# I am having trouble with these three problems that are in the attached document, thank you for your help. 9. Using your calculator, find the csc(-200). 10. Solve the following right triangle (see...

I am having trouble with these three problems that are in the attached document, thank you for your help.

9. Using your calculator, find the csc(-200).

10. Solve the following right triangle (see image.)

11. Find a trigonometric equation for the following graph (see image.)

*print*Print*list*Cite

### 1 Answer

**9.** To find csc(-200) using a calculator, first find sin(-200). Make sure your calculator is in "degrees" mode! Then, use the following reciprocal identity:

`csc(x) = 1/sin(x)`

Since cosecant is the reciprocal of sine, use "reciprocal" (1/x) button on a calculator to find csc(-200):

`csc(-200) = 1/sin(-200) = 2.9238`

10. In the right triangle on the picture, side a = 10 and the angle opposite to it,

`alpha = 50` , are given. Side b is the side adjacent to `alpha` and side c is the hypotenuse.

Use the definition of sine to find the hypotenuse c first:

`sin(alpha) = (opp)/(hyp) = a/c`

`sin(50) = 10/c`

Solving for c, we find `c = 10/sin(50) =13.054 `

Now the definition of tangent can be used to find the side b:

`tan(alpha) = (opp)/(adj) = a/b`

`tan(50)=10/b`

Solving for b, we find `b = 10/tan(50) =8.39 `

(Alternatively, side b can also be found from Pythagorean Theorem: `a^2 + b^2 = c^2` )

Lastly, we can find angle

`beta`

by using one of its trigonometric functions, for example, sine:

`sin(beta) = (opp)/(hyp) = b/c = 8.39/13.054=0.643`

Angle `beta` can now be found by taking inverse sine of the sine:

`beta = sin^(-1) (0.643) = 40`

Notice that the sum of the angles is 50 + 40 = 90, which is consistent with the fact that the sum of angles in a triangle is 180 degrees: 50 + 40 + 90 = 180.

**11.** The general equation for a trigonometric graph on the picture is

`y = A*sin(b(x -c))` , where A is the amplitude, `b = (2pi)/T` , where T is the period, and c is the shift.

From the graph we can see that amplitude is A = 4 (the highest point on the graph.)

Since the curve goes through complete cycle from `-pi` to `3pi` , the period is `4pi` .

So, `b=(2pi)/T = (2pi)/(4pi) = 1/2` .

Thus, `y=4sin(1/2 (x-c))`

To find c, plug in a point from a graph in the above equation. For example, for

`x=pi` , y = 0:

`0 = 4sin(1/2(pi-c)) = 4sin(pi/2 -c/2)`

Using cofunction identity `sin(pi/2 -x) = cos(x)` , we can rewrite

`sin(pi/2-c/2)=cos(c/2)`

Then, `4cos(c/2) = 0` , which means c/2 has to be `pi/2` or `-pi/2` and `c=pi` or `c=-pi` .

So, the equation of the graph on the picture is either

`y=4sin(1/2(x-pi))` or `y=4sin(1/2(x+pi))`

To determine which one it is, plug in another point. For `x=2pi` , y should be -4.

For the first equation, `x = 2pi` results in

`y=4sin(1/2(2pi-pi))=4sin(pi/2) =4`

For the second equation,

`y=4sin(1/2(2pi+pi))=4sin((3pi)/2) = -4`

So the second equation is the correct one.

**The equation for the graph on the picture is**

`y=4sin(1/2(x+pi))` .