With usual notations prove that $(a-b)^2\cos^2\dfrac{C}{2} + (a+b)^2\sin^2\dfrac{C}{2} = c^2$.

With usual notations prove that (a-b)²cos²(C)/(2) + (a+b)²sin²(C)/(2) = c².

Miscellaneous Exercise 3 · Q98

(a-b)^2\cos^2\dfrac{C}{2} + (a+b)^2\sin^2\dfrac{C}{2} = c^2

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Using $\cos^2\dfrac C2=\dfrac{1+\cos C}{2}$ and $\sin^2\dfrac C2=\dfrac{1-\cos C}{2}$ : L.H.S. $=(a-b)^2\cdot\dfrac{1+\cos C}{2}+(a+b)^2\cdot\dfrac{1-\cos C}{2}=\dfrac12\left[(a-b)^2+(a+b)^2\right]+\dfrac{\cos C}{2}\left[(a-b)^2-(a+b)^2\right]$ . Now $(a-b)^2+(a+b)^2=2a^2+2b^2$ and $(a-b)^2-(a+b)^2=-4ab$ . So L.H.S …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL2 marks
$\triangle ABC$
›Reveal solutionSolution

Substitute $\cos C$ and $\cos B$ from the cosine rule and simplify.

By the cosine rule:

$\cos C=\frac{a^2+b^2-c^2}{2ab}, \qquad \cos B=\frac{a^2+c^2-b^2}{2ac}$

LHS $=a(b\cos C-c\cos B) = ab\cos C - ac\cos B$

$ab\cos C = ab\cdot\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2-c^2}{2}$

$ac\cos B = ac\cdot\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+c^2-b^2}{2}$

…

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CBSE 2023Set ANNUAL2 marksMCQ
$\triangle ABC$
(a) $\dfrac{\pi}{4}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{2}$
(d) $\dfrac{\pi}{6}$
›Reveal solutionSolution

Compare with the cosine rule $b^2=a^2+c^2-2ac\cos B$ .

By the cosine rule, $b^2=a^2+c^2-2ac\cos B \Rightarrow a^2+c^2-b^2=2ac\cos B$ .

…

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CBSE 2022Set ANNUAL2 marksMCQ
$\triangle ABC$
(a) $\dfrac{\pi}{4}$
(b) $\dfrac{\pi}{3}$
(c) $\dfrac{\pi}{2}$
(d) $\dfrac{\pi}{6}$
›Reveal solutionSolution

Compare with the Cosine Rule formula for $\cos B$ .

By the Cosine Rule: $\cos B = \dfrac{a^2+c^2-b^2}{2ac}$

Given $c^2+a^2-b^2=ac$ : …

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CBSE 2017Set ANNUAL2 marks
$\triangle ABC$
›Reveal solutionSolution

Substitute $\cos B$ and $\cos C$ from the cosine rule and simplify.

By the cosine rule:

$\cos C = \dfrac{a^2+b^2-c^2}{2ab}, \qquad \cos B = \dfrac{a^2+c^2-b^2}{2ac}$

$a(b\cos C - c\cos B) = a\left[b\cdot\dfrac{a^2+b^2-c^2}{2ab} - c\cdot\dfrac{a^2+c^2-b^2}{2ac}\right]$

$= a\left[\dfrac{a^2+b^2-c^2}{2a} - \dfrac{a^2+c^2-b^2}{2a}\right] = \dfrac{1}{2}\Big[(a^2+b^2-c^2)-(a^2+c^2-b^2)\Big]$

…

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Q.With usual notations prove that $(a-b)^2\cos^2\dfrac{C}{2} + (a+b)^2\sin^2\dfrac{C}{2} = c^2$ .

Concept understanding — Cosine Rule

Cosine Rule

Where it comes from

How it is used

Q.With usual notations prove that (a−b)2cos⁡2C2+(a+b)2sin⁡2C2=c2(a-b)^2\cos^2\dfrac{C}{2} + (a+b)^2\sin^2\dfrac{C}{2} = c^2(a−b)2cos22C​+(a+b)2sin22C​=c2.

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Q.With usual notations prove that $(a-b)^2\cos^2\dfrac{C}{2} + (a+b)^2\sin^2\dfrac{C}{2} = c^2$ .

Unlock everything free for 14 days