Question

If A + B + C = 180° Prove that
$sin2a+sin2b+sin2c=2(1+cosacosbcosc)$
​

Answer 1

$\huge{\underline{\boxed{\red{\mathcal{QUESTION}}}}}$

If A + B + C = 180°. Prove that: sin²A + sin²B + sin²C = 2(1+cosA*cosB*cosC)

(the original question written was incorrect)

$\huge{\underline{\boxed{\red{\mathcal{ANSWER}}}}}$

Given:

• a + b + c = 180°

To Prove:

• $\sin^2A+\sin^2B+\sin^2C=2(1+\cos A*\cos B*\cos C)$

Proof:

$\text{Solving LHS,}\\ \text{We know that,}\\\longrightarrow \cos2\theta=1-2\sin^2\theta\\\\\implies2\sin^2\theta=1-\cos2\theta\\\\\implies\sin^2\theta=\dfrac{1-\cos2\theta}{2}$

$\text{So,}\\:\implies \sin^2A+\sin^2B+\sin^2C\\\\:\implies \dfrac{1-\cos2A}{2}+\dfrac{1-\cos2B}{2}+\dfrac{1-\cos2C}{2}\\\\:\implies \dfrac{1-\cos2A+1-\cos2B+1-\cos2C}{2}\\\\:\implies \dfrac{3-(\cos2A+\cos2B+\cos2C)}{2}\\\\\text{Also,}\\\\\longrightarrow \cos X+\cos Y = 2\cos\left(\frac{X+Y}{2}\right)\cos\left(\frac{X-Y}{2}\right)\\\\\text{So,}\\\\:\implies \dfrac{3-\left\{\left[2\cos\left(\frac{2A+2B}{2}\right)\cos\left(\frac{2A-2B}{2}\right)\right]+\cos2C\right\}}{2}$

$\text{Also,}\\\\\longrightarrow \cos2\theta=2\cos^2\theta-1\\\\:\implies \dfrac{3-\{2\cos(A+B)\cos(A-B)+2\cos^2C-1\}}{2}\\\\\text{We are given that, A+B+C=180 ^\circ .}\\\\\text{ \implies C = 180 - (A + B)}\\\\:\implies \dfrac{3-\{2\cos(A+B)\cos(A-B)+2\cos^2[180-(A+B)]-1\}}{2}\\\\:\implies \dfrac{4-\{2\cos(A+B)\cos(A-B)+2[-cos^2(A+B)]\}}{2} \\\\:\implies \dfrac{4-\{2\cos(A+B)\cos(A-B)+2cos^2(A+B)\}}{2} \\\\\text{Taking 2cos(A+B) as common}$

$:\implies \dfrac{4-\left\{\big(2\cos(A+B)\big)*\big(\cos(A-B)-cos(A+B)\big)\right\}}{2} \\\\\text{We saw above, that cosC = -cos(A+B). Hence,} \\\\:\implies \dfrac{4-\left\{\big(-2\cos C\big)*\big(\cos(A-B)+cos(A+B)\big)\right\}}{2} \\\\:\implies \dfrac{4+\left\{\big(2\cos C\big)*\left[2\cos\left(\frac{A-B+A+B}{2}\right)*\cos\left(\frac{A-B-A-B}{2}\right)\right]\right\}}{2} \\\\:\implies \dfrac{4+\left\{2\cos C*2\cos A*\cos B\right\}}{2} \\\\:\implies \dfrac{4+4\cos A*\cos B*\cos C}{2}$

$:\implies \dfrac{4\left(1+\cos A*\cos B*\cos C\right)}{2} \\\\:\implies 2(1+\cos A*\cos B*\cos C) = RHS \\\\\text{HENCE PROVED}$

Formulae Used:

• $\cos2\theta=1-2\sin^2\theta$

• $\cos X+\cos Y = 2\cos\left(\frac{X+Y}{2}\right)\cos\left(\frac{X-Y}{2}\right)$

• $\cos2\theta=2\cos^2\theta-1$