0
$\begingroup$

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!

Thanks

$\endgroup$
2
$\begingroup$
  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 $\sin^2(\frac{3\pi}8)=\frac{2+\sqrt2}4$.

$\endgroup$
  • $\begingroup$ 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." $\endgroup$ – Zelzy Jan 31 '14 at 2:58
  • $\begingroup$ Ah I see, thank you very much! $\endgroup$ – Terry Jan 31 '14 at 3:14
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.