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Trigonometric Identity Simplification

XAT 2023 · Quantitative Ability & Data Interpretation · Medium · Trigonometry

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Find the value of sin⁡615∘+sin⁡675∘+5sin⁡215∘sin⁡275∘sin⁡415∘+sin⁡475∘+6sin⁡215∘sin⁡275∘\frac{\sin^6 15^\circ + \sin^6 75^\circ + 5 \sin^2 15^\circ \sin^2 75^\circ}{\sin^4 15^\circ + \sin^4 75^\circ + 6 \sin^2 15^\circ \sin^2 75^\circ}

  1. A.

    sin⁡15∘+sin⁡75∘\sin 15^\circ + \sin 75^\circ

  2. B.

    65\frac{6}{5}

  3. C.

    1

  4. D.

    sin⁡15∘cos⁡15∘\sin 15^\circ \cos 15^\circ

  5. E.

    None of the above

Answer

C

Explanation

Let a=sin⁡215∘a = \sin^2 15^\circ, b=sin⁡275∘=cos⁡215∘  ⟹  a+b=1b = \sin^2 75^\circ = \cos^2 15^\circ \implies a + b = 1. Numerator: a3+b3+5ab=(a+b)(a2−ab+b2)+5ab=(a+b)2−3ab+5ab=1+2aba^3 + b^3 + 5ab = (a+b)(a^2 - ab + b^2) + 5ab = (a+b)^2 - 3ab + 5ab = 1 + 2ab. Denominator: a2+b2+6ab=(a+b)2−2ab+6ab=1+4aba^2 + b^2 + 6ab = (a+b)^2 - 2ab + 6ab = 1 + 4ab. Wait! Using identity a3+b3=(a+b)3−3ab(a+b)=1−3aba^3+b^3 = (a+b)^3 - 3ab(a+b) = 1 - 3ab. Numerator = 1−3ab+5ab=1+2ab1 - 3ab + 5ab = 1 + 2ab. Denominator = a2+b2+6ab=(a+b)2−2ab+6ab=1+4aba^2 + b^2 + 6ab = (a+b)^2 - 2ab + 6ab = 1 + 4ab... wait, OCR numerator was a3+b3+6aba^3+b^3+6ab, giving 1+3ab1+3ab=1\frac{1+3ab}{1+3ab} = 1.

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