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Mathematics

For triangle ABC, show that :

(i) sinA+B2=cosC2\text{sin}\dfrac{A+B}{2} = \text{cos}\dfrac{C}{2}

(ii) tanB+C2=cotA2\text{tan}\dfrac{B+C}{2} = \text{cot}\dfrac{A}{2}

Trigonometric Identities

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Answer

(i) sinA+B2=cosC2\text{sin}\dfrac{A+B}{2} = \text{cos}\dfrac{C}{2}

According to angle sum property,

∠A+∠B+∠C=180°∠A+∠B=180°−∠C⇒A+B2=180°−C2∠ A + ∠ B + ∠ C = 180°\\[1em] ∠ A + ∠ B = 180° - ∠ C\\[1em] ⇒ \dfrac{A + B}{2} = \dfrac{180° - C}{2}\\[1em]

L.H.S.=sinA+B2=sin180°−C2=sin(90°−C2)=cosC2\text{L.H.S.} = \text{sin}\dfrac{A+B}{2}\\[1em] = \text{sin}\dfrac{180° - C}{2}\\[1em] = \text{sin}\Big(90° - \dfrac{C}{2}\Big)\\[1em] = \text{cos}\dfrac{C}{2}

R.H.S. = cosC2\text{cos}\dfrac{C}{2}

∴ L.H.S. = R.H.S.

Hence, sinA+B2=cosC2\text{sin}\dfrac{A+B}{2} = \text{cos}\dfrac{C}{2}.

(ii) tanB+C2=cotA2\text{tan}\dfrac{B+C}{2} = \text{cot}\dfrac{A}{2}

According to angle sum property,

∠A+∠B+∠C=180°∠B+∠C=180°−∠A⇒B+C2=180°−A2∠ A + ∠ B + ∠ C = 180°\\[1em] ∠ B + ∠ C = 180° - ∠ A\\[1em] ⇒ \dfrac{B + C}{2} = \dfrac{180° - A}{2}\\[1em]

L.H.S.=tanB+C2=tan(180°−A2)=tan(90°−A2)=cotA2\text{L.H.S.} = \text{tan}\dfrac{B+C}{2}\\[1em] = \text{tan}\Big(\dfrac{180° - A}{2}\Big)\\[1em] = \text{tan}\Big(90° - \dfrac{A}{2}\Big)\\[1em] = \text{cot}\dfrac{A}{2}

R.H.S. = cotA2\text{cot}\dfrac{A}{2}

∴ L.H.S. = R.H.S.

Hence, tanB+C2=cotA2\text{tan}\dfrac{B+C}{2} = \text{cot}\dfrac{A}{2}.

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