0
$\begingroup$

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.

$\endgroup$
  • 6
    $\begingroup$ Hint: what do you get when you do $(\sin{\theta} + \cos{\theta})^2$ $\endgroup$ – Triatticus Mar 5 '18 at 2:19
5
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.