Calculus of One and Several Variables

2.4 Exercises

1. Find cosθ and sinθ using only knowledge of angles in the first quadrant for θ ∈
   { } 23π, 34π, 56π,π, 76π, 54π, 43π, 32π, 53π, 74π, 11π6-,2π
   .
2. Prove cos²θ =
   1+ cos2θ --------- 2
   and sin²θ =
   1− cos2θ --------- 2
   .
3. π∕12 = π∕3 −π∕4. Therefore, from Problem 2, cos
   (π∕12)
   =
   ∘----(√---) 1+---3∕2-- 2
   . On the other hand,

   cos(π∕12) = cos(π∕3− π∕4) = cosπ ∕3 cosπ∕4+ sin π∕3sin π∕4

   and so cos

   (π∕12)
   =
   √ - 2
   ∕4 +
   √ - 6
   ∕4. Is there a problem here? Please explain.

4. Prove 1 + tan²θ = sec²θ and 1 + cot²θ = csc²θ.
5. Prove that sinxcosy =
   1 2
   (sin(x+ y)+ sin(x− y))
   .
6. Prove that sinxsiny =
   1 2
   (cos(x− y)− cos(x+ y))
   .
7. Prove that cosxcosy =
   1 - 2
   (cos (x + y)+ cos(x − y))
   .
8. Using Problem 5, find an identity for sinx − siny.
9. Suppose sinx = a where 0 < a < 1. Find all possible values for
   1. tanx
   2. cotx
   3. secx
   4. cscx
   5. cosx
10. Solve the equations and give all solutions.
    1. sin
       (3x)
       =
       1 2
    2. cos
       (5x)
       =
       √- -3- 2
    3. tan
       (x)
       =
       √- 3
    4. sec
       (x)
       = 2
    5. sin
       (x+ 7)
       =
       √2 --- 2
    6. cos²
       (x)
       =
       1 2
    7. sin⁴
       (x)
       = 4
11. Sketch a graph of y = sinx.
12. Sketch a graph of y = cosx.
13. Sketch a graph of y = sin2x.
14. Sketch a graph of y = tanx.
15. Find a formula for sinxcosy in terms of sines and cosines of x + y and x − y.
16. Using Problem 2 graph y = cos²x.
17. If f
    (x)
    = Acosαx + B sinαx, show there exists ϕ such that

    ∘ --2---2- f (x) = A + B sin(αx + ϕ).

    Show there also exists ψ such that f

    (x)
    =
    √ -2---2- A + b
    cos
    (αx + ψ)
    . This is a very important result, enough that some of these quantities are given names.
    √ -2----2- A + B
    is called the amplitude and ϕ or ψ are called phase shifts.

18. Using Problem 17 graph y = sinx +
    √3
    cosx.
19. Give all solutions to sinx +
    √3-
    cosx =
    √3
    . Hint: Use Problem 18.
20. As noted above 45^o is the same angle as π∕4 radians. Explain why 90^o is the same angle as π∕2 radians. Next find a simple formula which will change the degree measure of an angle to radian measure and radian measure into degree measure.
21. Find a formula for tan
    (θ +β )
    in terms of tanθ and tanβ
22. Find a formula for tan
    (2θ)
    in terms of tanθ.
23. Find a formula for tan
    ( ) θ 2
    in terms of tanθ.
24. Show tan
    (4θ)
    =
    4tanθ − 4tan3θ 1−-6tan2θ+-tan4θ-
    . Now find x such that if tanθ = x, and tanβ =
    1 5
    , then 4β + θ =
    π 4-
    . This is the basis for a wonderful formula which has been used to compute π for hundreds of years.
25. The function, sin has domain equal to ℝ and range
    [− 1,1]
    . However, this function is not one to one because sin
    (x+ 2π)
    = sinx. Show that if the domain of the function is restricted to be
    [ ] − π, π 2 2
    , then sin still maps onto
    [− 1,1]
    but is now also one to one on this restricted domain. Therefore, there is an inverse function, called arcsin which is defined by arcsin
    (x)
    ≡ the angle whose sin is x which is in the interval,
    [ π- π] −2 ,2
    . Thus arcsin
    ( ) 1 2
    is the angle whose sin is
    1 2
    which is in
    [− π, π-] 2 2
    . This angle is
    π- 6
    . Suppose you wanted to find tan
    (arcsin(x))
    . How would you do it? Consider the following picture which corresponds to the case where x > 0.
    PICT

    Then letting θ = arcsin

    (x)
    , the thing which is wanted is tanθ. Now from the picture, you see this is
    √-x---- 1− x2
    . If x were negative, you would have the little triangle pointing down rather than up as in the picture. The result would be the same for tanθ. Find the following:
    1. cot
       (arcsin(x))
    2. sec
       (arcsin(x))
    3. csc
       (arcsin(x))
    4. cos
       (arcsin(x))
26. Using Problem 25 and the formulas for the trig functions of a sum of angles, find the following.
    1. cot
       (arcsin(2x))
    2. sec
       (arcsin(x+ y))
    3. csc
       ( ( )) arcsin x2
    4. cos
       (2arcsin (x))
    5. tan
       (arcsin (x) +arcsin (y))
    6. csc
       (arcsin(x)− arcsin(y))
27. The function, cos, is onto
    [− 1,1]
    but fails to be one to one. Show that if the domain of cos is restricted to be
    [0,π]
    , then cos is one to one on this restricted domain and still is onto
    [− 1,1]
    . Define arccos
    (x)
    ≡ the angle whose cosine is x which is in
    [0,π]
    . Find the following.
    1. tan
       (arccos(x ))
    2. cot
       (arccos (x))
    3. sin
       (arccos(x))
    4. csc
       (arccos(x))
    5. sec
       (arccos(x))
28. Using Problem 27 and the formulas for the trig functions of a sum of angles, find the following.
    1. cot
       (arccos (2x))
    2. sec
       (arccos(x + y))
    3. csc
       (arccos (x2))
    4. cos
       (arcsin(x)+ arccos(y))
    5. tan
       (arcsin (x) +arccos(y))
    6. csc
       (2arcsin(x)− arccos(y))
29. The function, arctan is defined as arctan
    (x)
    ≡ the angle whose tangent is x which is in
    ( − π-, π) 2 2
    . Show this is well-defined and is the inverse function for tan if the domain of tan is restricted to be
    ( π π ) − -,-- 2 2
    . Find
    1. cos
       (arctan (x ))
    2. cot
       (arctan (x ))
    3. sin
       (arctan (x))
    4. csc
       (arctan (x))
    5. sec
       (arctan (x))
30. Using the formulas for the trig functions of a sum of angles, find the following.
    1. cot
       (arctan (2x ))
    2. sec
       (arctan (x + y))
    3. csc
       (arccos (x2))
    4. cos
       (2arctan(x)+ arcsin(y))
    5. tan
       (arctan(x)+ 2arccos (y))
31. The graphs of tan and cot suggest that these functions are periodic of period π verify that this is indeed the case using the identities presented.
32. Give another argument which verifies the Pythagorean theorem by supplying the details for the following argument³ . Take the given right triangle and situate copies of it as shown below. The big four sided figure which results is a rectangle because all the angles are equal. Now from the picture, the area of the big square equals c², the area of each triangle equals ab∕2, since it is half of a rectangle of area ab, and the area of the inside square equals
    (b− a)
    ². Here a,b, and c are the lengths of the respective sides. Therefore,

    c2	= 4 (ab∕2) + (b− a) 2
    	= 2ab + b2 + a2 − 2ab = a2 + b2.

    PICT

33. Another very simple and convincing proof of the Pythagorean theorem⁴ is based on writing the area of the following trapezoid two ways. Sum the areas of three triangles in the following picture or write the area of the trapezoid as
    (a+ b)
    a +
    1 2
    (a+ b)
    (b− a)
    which is the sum of a triangle and a rectangle as shown. Do it both ways and see the Pythagorean theorem appear.
    PICT

34. Make up your own proof of the Pythagorean theorem based on the following picture.
    PICT

35. If A,B are two angles in a triangle, and cos(A) = cos(B), explain why A = B. Hint: Note that cos is one to one on
    [0,π]
    and all angles of a triangle have radian measure less than π.
36. An isosceles triangle is one which has two equal sides. For example the following picture is of an isosceles triangle
    PICT

    the two equal sides having length a. Show the “base angles” θ and α are equal. Hint: You might want to use the law of cosines and the observation in the above problem.

37. If x is small and positive, explain why tanx − x ≥ 0.