challenge trig question - no calculator

Question

The challenge trignometry question is: simplify $sin (80^\circ) + sin (40^\circ) $ using trignometric identities. All the angle values are in degrees.

This is what I did: Let $a=40^\circ$. So we have $sin(2a) +sin(a)$. Using the formula for $sin(2a)$, I have, $$2sin(a)cos(a) + sin(a)=sin(40)(2cos(40^\circ)+1)$$

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?

Answer 1

You may have been intended to write $$\begin {align}\sin (80)+\sin (40)&=2\sin\left(\frac {80+40}2\right)\cos\left(\frac {80-40}2\right)\\&=2\sin (60) \cos (20)\\&=\sqrt 3 \cos (20) \end {align}$$ but I don't have a nice value for $\cos(20)$ 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 $\cos(20)\approx 1-\frac 12(\frac {\pi}9)^2\approx1-\frac {10}{162}\approx 1-\frac 1{16}=0.9375$ which is very close.

Answer 2

Maybe we need to a single angle, in that case we have that

• ${{\sin(\theta + \varphi) + \sin(\theta - \varphi)} }=2\sin \theta \cos \varphi$

therefore

$$\sin (80)+\sin (40) = 2\sin (60)\cos (20)=\sqrt 3 \cos 20$$

Answer 3

You're supposed to find a numerical value.
sin 3x = 3.sin x - 4.sin$^4$ x
will give values for sin 40 deg.