August 19, 2019, 01:36:20 am

### AuthorTopic: 3U Maths Question Thread  (Read 538944 times) Tweet Share

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#### RuiAce

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##### Re: 3U Maths Question Thread
« Reply #4185 on: August 09, 2019, 07:43:37 pm »
+1
yes, but cos inverse 1 also equals 2pi, so how do you know which one to use?

for the second one, my question was how do you then isolate x from your last line?
This red bit is not true.
$\text{The function }f(x) = \cos^{-1}x\text{ has range }0\leq y \leq \pi.\\ \text{The value it returns }\textbf{must}\text{ be a value associated with the first or second quadrant.}$

Note that the $x$ can also be isolated in the second equation by brute force: $\displaystyle2x = 2n\pi \pm \frac\pi2 - \frac\pi4 \implies \boxed{x=n\pi \pm \frac\pi4 - \frac\pi8}$. Although I would favour fun_jirachi's answer over this notation.

#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4186 on: August 09, 2019, 08:56:51 pm »
0
$\text{The idea here is that for} \ \cos x = a \\ x= 2n\pi \pm \cos^{-1}a \ (\text{for} \ n \in \mathbb{Z})\\ \text{Now consider} \ \cos x = 1 \\ x=2n\pi \pm \cos^{-1} (1) = 2n\pi, n \in \mathbb{Z} \\ \text{or for 'using'} \ 2\pi \ x=2n\pi \pm 2\pi = (2n+1)\pi, n \in \mathbb{Z}$
So as you can see it doesn't actually matter if you we use 0 or 2pi since 2n and 2n+1 are both integers. Just looks nicer and cleaner to use 0 (EDIT: also it's correct unlike using 2pi [refer to next post ]).

To isolate x:
$\cos \left(2x+\frac{\pi}{4}\right)=0 \\ 2x+\frac{\pi}{4} = 2n\pi \pm \frac{\pi}{2} \\ 2x=2n\pi + \frac{\pi}{4}, 2n\pi - \frac{3\pi}{4} \\ x= n\pi + \frac{\pi}{8}, n\pi -\frac{3\pi}{8} \\ \text{Which combines to form} \ x=\frac{(4n+1)\pi}{8}, n \in \mathbb{Z}$

thank you how did you combine the two answers?

#### spnmox

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##### Re: 3U Maths Question Thread
« Reply #4187 on: August 09, 2019, 08:57:13 pm »
0
This red bit is not true.
$\text{The function }f(x) = \cos^{-1}x\text{ has range }0\leq y \leq \pi.\\ \text{The value it returns }\textbf{must}\text{ be a value associated with the first or second quadrant.}$

Note that the $x$ can also be isolated in the second equation by brute force: $\displaystyle2x = 2n\pi \pm \frac\pi2 - \frac\pi4 \implies \boxed{x=n\pi \pm \frac\pi4 - \frac\pi8}$. Although I would favour fun_jirachi's answer over this notation.

thank you, that clears it up!!