# Find $\sin(2\theta)$ from $\sin\theta + \cos\theta$ no calculator

If $\sin \theta + \cos \theta = 4/3$, what is the value of $\sin 2\theta$?

Solution: I am familiar with the trig identities, but I can't seem to apply it in this problem. Help would be appreciated, thank you.

• Hint: what do you get when you do $(\sin{\theta} + \cos{\theta})^2$ – Triatticus Mar 5 '18 at 2:19

Guide:

$$\sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta = \left(\frac43 \right)^2$$

Now use two trigonometry identities and the problem should be solved.

$$\sin \theta + \cos \theta = 4/3 \\ \implies (\sin \theta + \cos \theta)^2 = 4/3 \\ \implies \sin^2\theta +\cos^2\theta + 2\cos\theta\sin\theta=\frac{16}{9} \\ \implies 1+sin2\theta=\frac{16}{9} \\ \implies \sin 2\theta=\frac79$$

\begin{align*} sin\theta + cos\theta&=\frac{4}{3}\\ sin^{2}+cos^{2}\theta+2sin.cos\theta&=\frac{16}{9}\\ 1+2sin.cos\theta&=\frac{16}{9}\\ 1+2sin.cos\theta&=\frac{16}{9}\\2sin.cos\theta&=\frac{16}{9}-1\\2sin.cos\theta&=\frac{7}{9}\end{align*}

And now,use the identity of trigonometry $2sin.cos\theta=sin2\theta$. \begin{align*}2sin.cos\theta&=\frac{7}{9}\\sin2\theta&=\frac{7}{9}\end{align*}

• $\frac 4 3$ and not $\frac 3 4$. – Claude Leibovici Mar 5 '18 at 6:11