Question

How do you do this problem?

Answer 1

so we know that the angle "x" is in the 1st Quadrant, where cosine and sine are both positive, hmmm let's proceed.

`\sin(x )=\cfrac{\stackrel{opposite}{7}}{\underset{hypotenuse}{25}}\hspace{5em}\textit{let's find the \underline{adjacent side}} \\\\\\ \begin{array}{llll} \textit{using the pythagorean theorem} \\\\ a^2+o^2=c^2\implies a=√(c^2 - o^2) \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{25}\\ a=adjacent\\ o=\stackrel{opposite}{7} \end{cases} \\\\\\ a=\pm√( 25^2 - 7^2) \implies a=\pm√( 576 )\implies a=\pm 24\implies \stackrel{I~Quadrant }{a=+24} \\\\[-0.35em] ~\dotfill`

`\sin(2x)\implies 2\sin(x)\cos(x)\implies 2\left(\cfrac{7}{25} \right)\left( \cfrac{24}{25} \right)\implies \cfrac{336}{625} \\\\[-0.35em] ~\dotfill\\\\ \cos(2x)\implies 1-2\sin^2(x)\implies 1-2\left( \cfrac{7}{25} \right)^2\implies 1-\cfrac{98}{625}\implies \cfrac{527}{625}`