Problems in Calculus II

Chapter 1. Trigonometry Foundations.

    Problem 0.1 Give the 6-line definitions. List elementary properties.

    Ans.: Given h, what is o ? you say, o = h sinq.
Given h what is a? you say, a = h cosq.
Given a what is o? you say, o = a tanq.
Given a what is h? you say, h = a secq.
Given o what is h? you say, h = o cscq.
Given o what is a? you say, a = o cotq.

In the three-triangles diagram:
(1) All co-functions are on right side.
    Hence sinq = cos(90°-q), tanq = cot(90°-q),
    cotq = tan(90°-q), etc.
(2) Product of two non neighbors = middle guy.
    E.g., cosq tanq = sinq, sinq cscq = 1,
    cosq secq = 1, tanq cotq = 1.
(3) Dividing one guy by a neighbor = the other neighbor.
    E.g., sinq/cosq = tanq, tanq/secq = sinq,
    secq/cscq = tanq, cscq/secq = cotq,
    cotq/cosq = cscq.
(4) For each upside-down triangle, sum of squares of top two corners equals square of bottom corner.
    Thus sin²q + cos²q = 1, tan²q + 1 = sec²q, cot²q + 1 = csc²q.

    Problem 0.2 Simplify the following ratios to single trig function of an acute angle, assuming q is acute: cos (90°+q) , sin (90°+q), tan (p-q), sec(p+q), cos (-q), tan(270°-q), sin 1000°, cot 1000°, cos (-700°). sin(-q), sin(-180°-q).
    Ans.: Here are the rules:

(a) First, fix the ± sign by assuming q is acute, 
    and find out which quadrant it is at:
    First quadrant: all are positive.
    2nd quadrant: only sin (and csc) are positive.
    3rd quadrant: only tan (and cot) are positive.
    4th quadrant: only cos (and sec) are positive.
(b) Next, if the angle has an odd multiple of 90°,
    the answer is the co-function.

Thus cos(90°+q) = -sinq.    sin(90°+q) = cosq.
    tan(p-q) = -tanq.      sec(p+q) = -secq.
    cos(-q) = cosq.             tan(270°-q) = cotq.
    sin 1000° = sin(1000°-2×360°) = sin 280° = -cos10°.
    cot 1000° = cot 280° = -tan 10°.
    cos(-700°) = cos(-700°+720°) = cos 20°.
    sin(-q) = -sinq.          sin(-180°-q) = sinq. 

    Problem 1. Prove the 12-line poem of compound angles.
    Ans.:
In the first diagram, let OA = 1, and ÐOCA = 90°. Line ADB cuts OC at K, and BDCE is a rectangle.
Let a = ÐCOE, b = ÐAOC.
Notice DOBK and DACK are similar, hence ÐCAK = a.
Then sin(a+b) = AB = AD+CE = AC cosa + OC sina
= sinb cosa + cosb sina.
    Hence sin(a+b) = sina cosb + cosa sinb . . . . (i)
    Again, cos(a+b) = OB = OE - CD
= OC cosa - AC sina.
    Hence cos(a+b) = cosb cosa - sinb sina . . . . (ii)

    In the second diagram, let OA = 1, BDCE be rectangle, AB cuts line ODC at K, and ÐOCA = 90°, but let a = ÐAOB and b = ÐBOC.
Notice DOKB, AKC are similar, hence ÐCAK = b.
Then sin(a-b) = AC = AE - BD = AB cosb - OB sinb .
    Hence sin(a-b) = sina cosb - cosa sinb . . . . (iii)
    Again, cos(a-b) = OC = OD + BE
= OB cosb + AB sinb.
    Hence cos(a-b) = cosa cosb + sina sinb . . . . (iv)

    (i)+(iii) then gives 2 sina cosb = sin(a+b) + sin(a-b) . . . . (v)
    (i)-(iii) gives 2 cosa sinb = sin(a+b) - sin(a-b) . . . . (vi)
    (ii)+(iv) gives 2 cosa cosb = cos(a+b) + cos(a-b) . . . . (vii)
    (iv)-(ii) gives 2 sina sinb = cos(a-b) - cos(a+b) . . . . (viii)

    Let x = a+b, y = a-b, then (v) to (viii) become:
     sin x + sin y = 2 sin (½)(x+y) cos (½)(x-y) . . . . (ix)
     sin x - sin y = 2 cos (½)(x+y) sin (½)(x-y) . . . . (x)
     cos x + cos y = 2 cos (½)(x+y) cos (½)(x-y) . . . . (xi)
     cos y - cos x = 2 sin (½)(x+y) sin (½)(x-y) . . . . (xi)

     Problem 2. Prove (ix) and (xi) directly with geometry.
    Ans.: Let OAB be a triangle where OA = OB = 1, and ÐAOB = a+b. Let M be the midpoint of AB. Let K be the point on AB such that ÐAOK = a and ÐBOK = b. Let C, D be the projections of A,B upon line OK. Let BDCE be a rectangle.
Then ÐOAB = ÐOBA = 90°-(½)(a+b).
WLOC Assume a > b, then ÐMOK = ÐCAK = ÐDBK
= (½)(a+b)-b = (½)(a-b).
Hence sina+sinb = AC+BD = AE AB cos(½)(a-b)
= 2 AM cos(½)(a-b)= 2 sin(½)(a+b) cos(½)(a-b).

    In the second diagram,
let N be the midpoint of CD. Then MN ^ ON.
Then cosa+cosb = OC+OD = 2 ON = 2 OM cos ÐMON
= 2 cos(½)(a+b) cos(½)(a-b)
because ÐBOM = ÐAOM = (½)ÐAOB = (½)(a+b)
and ÐMON = ÐAOM-a = (½)(b-a). QED

    Exercise 2.1. Prove (x) and (xii) similarly.

    Exercise 2.2. Use the diagrams of Problem 2 , but rename a and b differently, to prove (v) and (vii) directly.

    Problem 3. Prove all the double angle formulas for sin and cos.
    Ans.: First Proof. Put a = b into (i) and (iii) in the 12-line poem.
Then sin 2a = 2 sina cosa, and
cos 2a = cos²a - sin²a = (1-sin²a)-sin²a = 1-2sin²a
Also cos2a = cos²a - sin²a = cos²a -(1-cos²a) = 2cos²a -1.

     Direct Proof. In the diagram, let OAB be a triangle where OA = OB = 1, and ÐAOB = 2a. Let M be the midpoint of AB, D the projection of A on OB. Let C,N be the midpoints of AD and BD. Then CDNM is a rectangle.
    Notice ÐBAD = 90°-ÐB = ÐBOM = a.
Then sin2a = AD = 2 AC = 2 AM cosÐBAD = 2 sina cosa
because AM = sina.
    Again, cos2a = OD = ON - CM = OM cosa -MA sinÐBAD = cos²a - sin²a.

    Problem 4. If s is semiperimeter, R and r are circumradius and inradius of a triangle ABC, a=A/2, b=B/2, g=C/2, prove:
(a) 2R = a/sinA = b/sinB = c/sinC.
(b)area of ABC = D = abc/4R
(c) s = 4R cosa cosb cosg
(d) r = 4R sina sinb sing
    Ans.: Let O be the circumcenter of ABC. Then OA=OB=OC. Let ÐOBC = ÐOCB = a, ÐOCA = ÐOAC = a, ÐOBA = ÐOAB = g. Then a+b = C, b+g = A, g+a = B.
Adding first two equations, then subtract third, we get 2b = A_C-B = (A+B+C)-2B = 180°-B. Hence b = 90°-B.
    Hence ÐAOC = 180°-2b = 2B. Let B' be the midpoint of AC, then Ds OAB' and OCB' are congruent by SAS, hence OB' bisects ÐAOC, giving ÐAOB' = B. Hence R = OA = AB' / sinB = b/2sinB, or 2R = b/sinB. Similarly, one can prove 2R = a/sinA and c/sinC. (a) is proved.
    (b) The length of the altitude AD is AD = AB sinB = c sinB.
Hence D = area of ABC = (½)AD . BC = (½) ac sinB.
But sinB = b/2R from law of sines, so D = (½)ac b/2R = a b c/4R.
    (c) s = (½)(a+b+c) = (½) 2R (sin A + sin B + sin C)
= R (2 sin(a+b) cos(a-b) + 2 sing cosg)
= R (2 cosg cos(a-b) + 2 sing cosg)
= 2 R cosg [cos(a-b) + cos(a+b)]
= 2 R cosg 2 cosa cosb = 4 R cosa cosb cosg
    (d) Let I be the incenter, which is at distance r from the sides of DABC.
Then abc/4R = D = area(AIB)+area(BIC)+area(CIA)
= (½) r(a + b + c) = r s.
\ r = D/s = abc/4Rs = 2R sinA 2R sinB 2R sinC/(4R . 4R cosa cosb cosg)
= (½)R 2 sina cosa 2 sinb cosb 2 sing cosg / cosa cosb cosg
= 4 R sina sinb sing.

    Problem 5. Given two angles and a side of a triangle, find all the other sides and altitudes.
    Ans.: Let B,C,a be given. A of course = 180°-B-C.
Law of sines says b/sinB = a/sinA, hence we can find b, and similarly, find c.
To find the altitude AD, let its length be h. Then a = BC = BD + CD = h cotB + h cotC.
Hence h = a/(cotB + cotC) (N.B. Memorize this ASA formula! )
= a sinB sinC / (cosB sinC + cosC sinB) = a sinB sinC / sin(B+C)
= a sinB sinC / sinA (because A = p-B-C)
    Another way of getting h in terms of B,C,a only, is:
h = AC sin C = b sin C = a (sinB/sinA) sinC = a sinB sinC / sin (B+C).
    (Personally, I like the first proof, just because I discovered it myself. But the 2nd proof makes the first proof looks like a dum-dum.) :(

    Problem 6. (A 3-D survey problem.) AB is a baseline on the Biola soccer field. CD is the La Mirada City Hall tower. C is level with AB. The elevation angles of D at A and B are a and b. The azimuthal angles CAB and CBA are g and d. Find the height h of the tower, interms of a, b, g, d, and AB.
    Ans.: AC = h cota, BC = h cotb. Let c = AB, then
c = AC cosg + BC cosd = h cota cosg + h cotb cosd.
Hence h = c / (cota cosg + cotb cosd). (Too easy a problem!)

     Problem 7. C is a Centauris, the closest star visible to human eyes. It is 4.4 light years away from us. Let A, B be the two points on earth's orbit such that AB goes through the sun S and AB is perpendicular to CS. Find ÐACS in seconds. Data: 60 seconds = 1 minute. 60 minutes = 1 degree. Speed of light = 186000 miles per second. 1 year = 365.256 days. Distance of earth from sun is 93 million miles.
    Ans.: AS = 9.3 × 10⁷ miles.
SC = 4.4 year × 365.256 days/yr × 24 hrs/day × 3600 seconds/hr × 186000 miles/sec.
= 2.583 ×10¹³ miles.
    Hence cotÐACS = SC/AS = 2.583×10¹³ / 9.3×10⁷ = 277711.5
Thus ÐACS = (1/277711.5) radians×57.2958°/radian×3600 secs/° = 0.74 seconds.

    Problem 8. Prove all the double and half angle formulas.
    Ans.: sin2q = sin(q+q) = sinq cosq + cosq sinq = 2 sinq cosq.
cos2q = cosq cosq - sinq sinq = cos²q-sin²q
= cos²q-(1-cos²\1) = 2cos²q-1
= (1-sin²q)-sin²q = 1 - 2sin²q.
tan2q = sin2q/cos2q = 2 sinq cosq / (cos²q-sin²q)
= 2 tanq / (1-tan²q).
    From cos2q formula, we can replace q by (q/2), then
cosq = 2 cos²(q/2)-1 = 1- 2sin²(q/2). Solving each, we get
cos(q/2) = Ö[(½)(1+cosq)], sin(q/2) = Ö[(½)(1-cosq)].
Hence
tan(q/2) = Ö[(1-cosq)/(1+cosq)]

    Problem 9. Express sin2q, cos2q, and tanq in terms of tanq only.
    Ans.: Let t = tanq.
Then sin2q = 2 sinq cosq = 2 t cos²q = 2 t/ sec²q = 2 t/(1+t²).
cos2q = cos²q-sin²q = cos²q(1-t²) = (1-t²)/sec²q
= (1-t²)/(1+t²)
tan2q = sin2q/cos2q = 2 t/(1-t²).

    Problem 10. Describe the process by which Zu ChongZhi in China (700+ a.d.) found p to 7 significant figures. (pronounced: dsoo choong jee)
    Ans.: It is possible to hand-calculate squareroots. E.g., for Ö2:

 
         1. 4  1  4  2  1

     1 | 2.00,00,00,00,00
       | 1
       |-----
    24 | 1 00
       | - 96
       |-----
   281 |    4 00
       |  - 2 81
       |    -------
  2824 |    1 19 00
       |  - 1 12 96
       |    ----------
 28282 |       6 04 00
       |     - 5 65 64
       |     ------------
282841 |         38 36 00
       |       - 28 28 41
       |       ------------- etc.

    Problem 11. Describe a process of computing p to as many places of decimals as we want.
    Ans.: tan(a+b) = sin(a+b)/cos(a+b)
= [sina cosb + cosa sinb] / [cosa cosb - sina sinb]
= [tana + tanb] / [1 - tana tanb] . . . . (11.1)
    Let x = tana, y = tanb, then a = arctan x, b = arctan y, and the graph of arctan function is the graph of tan function from -90° to 90°, reflected across the line y = x.
Then (11.1) can be rewritten as a + b = arctan [ (tana + tanb) / (1 - tana tanb)], or
arctan x + arctan y = arctan [(x+y)/(1-xy)].
Now tan(p/4) = 1, hence p/4 = arctan(1) = arctan [ (1/2 + 1/3)/(1-(1/2)(1/3))] (Clever!)
= arctan(1/2) + arctan(1/3).
= [(½) - (½)³/3 + (½)⁵/5 - (½)⁷/7 +- . . . ]
    + [(1/3) -(1/3)³/3 + (1/3)⁵/5 -+ . . . ] which can be hand-computed.

    Problem 12. Prove that 4 arctan(1/5) - arctan(1/239) = p/4.
    Ans.: arctan(1/5) + arctan(1/5) = arctan[(1/5 + 1/5)/(1-(1/5)(1/5))]
= arctan (10/24) = arctan(5/12).
    Hence 4 arctan(1/5) = arctan(5/12) + arctan(5/12)
= arctan[ 2× (5/12) / (1- (5/12)²)] = arctan ((5/6) / (119/144))
= arctan (720 / 714) = arctan (120/119)
     Hence 4 arctan(1/5) - arctan(1/239) = arctan(120/119)-arctan(1/239)
= arctan[ ((120/119)-(1/239)) / (1 + (120/119)(1/239)) ]
= arctan[ (120×239 -119) / (119×239 + 120) ]
= arctan[ 28561 / 28561 ] = p/4.