# challenge trig question - no calculator

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?

• MathJax hint: if you put a backslash before common functions you get the right font and spacing, so \sin x gives $\sin x$ Nov 13 '18 at 22:52
• A 'big red mark' should urge you to ask for the solution from your instructor. They will only be too happy to teach! Nov 13 '18 at 22:59

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.

• I think that you are right and $\sqrt 3 \cos (20)$ was the requested result.
– user
Nov 13 '18 at 23:05
• Thank you. I didn't recall that formula. Nov 13 '18 at 23:13
• @user163862: I had to look it up. I know the angle-sum formulas, but not the function-sum ones. Nov 13 '18 at 23:15
• After examining that formula, I can see that $sin (A+B) + Sin (A-B)=2Sin(A)Cos(B)$. But it would seem to me that when I divide by $2$ it divides the $sin$ term not the angles Nov 14 '18 at 20:24
• If you substitute $A=60,B=20$ into your formula you get mine exactly. This comes from solving $A+B=80,A-B=40$ and the divide by $2$ comes when you add and subtract the two equations. Nov 14 '18 at 20:53

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

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

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

• Actually the problem said to simplify; however, I'd be very interested in getting the numerical value without a calculator. How do I do that using the sin (3x) formula. Nov 14 '18 at 20:08
• @user163862. What is sin 3×40 degrees? Nov 15 '18 at 4:14