Question

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

Prove that :
sin²6x-sin²4x = sin2xsin10x.​​

Answer 1

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

To prove :

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

Solving L.H.S

$\color{red} \bigstar \boxed{\sf (a^2-b^2)=(a+b)(a-b)}$

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

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

= R.H.S

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

$\color{red}━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━$

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

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

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

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

$\color{red}━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━$

Answer 2

Answer:

:

To prove :

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

2

6x−sin

2

4x=sin2xsin10x

Solving L.H.S

$\color{red} \bigstar \boxed{\sf (a^2-b^2)=(a+b)(a-b)}★$

$\begin{gathered}\\ \sf (sin6x-sin4x)(sin6x+sin4x) \end{gathered} </p><p>(sin6x−sin4x)(sin6x+sin4x)$

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

= R.H.S

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

Hence proved

\color{red}━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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

Formulasused

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

2

x+y

)Cos(

2

x−y

)

$\dag \sf \color{red}{\sf Sinx-Siny=2Cos(\frac{x+y}{2})Sin(\frac{x-y}{2})}†Sinx−Siny=2Cos( </p><p>2</p><p>x+y[)tex]</p><p> </p><p> )Sin( </p><p>2</p><p>x−y</p><p> </p><p> )</p><p></p><p>[tex]\sf \dag \color{red}{Sin2x = 2SinxCosx}†Sin2x=2SinxCosx$