# Chapter 1 - Worksheet 5 - TRIGONOMETRY – PART 1 ... 1. sin( ```Chapter 1 - Worksheet 5 - TRIGONOMETRY – PART 1
1.
NAME_______________________________
Consider the transformation y  A  sin( Bx  C )  D where A, B, C , and D are positive constants.
How does the value of each constant affect the graph of the standard sine function? Be specific.
A is the amplitude – it is a vertical stretch or shrink and tells half the distance between the
maximum and minimum function values.
B is the angular frequency – it helps find the period of the function and is a horizontal stretch or
shrink. B indicates how many periods occur in 2π.
C indicates there is a horizontal shift (the actual horizontal shift is -C/B)
D is the midline – it is also a vertical shift up or down. – Average value
2.
The following function describes the air temperature in Fairbanks, Alaska as a function of time.
 2

T (t )  37 sin 
 t  1.7386   25
 365

A.
Without graphing this function, determine its period, amplitude, average value, max., and min. and include the units. What is the
phase shift what does this value represent?
2𝜋
2𝜋
Period is 365 days; 𝐵 = 𝑝𝑒𝑟𝑖𝑜𝑑 = 365; amplitude is 37 ( this is in degrees Fahrenheit);
Average value is 25 degrees. Max is 62 =(37 + 25). Min is -12 = (-37 +25)
There is a phase shift of
temperature.
365(1.7386 )
 101 days.
2
B. Graph one period of this function (beginning with
62
t  0 ). Label appropriate values on the axes.
T
62
25
T
25
101
-12
So in the middle of April is the average
466
days
t
365
days
t
-11.5
-12
The period starting at (101,25) 365/4=91.25
max is (91 days later) then (192, 62) (middle of the year)
Next midline again (91 days later) (283, 25)
Min value is (91 days later) about (374, -12) - in Jan of the next year
Midline again (466, 25)
end of one period is (466, 25)
When you start at t=0 then value would be (0, -11.5) SO days 365 to 466 would look like day 0 to 101. So on day (365, 11.5) as indicated on the second graph.
3.
The rate of intake during a respiratory cycle (liters/sec) for a person at rest is proportional to a
sine wave with period six seconds. Suppose the rate is 0.85 liters/sec when
t  1.5 sec.
A. Find an equation that describes the rate of intake as a function of time.
2 
 .
6
3
   3 
 
Rate  y  k sin  t  and 0.85  k sin      k  0.85
3 
 3  2 
 
Equation: R(t )  0.85 sin  t 
3 
B. Graph one cycle of your equation. Which part corresponds to inhaling? Exhaling?
Since the period is 6 seconds, the value for B 
B. Graph one cycle of your equation. Which part corresponds to inhaling? Exhaling?
.85
1.5
3
4.5
6
seconds
-.85
Inhale (0, 3) Exhale (3, 6)
4. Find the exact value, if possible:
1
1
 3 
  
A. cos   = 
B. tan 
 = 
3
2
 4 
 6 
sin(2)  0.9093
C.
5. State if the two expressions are equivalent, C for correct and I for incorrect.
I A. sin  1   tan( x)
C B. 1  tan( x)
I
C. sin 2 ( x)  sin( x 2 )
 cos(x) 


cot(x)
C D. sin ( x)  (sin( x))
2
2
I G. sin 1 ( x)  (sin( x)) 1
I
E. sin 2 ( x)  sin(sin( x))
I
H. tan(cos1 ( x))  sin( x)
I
C
1
sin( x)

I. sin x 1  sin  1 
 x
F. sin 1 ( x) 
6. Find exact values for each: These are inverse function notation.
A. cos1 (1) = 0 because cos0  1
 1    
B. sin 1   =   
 2   6
1
 

2
 6
because sin  
7. Evaluate each (your answer will be in terms of x ):
A. sin(arcsin ( x)) = x
B. cos  tan 1 x 
Hint: make a right triangle and fill in the missing side, also
x
x
1
Since tan 1 x represents and angle,  , we label the opposite side x and the adjacent side 1 and
1
use the Pythagorean theorem to find the hypotenuse which is x 2  1 . Therefore cos  
x2 1
8. An angle A is drawn in standard position with its terminal side in quadrant III.
If tan A 
3
, find the value of sin A . Hint: make a right triangle and fill in the missing side
5
Since given the tangent of angle A is 3/5 we know that the opposite side is 3 and the adjacent side is 5.
3
Using the Pythagorean theorem we find the hypotenuse to be 34 . So sin A  
(terminal side is
34
9. Solve for t : tan(3t )  1 where 0  t 

2
.
5
4
12 12
Must also find all the reference angles that fall into this interval.
arctan(tan (3t )  arctan 1  3t  arctan 1 

t 

,
```