# Where is the mistake in my solution? Trigonometry proof

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)

• Where's your proof? Jan 8 '16 at 22:52
• Perhaps the mistake isn't yours:$$1+2\sin 70^\circ = 2.879\dots \qquad \frac{1}{2\sin 20^\circ} =1.461\dots \qquad \frac{1}{2\sin 10^\circ} = 2.879\dots$$
– Blue
Jan 8 '16 at 22:57
• I TeX-ified your solution, fixing a typo (missing "sin"). Please double-check my work.
– Blue
Jan 8 '16 at 23:08
• 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 $\frac {1}{2\sin 10\circ}$ Jan 8 '16 at 23:34
• @Piquito: $\sin 30^\circ = \frac{1}{2}$. :)
– Blue
Jan 9 '16 at 2:56

• 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.