Math Help Forum: 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)).

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).

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.

First, about the negative sign.
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.
And those are 4pi/3 and 5pi/3 radians. -------answer.

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

======
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 -----***

Therefore, tan[arcsin(-4/5)] = +,-4/3 -----------answer.

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

Hello, Mike!

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

I'll assume the angle are between and .

Find the exact value of:

We want the angle whose sine is

You are expected to know that:

Sine is negative in quadrants 3 and 4.
So we have a reference angle of in quadrants 3 and 4.

Hence, the angles are: .

Inside, we have: .

Then we have: . . . . the angle whose cosine is

We know that: .
Cosine is negative in quadrants 2 and 3.

Therefore the angles are: .

Inside, we have: . . . . the angle whose sine is

Consider:
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

. . the answers are: .