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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?

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

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  • $\begingroup$ I think that you are right and $\sqrt 3 \cos (20)$ was the requested result. $\endgroup$
    – user
    Nov 13 '18 at 23:05
  • $\begingroup$ Thank you. I didn't recall that formula. $\endgroup$
    – user163862
    Nov 13 '18 at 23:13
  • $\begingroup$ @user163862: I had to look it up. I know the angle-sum formulas, but not the function-sum ones. $\endgroup$ Nov 13 '18 at 23:15
  • $\begingroup$ 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 $\endgroup$
    – user163862
    Nov 14 '18 at 20:24
  • $\begingroup$ 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. $\endgroup$ Nov 14 '18 at 20:53
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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

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

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

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  • $\begingroup$ 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. $\endgroup$
    – user163862
    Nov 14 '18 at 20:08
  • $\begingroup$ @user163862. What is sin 3×40 degrees? $\endgroup$ Nov 15 '18 at 4:14

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