 2022-01-26

The challenge trignometry question is: simplify $\mathrm{sin}\left({80}^{\circ }\right)+\mathrm{sin}\left({40}^{\circ }\right)$ using trignometric identities. All the angle values are in degrees.
This is what I did: Let $a={40}^{\circ }$. So we have $\mathrm{sin}\left(2a\right)+\mathrm{sin}\left(a\right)$. Using the formula for $\mathrm{sin}\left(2a\right)$, I have,
$2\mathrm{sin}\left(a\right)\mathrm{cos}\left(a\right)+\mathrm{sin}\left(a\right)=\mathrm{sin}\left(40\right)\left(2\mathrm{cos}\left({40}^{\circ }\right)+1\right)$
All I got was a big red mark through the question. I looked at co-functions and half-angles and couldn't find a solution that came out to something I could do without a calculator. Any thoughts? lilwhitelieyc

Expert

You may have been intended to write
$\mathrm{sin}\left(80\right)+\mathrm{sin}\left(40\right)=2\mathrm{sin}\left(\frac{80+40}{2}\right)\mathrm{cos}\left(\frac{80-40}{2}\right)$
$=2\mathrm{sin}\left(60\right)\mathrm{cos}\left(20\right)$
$=\sqrt{3}\mathrm{cos}\left(20\right)$
but I don't have a nice value for $\mathrm{cos}\left(20\right)$ and neither does Alpha, which finds about 0.93969. The first two terms of the Taylor series should be close, and if you know ${\pi }^{2}\approx 10$ you can write $\mathrm{cos}\left(20\right)\approx 1-\frac{1}{2}{\left(\frac{\pi }{9}\right)}^{2}\approx 1-\frac{10}{162}\approx 1-\frac{1}{16}=0.9375$ which is very close.

Do you have a similar question?