If A + B + C = 180° Prove that[tex]sin2a+sin2b+sin2c=2(1+cosacosbcosc)[/tex] About the author Claire
[tex]\huge{\underline{\boxed{\red{\mathcal{QUESTION}}}}}[/tex] If A + B + C = 180°. Prove that: sin²A + sin²B + sin²C = 2(1+cosA*cosB*cosC) (the original question written was incorrect) [tex]\huge{\underline{\boxed{\red{\mathcal{ANSWER}}}}}[/tex] Given: a + b + c = 180° To Prove: [tex]\sin^2A+\sin^2B+\sin^2C=2(1+\cos A*\cos B*\cos C)[/tex] Proof: [tex]\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}[/tex] [tex]\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}\\\\[/tex] [tex]\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}\\\\[/tex] [tex]:\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} \\\\[/tex] [tex]:\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}[/tex] Formulae Used: [tex]\cos2\theta=1-2\sin^2\theta[/tex] [tex]\cos X+\cos Y = 2\cos\left(\frac{X+Y}{2}\right)\cos\left(\frac{X-Y}{2}\right)[/tex] [tex]\cos2\theta=2\cos^2\theta-1[/tex] Reply
[tex]\huge{\underline{\boxed{\red{\mathcal{QUESTION}}}}}[/tex]
If A + B + C = 180°. Prove that: sin²A + sin²B + sin²C = 2(1+cosA*cosB*cosC)
(the original question written was incorrect)
[tex]\huge{\underline{\boxed{\red{\mathcal{ANSWER}}}}}[/tex]
Given:
To Prove:
Proof:
[tex]\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}[/tex]
[tex]\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}\\\\[/tex]
[tex]\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}\\\\[/tex]
[tex]:\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} \\\\[/tex]
[tex]:\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}[/tex]
Formulae Used: