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Mathematics Dictionary
Dr. K. G. Shih

Topics including trigonometric functions
Subjects


  • TR 14 01 | - Lim[sin(x)/x] = 1 as x tends to 0
  • TR 14 02 | - Find sin(18) and cos(18) without calculator
  • TR 14 03 | - Special angle 9° and 36°
  • TR 14 04 | - cos(2*x) + sin(x) = q has real roots
  • TR 14 05 | - x^2 - 2*x*sin(pi*x/2) + 1 = 0 has real roots
  • TR 14 06 | - Find minimum and maximum of (sic(A)^2 + tan(A))/(sec(A)^2 + tan(A))
  • TR 14 07 | - Find minimum of 3 - 2*cos(A) + cos(A)^2
  • TR 14 08 | - If cos(2*A) = Sqr(2)/3, find cos(A)^4 + sin(A)^4
  • TR 14 09 | - sin(1) + sin(2) + sin(3) + ... + sin(359) + sin(360) = ?
  • TR 14 10 | - Max. and min. of cos(x)^6 + sin(x)^6 if x between 0 and pi/2
  • TR 14 11 | - Max. and min. of cos(x)^2 -4*cos(x)*sin(x) -3*cos(x)^2. x between 0 & pi/2
    • Answers


    TR 14 01. Lim[sin(x)/x] = 1 as x tends to zero

    Text
    Go to Begin

    TR 14 02. Sin(18) and cos(18)

    Soultion
    • Let x = 18 and then 90 = 5*x.
    • 3*x = 90 - 2*x.
    • Take cosine on both sides we have cos(3*x) = cos(90-2*x).
    • Hence cos(3*x) = sin(2*x).
    • 4*cos(x)^3 - 3*cos(x) = 2*cos(x)*sin(x).
    • Factor out cos(x), we have 4*cos(x)^2 - 2*sin(x) - 3 = 0.
    • 4*(1-sin(x)^2) - 2*sin(x) - 3 = 0.
    • 4*sin(18)^2 + 2*sin(18) - 1 = 0.
    • Hence sin(18) = (Sqr(5) - 1))/4 since sin(18) is positive.
    Note
    • Both sides take sine : sin(3*x) = sin(90-2*x).
    • sin(3*x) = cos(2*x).
    • Hence 3*sin(x) - 4*sin(x)^3 = 1 - 2*sin(x)^2.
    • Hence 4*sin(x)^3 - 2*sin(x)^2 - 3*sin(x) + 1 = 0.
    • This is requiredd to solve a cubic equation.
    Find cos(18)
    • cos(18) = Sqr(1 - sin(18)^2)
    • cos(18) = Sqr(1 - (6 - 2*Sqr(5))/16)
    • cos(18) = Sqr((16 - 6 + 2*Sqr(5))/16)
    • cos(18) = Sqr(10+2*Sqr(5))/4
    Verify
    • sin(18)^2 + cos(18)^2 = (Sqr(5) - 1)^2/16 + (10 + 2*Sqr(5))/16
    • = (6 - 2*Sqr(5))/16 + (10 + 2*Sqr(5))/16
    • = 16/16
    • = 1.

    Go to Begin

    TR 14 03. Special angle 9° and 36°

    Find sin(9) and cos(9)
    • cos(9) = cos(18/2) = Sqr((1 + cos(18))/2) = ?
    • Sin(9) = sin(18/2) = Sqr((1 - cos(18))/2) = ?
    Find sin(36) and cos(36)
    • cos(36) = cos(2*18)
    • = 2*cos(18)^2 - 1
    • = 2*(10 + 2*Sqr(5))/16 - 1.
    • = (Sqr(5) +1)/4
    • Sin(36) = sin(2*18)
    • = 2*sin(18)*cos(18)
    • = 2*(Sqr(5) - 1)*(10 + 2*Sqr(5))/16.
    • = Sqr(10 - 2*sqr(5))/4
    Summary
    • sin(18) = (Sqr(5) - 1)/4.
    • cos(18) = Sqr(10+2*Sqr(5))/4.
    • sin(36) = cos(54) = Sqr(10 - 2*Sqr(5))/4.
    • cos(36) = sin(54) = (Sqr(5) + 1)/4

    Go to Begin

    TR 14 04. cos(2*x) + sin(x) = q has real roots

    Solution
    • 1 -2*sin(x)^2 + sin(x) - q = 0
    • 2*sin(x)^2 - sin(x) + (q - 1) = 0
    • Discriminant D = (-1)^2 - 4*2*(q-1) = 1 - 8*q + 8 = 9 - 8*q
    • Hence it has real roots if 9 - 8*q GE 0
    • Hence q LE 9/8
    • Case 1 :
      • sin(x) = (-(-1) + Sqr(D))/(2*2) = (1 + Sqr(9 - 8*q))/2
      • Since sin(x) GE -1 or LE 1
      • Hence (1 + Sqr(9 - 8*q))/4 GE -1 and Sqr(9 - 8*q) GE -5 and q LE 9/8
      • Hence (1 + Sqr(9 - 8*q))/4 LE +1 and Sqr(9 - 8*q) GE +3 and q GE 0
      • Hence 0 GE q LE 9/8
    • Case 2 :
      • sin(x) = (1 - Sqr(5 - 8*q))/2
      • Since sin(x) GE -1 or LE 1
      • Hence (1 - Sqr(9 - 8*q))/4 GE -1 and -Sqr(9 - 8*q) GE -5 and q GE -2
      • Hence (1 - Sqr(9 - 8*q))/4 LE +1 and -Sqr(9 - 8*q) GE +3 and q LE 9/8
      • Hence -2 GE q LE 9/8

    Go to Begin

    TR 14 05. x^2 - 2*x*sin(pi*x/2) + 1 = 0 has real roots

    Solution
    • Discriminant : D = 4*sin(pi*x/2)^2 - 4*1*1 = 4*(sin(pi*x)^2 - 1)
    • The equation has real roots if D GE 0
    • sin(pi*x)^2 GE 1
    • Since sin(A) GE -1 or LE 1
    • Hence sin(pi*x/2) = +1 0r -1
    • Hence x = +1 or -1, or x = +3 or -3, ....
    • If x = 3, we have (3)^2 - 2*3 + 1 = 4. Hence x = 3 is not a solution
    • Simlarly, x = -3, +5, -5, .... also not solution.
    • The real roots are x = 1 or -1.

    Go to Begin

    TR 14 06. Find minimum and maximum of (sic(A)^2 - tan(A))/(sec(A)^2 + tan(A))

    Keyword
    • 1 + tan(x)^2 = sec(x)^2
    Solution
    • Let y = (sic(A)^2 - tan(A))/(sec(A)^2 + tan(A))
    • Hence y*(1 + tan(A)^2 + tan(A)) = (1 - tan(A)^2 - tan(A))
    • (y - 1)*tan(A)^2 + (y + 1)*tan(A) + (y - 1)
    • This is a quadratic equation of tan(A) and tan(A) is real number
    • Hence (y + 1)^2 - 4*(y - 1)*(y - 1) GE 0
    • (y + 1)^2 - 4*(y - 1)^2 GE 0
    • (y + 1 - 2*(y - 1))*(y + 1 +2*(y - 1)) GE 0
    • (-y + 3)*(3*y - 1) GE 0
    • (y - 3)*(3*y - 1) LE 0
    • Hence y LE 3 and y GE 1/3

    Go to Begin

    TR 14 07. Find minimum of 3 - 2*cos(A) + cos(A)^2

    Keywords
    • Use vertex of a*x^2 + b*x + c : xv = -b/(2*a) and yv = (b^2 - 4*a*c)/(4*a)
    Solution
    • Let y = cos(A)^2 - 2*cos(A) + 3
    • This is a quadratic function of cos(A)
    • AT vertex has minimum : cos(A) = -(-2)/(2*1) = 1 and y = 1 - 2 + 3 = 2
    • Hence minimum is 2
    Derivative method
    • Let y = cos(A)^2 - 2*cos(A) + 3
    • y' = 2*cos(A)*(-sin(A) + 2*sin(A) = 0
    • Hence cos(A) = 1 and y = 2

    Go to Begin

    TR 14 08. If cos(2*A) = Sqr(2)/3, find cos(A)^4 + sin(A)^4
    Solution
    • cos(A)^4 + sin(A)^4 = (cos(A)^2 + sin(A)^2)^2 - 2*(cos(A)^2)*(sin(A)^2)
    • = 1 - 2*((sin(A)*cos(A))^2)
    • = 1 - 2*(sin(2*A)/2)^2)
    • = 1 - (sin(2*A)^2)/2
    • = 1 - (1 - cos(2*A)^2)/2
    • = 1/2 + cos(2*A)^2
    • = 1/2 + (Sqr(2)/3)
    • = 1/2 + 2/9
    • = 11/18

    Go to Begin

    TR 14 09. sin(1) + sin(2) + sin(3) + ... + sin(359) + sin(360) = ?

    Keyword
    • Method 1 : sin(359) = sin(360-1) = -sin(1)
    • Method 2 : sin(1) + sin(359) = 2*sin((1+359)/2)*cos(359-1)) = 0
    Solution : method 1
    • sin(359) = sin(360 - 1) = -sin(1)
    • sin(358) = sin(360 - 2) = -sin(2)
    • ...
    • sin(182) = sin(360 - 178) = -sin(178)
    • sin(181) = sin(360 - 179) = -sin(179)
    • Hence after addition : expression = sin(180) + sin(360)
    • Since sin(180) = 0 and sin(360) = 0
    • Hence Expression = 0
    Solution : method 2
    • sin(1) + sin(359) = 2*sin((1+359)/2)*cos(359-1)) = 0
    • sin(2) + sin(358) = 0
    • ...
    • sin(178) + sin(182) = 0
    • sin(179) + sin(181) = 0
    • Hence expression = sin(180) + sin(360)
    • Since sin(180) = 0 and sin(360) = 0
    • Hence Expression = 0

    Go to Begin

    TR 14 10. Maximum and minimum of cos(x)^6 + sin(x)^6 if x between 0 and pi/2

    Solution
    • y = cos(x)^6 + sin(x)^6
    • = (cos(x)^2 + sin(x)^2)^3 - 3*(cos(x)^4)*(sin(x)^2) - 3*(cos(x)^2)*(sin(x)^4)
    • = 1 - 3*(sin(x)^2)*(cos(x)^2)*(cos(x)^2 + sin(x)^2)
    • = 1 - 3*(sin(x)^2)*(cos(x)^2)
    • = 1 - 3*(sin(2*x)^2)/4
    • Since x between 0 and pi/2 Hence (2*x) is between 0 and pi
    • If x = 0, y = 1
    • if x = pi/4, y = 1 - 3/4 = 1/4
    • if x = pi/2, y = 1
    • Hence y is LE 1 and GE 1/4

    Go to Begin

    TR 14 11. Max. & min. of cos(x)^2 -4*cos(x)*sin(x) -3*cos(x)^2. x between 0 & pi/2

    Solution
    • y = cos(x)^2 - 4*cos(x)*sin(x) - 3*cos(x)^2
    • = (1 + cos(2*x))/2 - 2*sin(2*x) - 3*(1 - cos(2*x))/2
    • = 1 + cos(2*x)/2 - 2*sin(2*x) - 3/2 + 3*cos(2*x))/2
    • = -1 - 2*sin(2*x) + 2*cos(2*x)
    • = -1 - 2*Sqr(2)*(cos(2*x)*cos(pi/4) - sin(2*x)*sin(pi/4)
    • = -1 - 2*Sqr(2)*cos(2*x + pi/4)
    • Since 0 GE x LE pi/2
    • Hence 0 GE 2*x LE pi
    • Hence pi/4 GE (2*x + pi/4) LE 5*pi/4
    • Hence cos(2*x + pi/4) GE -1 and LE Sqr(2)/2
    • If cos(2*x + pi/4) = -1, then y = -1 + 2*Sqr(2)
    • if cos(2*x + pi/4) = Sqr(2)/2, then y = -1 - 2*Sqr(2)*Sqr(2)/2 = -3
    Why cos(2*x + pi/4) GE -1 and LE Sqr(2)/2 when x between 0 and pi/2
    • x = 0, 2x + pi/4 = pi/4 Hence cos(2*x + pi/4) = cos(pi/4) = Sqr(2)/2
    • x = pi/4, 2x + pi/4 = 3*pi/4 Hence cos(2*x + pi/4) = cos(3*pi/4) = -Sqr(2)/2
    • x = 3*pi/8, 2*x + pi/4 = pi, hence cos(2*x + pi/4) = cos(pi) = -1
    • x = pi/2 and 2x + pi/4 = 5*pi/4 Hence cos(2*x + pi/4) = cos(5*pi/4) = -Sqr(2)/2

    Go to Begin

    TR 14 12.


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