Inverse trig functions

[jonnburton]

In one worked example on this subject in my book, there is a step that I don't understand. I was wondering if somebody here could tell me how this works?

This is the question:
Given that $\displaystyle sin \frac{\pi}{3} = \frac{\sqrt3}{2}$, find the exact value of $\displaystyle sin^{-1}(sin\frac{4\pi}{3})$

The solution begins: $\displaystyle sin\frac{4\pi}{3} = sin(\pi + \frac{\pi}{3})$, which is clear to me.

I cannot, however, see the reasoning behind the next step:

$\displaystyle sin(\pi + \frac{\pi}{3}) = -sin\frac{\pi}{3}$

I would have thought it would have come out to $\displaystyle sin \pi + sin\frac{\pi}{3}$, rather than the above.

I'd be very grateful for any information as to how this works!

 

[Deleted member 4993]

In one worked example on this subject in my book, there is a step that I don't understand. I was wondering if somebody here could tell me how this works?

This is the question:
Given that $\displaystyle sin \frac{\pi}{3} = \frac{\sqrt3}{2}$, find the exact value of $\displaystyle sin^{-1}(sin\frac{4\pi}{3})$

The solution begins: $\displaystyle sin\frac{4\pi}{3} = sin(\pi + \frac{\pi}{3})$, which is clear to me.

I cannot, however, see the reasoning behind the next step:

$\displaystyle sin(\pi + \frac{\pi}{3}) = -sin\frac{\pi}{3}$

I would have thought it would have come out to $\displaystyle sin \pi + sin\frac{\pi}{3}$, rather than the above.

I'd be very grateful for any information as to how this works!

It comes from:

sin (A+B) = sin(A) * cos(B) + sin(B) * cos(A)

for quick review go to:

http://www.math10.com/en/geometry/trigonometry-and-geometry-conversions/trigonometry.html

 

[DrPhil]

In one worked example on this subject in my book, there is a step that I don't understand. I was wondering if somebody here could tell me how this works?

This is the question:
Given that $\displaystyle sin \frac{\pi}{3} = \frac{\sqrt3}{2}$, find the exact value of $\displaystyle sin^{-1}(sin\frac{4\pi}{3})$

The solution begins: $\displaystyle sin\frac{4\pi}{3} = sin(\pi + \frac{\pi}{3})$, which is clear to me.

I cannot, however, see the reasoning behind the next step:

$\displaystyle sin(\pi + \frac{\pi}{3}) = -sin\frac{\pi}{3}$

I would have thought it would have come out to $\displaystyle sin \pi + sin\frac{\pi}{3}$, rather than the above.

I'd be very grateful for any information as to how this works!

Or, instead of invoking the sum-of-angles formula explicitly, you can see it by looking at the periodicity and symmetry of the sine function. The sine of an angle in the first quadrant is equal to that of an angle in the 2nd quadrant which is the same elevation above the -x axis. That is,
$\displaystyle \sin (\theta) = \sin (\pi - \theta)$

The corresponding angles reflected into the 4th and 3rd quadrants have the negative of that sine:
$\displaystyle \sin(-\theta) = \sin (\pi + \theta) =-\sin (\theta)$

The sine is positive in the 1st & 2nd quadrants, and negative in the 3rd & 4th. The magnitude of the sine is the y-coordinate on the unit circle.

 

[jonnburton]

Thankyou very much for both of your replies, Subhotosh and DrPhil. I am going to go through both carefully to make sure I understand it properly.