Question

pls solve this trigonometry question

Prove the following

$( cosx + cosy) {}^{2} + ( sinx - siny) {}^{2} = 4 \cos {}^{2} \frac{x + y}{2}$
Please answer

Answer 1

Hope this helps you ☺☺

Answer 2

$\large{\mathbf{HELLO\:\: DEAR,}}$

$\mathit{WE \:\:KNOW \:\:THAT:-} \\ \\ \mathit{sinA - sinB = 2cos\frac{A + B}{2}sin\frac{A - B}{2}}$

$\mathit{cosA + cosB = 2cos\frac{A + B}{2}cos\frac{A - B}{2}}\\ \\ \mathit{PROVE\:\: THAT:-} \\ \\ \mathit{(cosx + cosy)^2 + (sinx - siny)^2 = 4cos^2\frac{x + y}{2}}\\ \\ \mathit{NOW,} \\ \\ \mathit{[2cos(\frac{x + y}{2})*cos(\frac{x - y}{2})]^2 + [2cos(\frac{x + y}{2})*sin(\frac{x - y}{2})]^2}$

$\mathit{[4cos^2(\frac{x + y}{2})* cos^2(\frac{x - y}{2}) + 4cos^2(\frac{x + y}{2})*sin^2(\frac{x - y}{2})]}$

$\mathit{[4cos^2(\frac{x + y}{2})]*[cos^2(\frac{x - y}{2}) + sin^2(\frac{x - y}{2})]}$

$\mathit{4cos^2(\frac{x + y}{2}) * ( 1 ) } \\ \mathit{\to\to\to\to\to\to\to\therefore\boxed{sin^2\theta + cos^2\theta = 1}}$

$\large{\mathbf{\underline{I\:\: HOPE\:\: ITS \:\:HELP\:\: YOU\:\: DEAR,\:\: THANKS}}}$