B2)cos⁡(A−B2)\sin(A) + \sin(B) = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)

  • Sum-to-Product Identity for Cosine:

    • cos⁡(A)+cos⁡(B)=2cos⁡(A+B2)cos⁡(A−B2)\cos(A) + \cos(B) = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)

  • Tangent Quotient Identity:

    • tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

Detailed Step-by-Step Proof

  • Step 1: Write down the Left-Hand Side (LHS) of the trigonometric equation:

    • LHS=sin⁡(θ+15∘)+sin⁡(θ−15∘)cos⁡(θ+15∘)+cos⁡(θ−15∘)\text{LHS} = \frac{\sin(\theta + 15^\circ) + \sin(\theta - 15^\circ)}{\cos(\theta + 15^\circ) + \cos(\theta - 15^\circ)}

  • Step 2: Define angle variables AA and BB for applying the sum-to-product formulas:

    • Let A=θ+15∘A = \theta + 15^\circ
    • Let B=θ−15∘B = \theta - 15^\circ

  • Step 3: Apply the sum-to-product identities to the numerator and denominator:

    • LHS=2sin⁡(θ+15∘+θ−15∘2)cos⁡(θ+15∘−(θ−15∘)2)2cos⁡(θ+15∘+θ−15∘2)cos⁡(θ+15∘−(θ−15∘)2)\text{LHS} = \frac{2 \sin\left(\frac{\theta + 15^\circ + \theta - 15^\circ}{2}\right) \cos\left(\frac{\theta + 15^\circ - (\theta - 15^\circ)}{2}\right)}{2 \cos\left(\frac{\theta + 15^\circ + \theta - 15^\circ}{2}\right) \cos\left(\frac{\theta + 15^\circ - (\theta - 15^\circ)}{2}\right)}

  • Step 4: Expand the inner angle expressions:

    • Calculate angle sum A+B=θ+15∘+θ−15∘=2θA + B = \theta + 15^\circ + \theta - 15^\circ = 2\theta
    • Calculate angle difference A−B=θ+15∘−(θ−15∘)=θ+15∘−θ+15∘A - B = \theta + 15^\circ - (\theta - 15^\circ) = \theta + 15^\circ - \theta + 15^\circ
    • Substitute the expanded angle arguments back into the expression:
    • LHS=2sin⁡(2θ2)cos⁡(θ+15∘−θ+15∘2)2cos⁡(2θ2)cos⁡(θ+15∘−θ+15∘2)\text{LHS} = \frac{2 \sin\left(\frac{2\theta}{2}\right) \cos\left(\frac{\theta + 15^\circ - \theta + 15^\circ}{2}\right)}{2 \cos\left(\frac{2\theta}{2}\right) \cos\left(\frac{\theta + 15^\circ - \theta + 15^\circ}{2}\right)}

  • Step 5: Evaluate the simplified angle terms inside each trigonometric function:

    • First angle term: 2θ2=θ\frac{2\theta}{2} = \theta
    • Second angle term: θ+15∘−θ+15∘2=30∘2=15∘\frac{\theta + 15^\circ - \theta + 15^\circ}{2} = \frac{30^\circ}{2} = 15^\circ
    • Substitute evaluated angles into the fraction:
    • LHS=sin⁡(θ)cos⁡(15∘)cos⁡(θ)cos⁡(15∘)\text{LHS} = \frac{\sin(\theta) \cos(15^\circ)}{\cos(\theta) \cos(15^\circ)}

  • Step 6: Cancel common factors in numerator and denominator:

    • Since cos⁡(15∘)≠0\cos(15^\circ) \neq 0, cancel the non-zero factor cos⁡(15∘)\cos(15^\circ) as well as the constant factor 22:
    • LHS=sin⁡(θ)cos⁡(θ)\text{LHS} = \frac{\sin(\theta)}{\cos(\theta)}

  • Step 7: Apply the quotient identity to complete the proof:

    • LHS=tan⁡(θ)=RHS\text{LHS} = \tan(\theta) = \text{RHS}
    • The identity is successfully proven.