I have a few questions on trig/calc stuff I am having trouble with, for some reason I am just not getting the concept.

1: What happens if you take $B=2\pi$ in the addition formula? Do the results agree with something you already know?

so... $\sin A \cos B + \cos A \sin B = \sin A \cos 2\pi + \cos A \sin 2\pi$, without $A$, we cannot get further?

2: Find the function values. $\displaystyle \sin^2(\frac{3\pi}{8})$

Thank you, I would really appreciate it if someone could help me figure these out, and learn the concept!


  1. You can get further without $A$, since you know that $\cos2\pi = 1$ and $\sin2\pi = 0$. (These are really fundamental properties of $\cos$ and $\sin$.) Therefore you have $\sin(A+2\pi) = \sin(A)$, as expected.

  2. We know that $\sin^2(\frac{3\pi}8) = (\sin(\frac{3\pi}8))^2$. We can simply plug in $\sin(\frac{3\pi}8)$ into a calculator and square it. If we want to find that out by hand, we can use an identity and find $\sin(\frac{3\pi}8) = \pm \sqrt\frac{1-\cos(\frac{3\pi}4)}2$. We know $\cos(\frac{3\pi}4)=-\frac{\sqrt2}2$. Anyway, plug in everything and you end up with $\sin^2(\frac{3\pi}8)=\frac{2+\sqrt2}4$.

  • $\begingroup$ how did we know sin3π/8= that identity tho? $\endgroup$ – Terry Jan 31 '14 at 2:55
  • $\begingroup$ We know that $\sin(\frac{\theta}2) = \pm\sqrt\frac{1-\cos(\theta)}2$. It's a trigonometric identity. It's called the "half-angle formula." $\endgroup$ – trystero Jan 31 '14 at 2:58
  • $\begingroup$ Ah I see, thank you very much! $\endgroup$ – Terry Jan 31 '14 at 3:14

So you're asking what happens if we substitute $B = 2\pi$ in $\sin(A+B)$. Well the trig functions are $2\pi$-periodic so by adding $2\pi$, you get the same result. You can check this with the sum formulas:

$$\sin(A+2\pi) = \sin A\cos 2\pi + \cos A\sin 2\pi = \sin A + 0 = \sin A$$

since $\sin2\pi = 0$ and $\cos 2\pi = 1$. This matches our intuition about how trig functions should work.

As for the second one, we know that $\cos(2\theta) = 1-2\sin^2 \theta$. So if $\theta = \frac{3\pi}{8}$, what do you get?

  • $\begingroup$ i see what you mean for the first one. For the second, i think i did it wrong but i got, Cos2(3π/8)=1-2Sin^2*(3π/8)= -cos(3π/4)+1=2Sin^2(3π/8) === (-Cos.75π+1)/(2)=Sin^2(3π/8) $\endgroup$ – Terry Jan 31 '14 at 2:42
  • $\begingroup$ This is fine. You can evaluate $\cos\left(\frac{3\pi}{4}\right)$ to get a nice answer. $\endgroup$ – Cameron Williams Jan 31 '14 at 2:45
  • $\begingroup$ wait, (-Cos.75π+1)/(2)=Sin^2(3π/8) is the answer? I did it right? :)))) $\endgroup$ – Terry Jan 31 '14 at 2:45
  • $\begingroup$ It can be simplified and you probably should. $\endgroup$ – Cameron Williams Jan 31 '14 at 2:46
  • $\begingroup$ ok I think i simplified right. -Cos(6π/16)+1=Sin^2(3π/16) $\endgroup$ – Terry Jan 31 '14 at 2:51

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