1. Exact values

Could someone help me with the following three problems?

1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).

2) cos[to the -1 power](cos(4[pi]/3)).

3) tan(sin[to the -1 power](-4/5)).

2. Originally Posted by mike1
Could someone help me with the following three problems?

1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).
You are expected to know some special triangles, one of which is the
30, 60, 90 degree triangle, where if the side opposite the 30 degree angle
is 1 unit the side opposite the 60 degree angle is sqrt(3), and the other side
is 2. So using this right triangle we see that sin(30)=sqrt(3)/2, hence if:

x=asin(-sqrt(3)/2),

then

sin(x)=-sqrt(3)/2,

so:

sin(-x)=sqrt(3)/2.

Now there are multiple solutions to this, but we want x to be in the range
[-90,90], so we want -x=60, or x=-60 degrees, or -pi/3 radian.

Check: asin(-sqrt(3)/2)~=-1.57, -pi/2~=-1.57

2) cos[to the -1 power](cos(4[pi]/3)).
Now 4 pi/3= pi+pi/3, so cos(4 pi/3)=cos(pi)cos(pi/3)-sin(pi)sin(pi/3)=-cos(pi/3)
Now pi/3 radian is 60 degrees so we are back to our special triangle, and cos(pi/3)=1/2.

Now acos(-1/2)=pi-pi/3=2pi/3.

Check acos(cos(4pi/3))~=2.094, 2*pi/3~=2.094

3) tan(sin[to the -1 power](-4/5)).
Here we need to think 3-4-5 triangle, first:

tan(asin(-4/5))=-tan(asin(4/5))

(draw the triangle identify the angle whose sin is 4/5, then find its
tangent), so:

tan(asin(-4/5))=-4/3

Check: tan(asin(-4/5))~=-1.33.., -4/3~=1.33..

RonL

Note: when evaluating the inverse trig functions it is usual to give the answer
in a fundamental range such that the function has a unique value (the principle value).

That is evaluating x=asin(y) is not the same as finding x such that sin(x)=y,
since asin(y) gives a single value, while solving sin(x)=y has multiple solutions.

For asin the range is usually -pi/2 to pi/2 (or in degrees -90 to 90)
For acos the range is usually 0 to pi (0 to 180 degrees)
For atan the range is usually -pi/2 to pi/2 (or sometimes 0 to pi) (-90 to 90 degrees).

3. Originally Posted by mike1
Could someone help me with the following three problems?

1) Find the exact value of sin[to the -1 power](-[sqrt]3/2).

2) cos[to the -1 power](cos(4[pi]/3)).

3) tan(sin[to the -1 power](-4/5)).
I assume sin[to the -1 power] is arcsine.

1) arcsin[-sqrt(3) /2]
That is an angle whose sine is -sqrt(3) /2. Let's call it theta.

In what quadrants is the sine negative? Since sine is y/r, and r is always positive, then sine is negative wherever y is negative. Below the origin, or below the x-axis. In the 3rd and 4th quadrants.
So theta is between pi and 2pi radians.

Then the angle.
If sin(theta) = sqrt(3) /2, then theta is 60degrees or pi/3 radians. What are the angles in the 3rd and 4th quadrants that are pi/3 from the x-axis?
Why, (pi +pi/3) in the 3rd quadrant and (2pi -pi/3) in the 4th quadrant.

----------------------
2) cos[to the -1 power](cos(4[pi]/3)).
That is "the cosine of an angle whose cosine is 4pi/3".

======
Ooppss, my mistake.
That is actually arccosine of cos(4pi/3).
An angle whose cosine is cos(4pi/3).

I do not know how to do that.

You sure you typed it right?
-------------------------------------
3) tan(sin[to the -1 power](-4/5)).
That is "the tangent of an angle whose sine is -4/5 "

First, about the negative sine value.
As in part (i) above, theta is in the 3rd or 4th quadrant, or theta between pi and 2pi radians.

Then the angle.
If sin(alpha) = 4/5, then y=4 and r=5 since sine = y/r.
tan is y/x. So we need to find x.
By Pythagorean theorem, x = sqrt(r^2 -y^2) = sqrt(5^2 -4^2) = 3.
Hence tan(alpha) = y/x = 4/3.

In the 3rd quadrant, x is negative, and y is negative also, hence,
tan(theta) = y/x = -4/ -3 = 4/3 ------***

In the 4th quadrant, x is positive, and y is negative, hence,
tan(theta) = y/x = -4/ 3 = -4/3 -----***

-----------------------
Note, in the reference triangle,
opposite side = y
hypotenuse = r --------always positive.

4. Hello, Mike!

ticbol did an excellent job of explaining.
Here are my versions . . .

I'll assume the angle are between $\displaystyle 0$ and $\displaystyle 2\pi$.

Find the exact value of:

$\displaystyle (1)\;\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

We want the angle whose sine is $\displaystyle -\frac{\sqrt{3}}{2}$

You are expected to know that: $\displaystyle \sin\left(\frac{\pi}{3}\right) \,=\,\frac{\sqrt{3}}{2}$

Sine is negative in quadrants 3 and 4.
So we have a reference angle of $\displaystyle \frac{\pi}{3}$ in quadrants 3 and 4.

Hence, the angles are: .$\displaystyle \frac{4\pi}{3},\:\frac{5\pi}{3}$

$\displaystyle (2)\;\cos^{-1}\left[\cos\left(\frac{4\pi}{3}\right)\right]$

Inside, we have: .$\displaystyle \cos\frac{4\pi}{3} \:=\:\text{-}\frac{1}{2}$

Then we have: .$\displaystyle \cos^{-1}\left(\text{-}\frac{1}{2}\right)$ . . . the angle whose cosine is $\displaystyle \text{-}\frac{1}{2}$

We know that: .$\displaystyle \cos\left(\frac{\pi}{3}\right) \,=\,\frac{1}{2}$
Cosine is negative in quadrants 2 and 3.

Therefore the angles are: .$\displaystyle \frac{2\pi}{3},\:\frac{4\pi}{3}$

$\displaystyle (3)\;\tan\left[\sin^{-1}\left(\text{-}\frac{4}{5}\right)\right]$

Inside, we have: .$\displaystyle \sin^{-1}\left(\text{-}\frac{4}{5}\right)$ . . . the angle whose sine is $\displaystyle \text{-}\frac{4}{5}$

Consider: $\displaystyle \sin\theta = \frac{4}{5}$
We don't know the exact value of this angle,
. . but we know it comes from this triangle:
Code:
                  *
* |
*   |
5 *     | 4
*       |
* θ       |
* - - - - - *
3

Since sine is negative in quadrants 3 and 4,
. . the two possible angles look like this:
Code:
              |                         |
|                         |
-3   |                         |   3
- + - - - + - - - - -       - - - - + - - - + -
:   θ / |                         | \ θ   :
-4:   /5  |                         |  5\   :-4
: /     |                         |     \ :
*       |                         |       *

Since $\displaystyle \tan\theta \:=\:\frac{\text{-}4}{\text{-}3}\text{ or }\frac{\text{-}4}{3}$

. . the answers are: .$\displaystyle \pm\frac{4}{3}$

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find principal value of cos^-1(cos4pi/3)

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