Shamar Padilla

Answered

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,

$2\mathrm{sin}\left(a\right)\mathrm{cos}\left(a\right)+\mathrm{sin}\left(a\right)=\mathrm{sin}\left(40\right)(2\mathrm{cos}\left({40}^{\circ}\right)+1)$

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?

This is what I did: Let

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?

Answer & Explanation

lilwhitelieyc

Expert

2022-01-27Added 10 answers

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.

but I don't have a nice value for

Most Popular Questions