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.)
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`
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
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
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
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