Question

$\huge \mathfrak \color{green}{question} :$

Prove that :
sin²6x-sin²4x = sin2xsin10x.​

Answer 1

Answer:

={2cos(6x+4x)/2.sin(6x-4x)/2}{2sin(6x+4x)/2.cos(6x-4x)/2}

={2cos5x.sinx}{2sin5x.cosx}

={2sin5x.cos5x}{2sinx.cosx}

Use the formula,

sin2A = 2sinA.cosA

= sin2(5x).sin2(x)

= sin10x.sin2x

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Step-by-step explanation:

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Answer 2

$\huge \sf \color{red} \underline{Solution} :$

To prove :

$\bigstar \sf sin^26x-sin^24x = sin2xsin10x$

Solving L.H.S

$\colorbox{skyblue}{\sf (a^2-b^2)=(a+b)(a-b)}$

$\\ \sf (sin6x-sin4x)(sin6x+sin4x)$

$\\ \sf [2cos(\frac{6x+4x}{2})sin(\frac{6x-4x}{2}) ] [2sin(\frac{6x+4x}{2})Cos(\frac{6x-4x}{2})] \\ \sf \implies [2cos(5x)sin(x)][2sin5xCosx] \\ \sf \implies [2sin5xCos5x][2sinxCosx] \\ \sf \implies Sin10xSin2x$

= R.H.S

$\huge \underline{ Hence \ \ \ proved}$

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$\large \sf \color{brown}{\underbrace{Formulas \: used}}$

$\sf \dag \color{red} {\sf Sinx+Siny = 2sin(\frac{x+y}{2})Cos(\frac{x-y}{2})}$

$\dag \sf \color{red}{\sf Sinx-Siny=2Cos(\frac{x+y}{2})Sin(\frac{x-y}{2})}$

$\sf \dag \color{red}{Sin2x = 2SinxCosx}$

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