Use this document for some of the problems below. 

https://myfiles.st-agnes.org/Users/Students/cdunn/POST%201.tii

1.  Find sin(pi/4)
this is also (1/4)pi

=(sqrt2)/2

2. cos(1.5pi)
this is also (3/2)pi

=0

3. tan(pi/6)
this is also (1/6)pi
sin/cos
((1/2)/((sqrt3)/2))

=1/sqrt3

4. (3,-4) is a point on the terminal side of an angle theta.  Find 6 trigonometric functions.

sin= -4/5           cos=3/5
tan=-4/3            sec=5/3
csc=-5/4           cot=-3/4

Refer to the reference angles and quadrantal angles on page 785 in the textbook.

5. Find tan 225 degrees
We know the angle is a little more than 180 degrees.  pi+(1/4)pi is ((-sqrt2)/2,(-sqrt2)/2)
The reference angle is 225-180=45 degrees

=1

6. Find csc((11pi)/4)
This is the reciprocal of sin.
Theta prime is pi-(3/4)pi which is (1/4)pi
Use the point ((sqrt2)/2,(sqrt2)/2) which is (1/4)pi.
1/((sqrt2)/2) is 2/(sqrt2).  When rationalized, this =sqrt2

7. sin-210 degrees
Use the point ((sqrt3)/2,1/2).
The second number in this ordered pair is 1/2.

=1/2

8. sec((7pi)/6)
This is hte reciprocal of cosine
Use the point ((sqrt3)/2,(1/2)).
1/(-(sqrt3)/2)= -2/(sqrt3) and when rationalized, equals (-2sqrt3)/3

9. Try this problem for homework tonight:
A wheel in a clock has a tooth that is farthest right 4 units and is 10 units from the base of teh clock.  If the wheel rotates an angle of 240 degrees, how far away is the tooth from the base of the clock?

Answers to the problems on ExSet4

3b.) -1/2

3g.) (sqrt3)/3

3h.) -1

4.) sin=(sqrt15)/4
      tan= -sqrt15

6a.) .8415

6c.) -.542

8.) x= +/- ((pi/9)+2*pi*k) where k is any integer

Notes for Thursday, February 28, 2008

Power Point Worksheet

1. Find theta when:
• sin Ө = .5

30 + 360(K)

• cos Ө = .5

(+/–60) + 360(K)

• tan Ө = -1

-45+180(K)

2. Find the Reciprocal of:
• sin Ө    à   csc Ө
• cos Ө   à   sec Ө
• tan Ө    à   cot Ө

1. Find an inverse function for:
• y=2x+3

•  The graph of this equation doesn’t pass the horizontal line test, so it doesn’t have an inverse function unless we restrict the domain: x > -3

One to One function

1. y=(x+3)^2 + 5

-         you can tell this function is not one to one because the y values repeat

-         you can make it one to one by restricting the domain so that each y value shows only once (x > -3)

2. y = sin(x)

      - this function isn’t one to one because the y values repeat

      - by making the x values -90 < x < 90 the function becomes one to one

3.

      a.  y1 = cos(x)  à     0 < x > 180

      b.  y1 = tan(x)   à     -90 < x > 90

Inverse Trig Functions:

Function           Domain                        Range

sin θ = #           -90º≤ θ≤ 90º                1≤ #≤ 1

sin^-1# = θ          -1≤ #≤ 1                      -90º≤ θ≤ 90º

Evaluate the following with a calculator in both degrees and radians:

-sin^-1.5  = 30º = .524(radians)  

cos^-1.5 = 60º = 1.047

tan^-11= 45º = .785

Note that the radian answer must be between -1.57(-π / 2) and 1.57 (π / 2).

Inverse…

-sin^-1.5  is in the 1^st or 4^th quadrants

cos^-1.5 is in the 1^st or 2^nd quadrants

tan^-11 is in the 1^st or 4^th quadrants

Find:

sin^-1(sqrt2 / 2)   = 45º - only one answer, 1^st quadrant 

cos^-1(3) = no solution/undefined. The input must be between -1 and 1 for sine and cosine.

tan^-1(-1) = - 45º

Solving Trig Equations:

cosθ  = -2/3  for -180º≤ θ≤ 270º

-2/3 is not a special angle so don’t use the unit circle, use your calculator!

-2/3 is in the third quadrant so get the reference angle by getting rid of the negative and just putting 2/3, so cos^-1(2/3) = 48.190º (if you put cos^-1(-2/3) you get 131.810º which is far to large.)  48.190º + 180 = 228.190º

tanθ= -3/2  (notice there aren’t any limitations)

tan^-1(3/2) = 56.310 which is in the first quadrant so the same angle in the 4^th quadrant is -56.310. Remember no restrictions, so you have to include all co-terminal angles!

 θ = -56.310 º + 180º K

Warm-up/Classwork

1. What is the difference b/w sin^-1 θ and (sin θ) ^-1?

    Ans: sin^-1 θ gives the angle and (sin θ) ^-1 gives the reciprocal

2. A crane whose lower end is 4 ft off the ground has a 100ft arm.  The arm has to reach the top of a building 80 ft high. At what angle should the arm be set?

    Ans: θ =49.464 º

Also Ex Set 4 # 5, 9, 10, 11, 13