# Thread: arcos(sinθ)

1. ## arcos(sinθ)

If $\displaystyle 3 \pi < \theta < \frac{7 \pi}{2}$, what is $\displaystyle arccos(sin\theta)$ equal to?

Attempt:

Since $\displaystyle 3 \pi < \theta < \frac{7 \pi}{2}$, $\displaystyle \theta = \frac{\pi}{2}$ so:

$\displaystyle arccos(1)$

$\displaystyle = 0$

but think I made a mistake without considering the interval it in. What should I do

2. Originally Posted by SyNtHeSiS
If $\displaystyle 3 \pi < \theta < \frac{7 \pi}{2}$, what is $\displaystyle arccos(sin\theta)$ equal to?

Attempt:

Since $\displaystyle 3\pi < \theta < \frac{7 \pi}{2}$, $\displaystyle \theta = \frac{\pi}{2}$ so:

$\displaystyle arccos(1)$

$\displaystyle = 0$

but think I made a mistake without considering the interval it in. What should I do
$\displaystyle 3{\pi}\rightarrow\ 3.5{\pi}$ is the third quadrant, $\displaystyle 2{\pi}+{\pi}\rightarrow\ 2{\pi}+{\pi}+\frac{\pi}{2}$

$\displaystyle 360^o+180^o\rightarrow\ 360^o+180^o+90^o$

If you draw a unit-radius circle, then in that 3rd quadrant, pick a point on the circumference.
draw a right-angled triangle by drawing lines from the point directly up to the x-axis
and directly from the point to the circle centre.

For this triangle, $\displaystyle \alpha=$angle at the centre inside the triangle.

adjacent$\displaystyle =|x|$

opposite$\displaystyle =|y|$

hypotenuse$\displaystyle =1$

$\displaystyle sin\theta=sin\alpha=-y$

Draw a second right-angled triangle, whose base is $\displaystyle |y|$ and whose opposite is $\displaystyle |x|$ in that same 3rd quadrant

$\displaystyle arccos(-y)=3{\pi}+\frac{\pi}{2}-\alpha$

which is the angle after one revolution in the 3rd quadrant.
There is another angle in the 2nd quadrant, as cosine is negative there also.
From there, the range of angles can be deduced from $\displaystyle 0\rightarrow\ 2{\pi}$

3. Originally Posted by SyNtHeSiS
If $\displaystyle 3 \pi < \theta < \frac{7 \pi}{2}$, what is $\displaystyle arccos(sin\theta)$ equal to?
you can only determine the range of values for $\displaystyle \arccos(\sin{\theta})$ .

if $\displaystyle \displaystyle 3 \pi < \theta < \frac{7 \pi}{2}$ , then $\displaystyle -1 < \sin{\theta} < 0$ . Subsequently ...

$\displaystyle \displaystyle \frac{\pi}{2} < \arccos(\sin{\theta}) < \pi$ .

4. Originally Posted by Archie Meade
Draw a second right-angled triangle, whose base is $\displaystyle |y|$ and whose opposite is $\displaystyle |x|$ in that same 3rd quadrant

$\displaystyle arccos(-y)=3{\pi}+\frac{\pi}{2}-\alpha$
Why did you swap the x and y around on the triangle when working out the arccos? Also can you say $\displaystyle arccos(-y) = \pi - \beta$, since its in the 3rd quadrant?

5. Originally Posted by SyNtHeSiS
If $\displaystyle 3 \pi < \theta < \frac{7 \pi}{2}$, what is $\displaystyle arccos(sin\theta)$ equal to?

Attempt:

Since $\displaystyle 3 \pi < \theta < \frac{7 \pi}{2},\;\;range\;\;of\;\;\theta = \frac{\pi}{2}$ so:

$\displaystyle arccos(1)$

$\displaystyle = 0$

but think I made a mistake without considering the interval it in. What should I do
Better to start from scratch,

we can ascertain the graph of $\displaystyle acos(sinx)$

$\displaystyle u=sinx$

Notice that $\displaystyle \frac{d}{dx}acos(sinx)=\frac{d}{dx}acos(u)=\frac{d }{du}acos(u)\frac{d}{dx}sinx$

$\displaystyle \displaystyle\ =-\frac{1}{\sqrt{1-u^2}}cosx=-\frac{1}{\sqrt{1-sin^2x}}cosx=-\frac{cosx}{|cosx|}$

This means the slope of the graph is $\displaystyle \pm1$, depending on whether $\displaystyle cosx$ is positive or negative.

Hence, it is a periodic triangle wave.

For $\displaystyle 3{\pi}<\theta<3.5{\pi}$

$\displaystyle \displaystyle\ sin\theta$ goes from $\displaystyle 0\rightarrow-1$

Therefore $\displaystyle acos(sin\theta)$ goes from $\displaystyle acos(0)\rightarrow\ acos(-1)$ which is from $\displaystyle \frac{\pi}{2}\rightarrow\ {\pi}$

If you have an exact value for $\displaystyle \theta$, then since the graph is rising linearly from $\displaystyle 3{\pi}\rightarrow\ 3.5{\pi}$, then

$\displaystyle acos\left(sin\theta}\right)=\frac{\pi}{2}+\theta-3{\pi}$

### arcos-1 geom

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