1
$\begingroup$

To prove: $$1 + 2 \sin 70^\circ = \frac{1}{2\sin 20^\circ}$$

My attempt: $$\begin{align} 1 + 2 \sin 70^\circ &= 1 + \frac{\sin 140^\circ}{\cos 70^\circ} \\[6pt] &= 1 + \frac{\sin 40^\circ}{\sin 20^\circ} \\[6pt] &= \frac{\sin 20^\circ + \sin 40^\circ}{\sin 20^\circ} \\[6pt] &= \frac{2\sin 30^\circ \cos 10^\circ}{\sin 20^\circ} \\[6pt] &= \frac{\cos 10^\circ}{2\sin 10^\circ \cos 10^\circ} \\[6pt] &= \frac{1}{2\sin 10^\circ} \\ \end{align}$$

Can anyone explain where my mistake is?

(original solution image)

$\endgroup$
6
  • $\begingroup$ Where's your proof? $\endgroup$
    – mrnld
    Jan 8 '16 at 22:52
  • 1
    $\begingroup$ Perhaps the mistake isn't yours:$$1+2\sin 70^\circ = 2.879\dots \qquad \frac{1}{2\sin 20^\circ} =1.461\dots \qquad \frac{1}{2\sin 10^\circ} = 2.879\dots$$ $\endgroup$
    – Blue
    Jan 8 '16 at 22:57
  • $\begingroup$ I TeX-ified your solution, fixing a typo (missing "sin"). Please double-check my work. $\endgroup$
    – Blue
    Jan 8 '16 at 23:08
  • $\begingroup$ You have forget the factor $2\sin 30\circ$ in the 5th line. You must end with $\frac{\sin 30\circ}{\sin 10\circ}$ instead of $\frac {1}{2\sin 10\circ}$ $\endgroup$
    – Piquito
    Jan 8 '16 at 23:34
  • $\begingroup$ @Piquito: $\sin 30^\circ = \frac{1}{2}$. :) $\endgroup$
    – Blue
    Jan 9 '16 at 2:56
0
$\begingroup$

As noted in the comments:

  • The original identity is wrong (LHS $\approx 2.88$, RHS $\approx 1.46$).
  • Your derivation of $1 + 2 \sin 70^\circ = \dfrac{1}{2\sin 10^\circ}$is correct.
$\endgroup$

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.