Trig question(s)

[abhi223]

Hey guys, I am struggling with the question number 2.b)i) which is given that 0<x<(pi/4) and sin(x) + cos(x) = (4/3), 'use algebra' to show that
 cos(2x)=(4(2)^(1/2))/9. Thanks for helping!

[mzhao]

Hey guys, I am struggling with the question number 2.b)i) which is given that 0<x<(pi/4) and sin(x) + cos(x) = (4/3), 'use algebra' to show that
 cos(2x)=(4(2)^(1/2))/9. Thanks for helping!

Hey abhi223,

\begin{align*}
\text{Square both sides to get:}\\
(\sin(x) + \cos(x))^2 &= \frac{16}{9}\\
\implies \sin^2(x) + \cos^2(x) + 2\sin(x)\cos(x) &= \frac{16}{9}\\
\implies 1 + 2\sin(x)\cos(x) &= \frac{16}{9} \text{ (by Pythagorean identity)}\\
\implies 2\sin(x)\cos(x) &= \frac{7}{9}\\
\implies \sin(2x) &= \frac{7}{9}\\
\end{align*}
\begin{align*}
\text{Using the Pythagorean identity again:}\\
\implies \cos(2x) &= +\sqrt{1-\sin^2(2x)} \text{ (positive because } 2x \text{ is in 1st quadrant)}\\
&= \sqrt{1-\frac{49}{81}}\\
&= \sqrt{\frac{32}{81}}\\
\therefore \cos(2x) &= \frac{4\sqrt{2}}{9} \text{, as required}\\
\end{align*}

[abhi223]

Thank you!