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Mathematics Dictionary
Dr. K. G. Shih
Examples in Trigonometry
Questions


  • Q01 | - Bisector AD of triangle ABC : AD = 2*b*c*cos(A/2)/(b + c)
  • Q02 | - Bisectors make angles x,y,z with sides. a*sin(2*x) +b*sin() + c*... = 0
  • Q03 | - Unit circle : Radius AB and P a point on circle. PA^2 + PB^2 = ?
  • Q04 | - Unit circle : Equilateral triangle ABC and P a point on circle.
  • Q05 | - Unit circle : Square ABCD and P a point on circle.
  • Q06 | - Unit circle : Regular hexagon ABCDEF and P a point on circle.
  • Q07 | - Solve equations : sin(x)^3 = sin(y) and cos(x)^3 = cos(y)
  • Q08 | - Solve equations : 2^(sin(x)+cos(y)) =0 and 16^(sin(x)^2 + cos(y)^2) =4
  • Q09 | - Sketch y = cos(x) + sin(x) + Abs(cos(x) - sin(x))
  • Q10 | - If cos(2*x) = sin(x) + Abs(sin(x)), find sin(x)
  • Q11 | - Find (1 + tan(15))/(1 - tan(15)) without calculator
  • Q12 | - Sketch y = cos(x) - Abs(cos(x)) between 0 and 2*pi
  • Q13 | - Find S(n) = Sum[((1/2)^n)*cos(n*pi/2)] for n = 1 to infinite
  • Q14 | - Find S(n) = Sum[((1/2)^n)*sin(n*pi/2)] for n = 1 to infinite
  • Q15 | -
    • Answers


    Q01. Bisector AD of triangle ABC : AD = 2*b*c*cos(A/2)/(b + c)
    Hint
    • Use area of triangle = b*c*sin(A)/2
    Solution
    • Area ABC = b*c*sin(A)/2
    • Area ABD = c*AD*sin(A/2)/2
    • Area ACD = b*AD*sin(A/2)/2
    • Area ABC = Area ABD + area ACD
    • b*c*sin(A)/2 = c*AD*sin(A/2)/2 + b*AD*sin(A/2)/2
    • sin(A) = 2*sin(A/2)*cos(A/2)
    • b*c*sin(A/2)*cos(A/2) = AD*(c + b)*sin(A/2)/2
    • AD = 2*b*c*cos(A/2)/(b + c)

    Go to Begin

    Q02. Bisectors make angle x,y,z with sides.
    Construction
    • Draw triangle ABC
    • Bisector AD make angle ADB = x
    • Bisector BE make angle BEC = y
    • Bisector CF make angle CFA = y
    Keywrods
    • sin(A+B)*sin(A-B) = cos(2*B) - cos(2*A)
    Proof : a*sin(2*x) + b*sin(2*y) + c*sin(2*z) = 0
    • In triangle ABD
      • Angle x = 180 - B - A/2
      • Angle x = C + A/D
      • Hence 2*x = 180 - (B - C)
    • Similarly : 2*y = 180 - (C - A) and 2*z = 180 - (A - B)
    • a*sin(2*x) + b*sin(2*y) + c*sin(2*z)
    • = 2*R*sin(A)*sin(B-C) + 2*R*sin(B)*sin(C-A) + 2*R*sin(C)*sin(A-B)
    • = 2*R*sin(B+C)*sin(B-C) + 2*R*sin(C-A)*sin(C-A) + 2*R*sin(A+B)*sin(A-B)
    • = R*(cos(2*C)-cos(2*b)+cos(2*A)-cos(2*C)+cos(2*B)*cos(2*A))
    • = R*0
    • = 0

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    Q03. Unit circle : AB is diameter and P on circle, PA^2 + PB^2 = 2*2
    Method 1 : Triangle APB is right triangle
    • PA^2 + PB^2 = AB^2
    • Since AB = 2
    • Hence PA^2 + PB^2 = 4
    Method 2 : Using cosine law
    • Let O be the center. Then AO = PO = 1
    • Triangle POA
      • Let angle POA = x
      • PA^2 = AO^2 + PO^2 - 2*AO*PO*cos(x) = 2 - 2*cos(x)
    • Triangle POB
      • Let angle POB = 180 - x
      • PB^2 = BO^2 + PO^2 - 2*BO*PO*cos(180-x) = 2 + 2*cos(x)
    • Hence PA^2 + PB^2 = 2*2

    Go to Begin

    Q04. Unit circle : Inscribed triangle ABC and P on circle, PA^2 +PB^2 +PC^2 =2*3

    Using cosine law
    • Let O be the center. Then AO = PO = 1
    • Triangle POA
      • Let angle POA = x
      • PA^2 = AO^2 + PO^2 - 2*AO*PO*cos(x) = 2 - 2*cos(x)
    • Triangle POB
      • Let angle POB = 120 - x
      • PA^2 = BO^2 + PO^2 - 2*BO*PO*cos(120-x) = 2 - 2*cos(120 - x)
    • Triangle POC
      • Let angle POC = 240 - x or 120 + x
      • PC^2 = CO^2 + PO^2 - 2*CO*PO*cos(120+x) = 2 - 2*cos(120 + x)
    • Hence PA^2 + PB^2 + PC^2 = 2*3 - 2*(cos(x) + cos(120-x) + cos(120+x))
    • = 2*3 - 2*(cos(x) + 2*cos(120)*cos(x))
    • = 2*3

    Go to Begin

    Q05. Unit circle : Inscribed Square ABCD and P on circle, PA^2 +PB^2 +PC^2 =2*3

    Using cosine law
    • Let O be the center. Then AO = PO = 1
    • Triangle POA
      • Let angle POA = x
      • PA^2 = AO^2 + PO^2 - 2*AO*PO*cos(x) = 2 - 2*cos(x)
    • Triangle POB
      • Let angle POB = 90 - x
      • PA^2 = BO^2 + PO^2 - 2*BO*PO*cos(90-x) = 2 - 2*sin(x)
    • Triangle POC
      • Let angle POC = 180 - x
      • PC^2 = CO^2 + PO^2 - 2*CO*PO*cos(180+x) = 2 + 2*cos(x)
    • Triangle POD
      • Let angle POD = 270 - x or 90 + x
      • PC^2 = CO^2 + PO^2 - 2*CO*PO*cos(90+x) = 2 + 2*sin(x)
    • Hence PA^2 + PB^2 + PC^2 + PD^2
    • = 2*4 - 2*(cos(x) + sin(x) - cos(x) - sin(x))
    • = 2*4

    Go to Begin

    Q06. Unit circle : Regular hexagon ABCDEF and P a point on circle

    Using cosine law
    • Let O be the center. Then AO = PO = 1
    • Triangle POA
      • Let angle POA = x
      • PA^2 = AO^2 + PO^2 - 2*AO*PO*cos(x) = 2 - 2*cos(x)
    • Angle POB = 1*60 - x, PB^2 = 2 - 2*cos(60-x)
    • Angle POC = 2*60 - x, PC^2 = 2 - 2*cos(120-x)
    • Angle POD = 3*60 - x, PD^2 = 2 - 2*cos(180-x)
    • Angle POE = 4*60 - x, PE^2 = 2 - 2*cos(240-x)
    • Angle POF = 5*60 - x, PF^2 = 2 - 2*cos(300-x)
    • using sum to product formula
    • PA^2 + PB^2 + PC^2 + PD^2 + PE^2 + PF^2 = 2*6 - 2(....) = 2*6

    Go to Begin

    Q07. Solve equations : sin(x)^3 = sin(y) and cos(x)^3 = cos(y)
    Solution
    • Squre both sides of two equations
    • sin(x)^6 = sin(y)^2
    • cos(x)^6 = cos(y)^2
    • Add them : sin(x)^6 + cos(x)^6 = sin(y)^2 + cos(y)^2 = 1
    • (sin(x)^2 +cos(x)^2)^3 = cos(x)^6 +sin(x)^6 +3*sin(x)*cos(x)*(cos(x)+sin(x))
    • 1 = 1 + 3*sin(x)*cos(x)*(cos(x)+sin(x))
    • 3*sin(x)*cos(x)*(cos(x)+sin(x)) = 0
    • sin(x) = 0 and x = n*pi
    • cos(x) = 0 and x = n*pi+pi/2
    • sin(x) + cos(x) = 0
      • tan(x) = -1.
      • Hence x = (2*n+1)*pi - pi/4 in 2nd quadrant
      • Hence x = (2*n)*pi - pi/4 in 4th quadrant
      • Or combine above, we have x = n*pi - pi/4

    Go to Begin

    Q08. Solve equations : 2^(sin(x)+cos(y)) = 0 and 16^(sin(x)^2 + cos(y)^2) = 4
    • Sine 2^0 = 1 and 16^2 = 4^2
    • Hence sin(x) + cos(y) = 0 ...... (1)
    • 2*(sin(x)^2 + cos(y)^2) = 1 .... (2)
    • (2) - equare of (1) : (sin(x) - cos(y))^2 = 1
    • sin(x) - cos(y) = +1 ... (3)
    • sin(x) - cos(y) = -1 ... (4)
    • Solve (1) and (3)
      • sin(x) = +1/2 and x = 2*n*pi + pi/6 or x = (2*n+1)*pi - pi/6
      • cos(y) = -1/2 and y = (2*n+1)*pi - pi/6 or y = (2*n+1)*pi + pi/6
    • Solve (1) and (4)
      • sin(x) = -1/2
      • cos(y) = +1/2 and y = 2*n*pi + pi/6 or y = 2*n*pi - pi/6

    Go to Begin

    Q09. Sketch y = (cos(x) + sin(x))/2 + Abs(cos(x) - sin(x))/2

    Hint
    • Abs(+sin(x)) = +sin(x)
    • Abs(-sin(x)) = -(-sin(x)) = sin(x)
    Sketch notes
    • If 0 GT x LT pi/4, cos(x) GT sin(x)
      • Sqr(cos(x) - sin(x)) = +(cos(x) - sin(x))
      • y = (cos(x) + sin(x))/2 + (cos(x) - sin(x))/2 = cos(x)
      • Hence the sketch is y = cos(x)
    • If pi/4 GT x LT 5*pi/4, cos(x) LT sin(x)
      • Sqr(cos(x) - sin(x)) = -(cos(x) - sin(x))
      • y = (cos(x) + sin(x))/2 - (cos(x) - sin(x))/2 = sin(x)
      • Hence the sketch is y = sin(x)
    • If 5*pi/4 GT x LT 2*pi, cos(x) GT sin(x)
      • Sqr(cos(x) - sin(x)) = +(cos(x) - sin(x))
      • y = (cos(x) + sin(x))/2 + (cos(x) - sin(x))/2 = cos(x)
      • Hence the sketch is y = cos(x)

    Go to Begin

    Q10. If cos(2*x) = sin(x) + Abs(sin(x)), find sin(x)

    Solution
    • 0 GT x LT pi and sin(x) GT 0 : Sqr(sin(x)) = +sin(x)
      • cos(2*x) = sin(x) + sin(x)
      • 1 - 2*sin(x)^2 = 2*sin(x)
      • 2*sin(x)^2 + 2*sin(x) - 1 = 0
      • Hence sin(x) = (-2 + Sqr(2^2 - 4*2*(-1))/(2*2) = (-1 + Sqr(3))/2
      • Or sin(x) = (-2 - Sqr(2^2 - 4*2*(-1))/(2*2) = (-1 - Sqr(3))/2
    • pi GT x LT 2*pi and sin(x) LT 0 : Sqr(sin(x)) = -sin(x)
      • cos(2*x) = sin(x) - sin(x)
      • 1 - 2*sin(x)^2 = 0
      • 2*sin(x)^2 = 1
      • Hence sin(x) = 1/Sqr(2) = Sqr(2)/2
      • Or sin(x) = -Sqr(2)/2

    Go to Begin

    Q11. Find (1 + tan(15))/(1 - tan(15)) without calculator

    Keyword
    • tan(A+B) = (tan(A)+tan(B))/(1 - tan(A)*tan(B))
    • tan(45) = 1
    Proof
    • (1 + tan(15))/(1 - tan(15))
    • = (tan(45) + tan(15))/(1 - tan(45)*tan(15))
    • = tan(45 + 15)
    • = tan(60)
    • = Sqr(3)

    Go to Begin

    Q12. Sketch y = cos(x) - Abs(cos(x)) between 0 and 2*pi

    Keyword
    • Abs(-x) = -(-x) = x
    Sketch notes
    • if cos(x) GT 0, then Abs(cos(x)) = cos(x)
    • If cos(x) LT 0, then Abs(cos(x)) = -cos(x)
    • 0 to pi/2, cos(x) is positive
      • y = cos(x) - Abs(cos(x)) = cos(x) - cos(x) = 0
      • Hence between 0 and pi/2, the sketch is y = 0
    • pi/2 to 3*pi/2, cos(x) is negative
      • y = cos(x) - Abs(cos(x)) = cos(x) - (-cos(x)) = 2*cos(x)
      • Hence between pi/2 and 3*pi/2, the sketch is y = 2*cos(x)
    • 3*pi/2 to 2*pi, cos(x) is positive
      • y = cos(x) - Abs(cos(x)) = cos(x) - cos(x) = 0
      • Hence between 3*pi/2 and 2*pi, the sketch is y = 0

    Go to Begin

    Q13. Find S(n) = Sum[((1/2)^n)*cos(n*pi/2)] for n = 1 to infinite

    Keywords
    • cos(n*pi/2) = +0, if n is odd
    • cos(n*pi/2) = -1, if n is 2,6,10, ...
    • cos(n*pi/2) = +1, if n is 4,8,12, ...
    • GP : S(n) = a/(1 - r) if n goes to infinite and a is 1st term
    Solution
    • S(n) = Sum[((1/2)^n)*cos(n*pi/2)]
    • = 0 - (1/2)^2 + 0 + (1/2)^4 + 0 - (1/2)^6 + ....
    • = -((1/2)^2)*(1 - (1/2)^2 + (1/2)^4 - .....)
    • Since common ratio = -(1/2)^2
    • Hence S(n) = (-1/4)*(1/(1 + (1/2)^2))
    • = (-1/4)*(4/5)
    • = -1/5

    Go to Begin

    Q14. Find S(n) = Sum[((1/2)^n)*sin(n*pi/2)] for n = 1 to infinite

    Keywords
    • sin(n*pi/2) = +1, if n is 1, 5, 9, ....
    • sin(n*pi/2) = -1, if n is 3, 7, 11, ...
    • sin(n*pi/2) = +0, if n is even
    • GP : S(n) = a/(1 - r) if n goes to infinite and a is 1st term
    Solution
    • S(n) = Sum[((1/2)^n)*sin(n*pi/2)]
    • = (1/2) + 0 - (1/2)^3 + 0 + (1/2)^5 + 0 - (1/2)^7 + ....
    • = (1/2)*(1 - (1/2)^2 + (1/2)^4 - .....)
    • Since common ratio = -(1/2)^2
    • Hence S(n) = (1/2)*(1/(1 + (1/2)^2))
    • = (1/2)*(4/5)
    • = 2/5

    Go to Begin

    Q15.


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