Counter
Mathematics Dictionary
Dr. K. G. Shih

Trigonometric Functions of Compound Angles
Subjects


  • Q01 | - A+B+C=pi : cos(2*A)^2 +cos(2*B) +cos(2*C)=-1 -4*cos(A)*cos(B)*cos(C)
  • Q02 | - A+B+C=pi : sin(A)^2 +sin(B)^2 +sin(C)^2=2 +2*cos(A)*cos(B)*cos(C)
  • Q03 | - A+B+C=pi : cos(A) +cos(B) +cos(C) =1 +4*sin(A/2)*sin(B/2)*sin(C/2)
  • Q04 | - sin(A)+sin(B) = p and cos(A)+cos(B) = q, find sin(A+B) and cos(A+B)
  • Q05 | - sin(A) and cos(A) has mid AP term sin(x) and GP mid term sin(y)
  • Q06 | - Show that 5 + 8*cos(x) + 4*cos(2*x) + cos(3*x) GE 0
  • Q07 | - If x and y are acute angles, then sin(x + y) LT (sin(x) + sin(y))
  • Q08 | - If sin(x+y) = sin(x) + sin(y), find conditions
  • Q09 | - Find sin(20)*sin(40)*sin(60)*sin(80) without calculater
  • Q10 | - cos(20)*cos(40)*cos(60)*cos(80)
  • Q11 | - tan(20)*tan(40)*tan(60)*tan(80)
  • Q12 | - S(n) = cos(x) + cos(3*x) + cos(5*x) + ...
  • Q13 | - S(n) = sin(x)^2 + sin(2*x)^2 + sin(3*x)^2 + ... + sin(n*x)^2
  • Q14 | - Quiz
  • Q15 | - Quiz answer
  • Q16 | - If A,B,C in GP with r = 2, cos(A)*cos(B)+cos(B)*cos(C)+cos(C)*cos(A) = ?
  • Q17 | - sin(pi/16)^4 + sin(3*pi/16)^4 + sin(5*pi/16)^4 + sin(7*pi/16)^4 = 3/2
  • Q18 | - cos(pi/16)^4 + cos(3*pi/16)^4 + cos(5*pi/16)^4 + cos(7*pi/16)^4 = 3/2
  • Answers


    Q01. Triangle ABC : cos(2*A) + cos(2*B) + cos(2*C) = -1 + 4*cos(A)*cos(B)*cos(C)

    Keywords
    • cos(360-x) = cos(x)
    • cos(2*x) = 2*cos(x)^2 - 1
    • cos(x) + cos(y) = 2*cos((x+y)/2)*cos((x-y)/2)
    Proof
    • cos(2*A) + cos(2*B) + cos(2*C)
    • = 2*cos(A+B)*cos(A-B) + cos(360 - 2*(A+B))
    • = 2*cos(A+B)*cos(A-B) + cos(2*(A+B))
    • = 2*cos(A+B)*cos(A-B) + 2*cos(A+B)^2 - 1
    • = 2*cos(A+B)*(cos(A-B) + cos(A+B)) - 1
    • = 2*cos(C)*(2*cos(A)*cos(B)) - 1
    • = 4*cos(A)*cos(B)*cos(C) - 1

    Go to Begin

    Q02. Triangle ABC : sin(A)^2 + sin(B)^2 + sin(C)^2 = 2 + 2*cos(A)*cos(B)*cos(C)

    Keywords
    • cos(180-x) = -cos(x)
    • cos(2*x) = 1 - 2*sin(x)^2
    • cos(x) + cos(y) = 2*cos((x+y)/2)*cos((x-y)/2)
    Proof : Use Q1 example
    • sin(A)^2 + sin(B)^2 + sin(C)^2
    • = (1-cos(2*A))/2 + (1-cos(2*B))/2 + (1-cos(2*C))/2
    • = 3/2 - (cos(2*A) + cos(2*B) + cos(2*C))/2
    • = 3/2 - (4*cos(A)*cos(B)*cos(C) - 1)/2
    • = 2 + 2*cos(A)*cos(B)*cos(C)

    Go to Begin

    Q03. Triangle ABC : cos(A) + cos(B) + cos(C) = -1 + 4*sin(A/2)*sin(B/2)*sin(C/2)

    Proof
    • cos(A) + cos(B) + cos(C)
    • = cos(A) + cos(B) + cos(180-(A+B))
    • = 2*(cos((A+B)/2)*cos((A-B)/2) + 2*cos((A+B)/2)^2 - 1
    • = 2*cos((A+B)/2)*(cos((A-B)/2) - cos((A+B)/2)) - 1
    • = 2*sin(C/2)*(2*sin(B/2)*sin(C/2)) - 1
    • = -1 + 4*sin(A/2)*sin(B/2)*sin(C/2)

    Go to Begin

    Q04. sin(A)+sin(B) = p and cos(A)+cos(B) = q, find sin(A+B) and cos(A+B)

    Keywords
    • sin(A) + sin(B) = (sin((A+B)/2)*cos((A-B)/2)
    • cos(A) + cos(B) = (cos((A+B)/2)*cos((A-B)/2)
    • tan(2*x) = 2*tan(x)/(1 - tan(x)^2)
    Proof
    • (sin(A)+sin(B))/(cos(A)+cos(B)) = p/q
    • (sin((A+B)/2)*cos((A-B)/2)/(cos((A+B)/2)*cos((A-B)/2) = p/q
    • (sin((A+B)/2)/(cos((A+B)/2) = p/q
    • Hence tan((A+B)/2) = p/q
    • Find tan(A+B)
      • tan(A+B) = tan(2*(A+B)/2)
      • = 2*tan((A+B)/2)/(1 - tan((A+B)/2)^2)
      • = 2*(P/q)/(1 - (p^2)/(q^2))
    • Use triangle method to find sin(A+B) and cos(A+B) using tan(A+B)
      • From tan(A+B) : Opp = 2*(p/q) and Adj = 1 - (p^2)/(q^2)
      • Hence hyp^2 = Opp^2 + Adj^2 = (1 + (p^2)/(q^2))^2
      • Hence hyp = 1 - (p^2)/(q^2))
    • sin(A+B) = Opp/hyp = (2*p/q)/(1 + (p^2)/(q^2))
    • cos(A+B) = Adj/hyp = (1 - (p^2)/(q^2))/(1 + (p^2)/(q^2))

    Go to Begin

    Q05. sin(A) and cos(B) has AP mid term sin(x) and GP mid term sin(y)

    Question
    • Prove that 2*cos(2*x) = cos(2*y)
    Keywords
    • sin(2*x) = 2*sin(x)*cos(x)
    • cos(2*x) = 1 - 2*sin(x)^2
    • a,b,c in AP : b = (a+c)/2
    • a,b,c in GP : b^2 = a*c
    Proof
    • Use AP and GP
      • sin(x) = (sin(A) + cos(A))/2
      • sin(y)^2 = sin(A)*cos(A)
    • sin(x)^2 = (1 + 2*sin(x)*cos(y))/4 = (1 + 2*sin(y))/4
    • (1 - cos(2*x))/2 = (1 + 2*(1 - 2(1 - cos(2*y))/2)/4
    • Hence 2*cos(2*x) = cos(2*y)

    Go to Begin

    Q06. Show that 5 + 8*cos(x) + 4*cos(2*x) + cos(3*x) GE 0

    Keywords
    • cos(2*x) = 2*cos(x)^2 -1
    • cos(3*x) = 4*cos(x)^3 - 3*cos(x)
    • Find factors of cubic equation
    Proof
    • y = 5 + 8*cos(x) + 4*cos(2*x) + cos(3*x)
    • = 5 + 8*cos(x) + 4*(2*cos^2 - 1) + (4*cos(x)^3 - 3*cos(x))
    • = 1 + 5*cos(x) + 8*cos(x)^2 + 4*cos(x)^3
    • = (1 + cos(x))*(1 + 4*cos(x) + 4*cos(x)^2)
    • = (1 + cos(x))*((1 + 2*cos(x))^2)
    • Since cos(x) is GE -1 and LE 1 hence(1 + cos(x)) is positive
    • Also (1 + 2*cos(x))^2 is positive
    • Hence y is positive

    Go to Begin

    Q07. If x and y are acute angles, then sin(x + y) LT (sin(x) + sin(y))

    Keyword
    • sin(x + y) = sin(x)*cos(y) + cos(x)*sin(y)
    Proof
    • x and y are acute angles, hence
      • cos(x) LT 1
      • cos(y) LT 1
    • Both sides of above multiply sin(y) or sin(x)
    • cos(x)*sin(y) LT sin(y)
    • cos(y)*sin(x) LT sin(x)
  • Add left side and right side of above
  • sin(x)*cos(y) + cos(x)*sin(y) LT (sin(x) + sin(y))
  • Hence sin(x + y) LT (sin(x) + sin(y)) Exercise
    • Verify using x = 30 and y = 60

    Go to Begin

    Q08. sin(x+y) = (sin(x) + sin(y)), find condtions

    Keywords
    • sin(2*x) = 2*sin(x)*cos(x)
    • sin(x) + sin(y) = 2*siin((x+y)/2)*cos((x-y)/2)
    • cos(A+B) - cos(A-B) = -2*sin(A)*sin(B)
    Solution
    • LHS = sin(x + y) = 2*sin((x+y)/2)*cos((x+y)/2)
    • RHS = sin(x) + sin(y) = 2*sin((x+y)/2)*cos((x-y)/2)
    • Hence 2*sin((x+y)/2)*cos((x+y)/2) = 2*sin((x+y)/2)*cos((x-y)/2)
    • sin((x+y)/2)*(cos((x+y)/2) - cos((x-y)/2)) = 0
    • sin((x+y)/2)*sin(x/2)*sin(y/2) = 0
    • Hence the conditions are
      • sin((x+y)/2) = 0 and (x + y) = 2*n*pi
      • sin(x/2) = 0 and x = 2*n*pi
      • sin(y/2) = 0 and y = 2*n*pi

    Go to Begin


    Q09. Find sin(20)*sin(40)*sin(60)*sin(80) without calculater

    keyword
    • sin(A)*sin(B) = (cos(A-B) - cos(A+B))/2
    • Sin(A)*cos(B) = (sin(A+B) - sin(A-B))/2
    • Expression = sin(60)*sin(80)*(cos(40-20) - cos(40+20))/2
    • = Sqr(3)*sin(80)*(cos(20) - cos(60))/4
    • = (Sqr(3)/4)*(sin(80)*cos(20) - sin(80)/2)
    • = (Sqr(3)/4)*(sin(80+20) + sin(80-20) - sin(80)/2)
    • = (Sqr(3)/8)*(cos(100) + sin(60) - sin(80))
    • = (Sqr(3)/8)*(sin(80) + sin(60) - sin(80))
    • = (Sqr(3)/8)*sin(60)
    • = 3/16

    Go to Begin

    Q10. Find cos(20)*cos(40)*cos(60)*cos(80) without calculater

    keyword
    • sin(A)*sin(B) = (cos(A-B) - cos(A+B))/2
    • Sin(A)*cos(B) = (sin(A+B) - sin(A-B))/2
    • Expression = cos(60)*cos(80)*(cos(40-20) + cos(40+20))/2
    • = cos(80)*(cos(20) + cos(60))/4
    • = (1/4)*(cos(80)*cos(20) + cos(80)/2)
    • = (1/8)*(cos(80+20) + cos(80-20) - cos(80))
    • = (1/8)*(cos(100) + cos(60) - cos(80))
    • = (1/8)*(cos(80) + cos(60) - cos(80))
    • = (1/8)*cos(60)
    • = 1/16

    Go to Begin

    Q11. Find tan(20)*tan(40)*tan(60)*tan(80) without calculater

    Method 1
    • Expression = sin(20)*sin(40)*sin(60)*sin(80)/cos(20)*cos(40)*cos(60)*cos(80)
    • = (3/16)/(1/16) = 3
    Method 2
    • Expression = (sin(20)*sin(40)*sin(60)/cos(20)*cos(40)*cos(60))*tan(80)
    • Free Q09 and Q10, we have
    • Sin(20)*sin(40)*sin(80) =(1/8)*sin(60)
    • cos(20)*cos(40)*cos(80) =(1/8)*cos(60)
    • Hence Expression = (sin(60)/cos(60))*tan(60)
    • = tan(60)^2
    • = 3

    Go to Begin

    Q12. S(n) = cos(x) + cos(3*x) + cos(5*x) + ...

    Keywords : Prodcut of functions to sum
    • Multiply by sin(x)/(sin(x))
    • Then sin(A)*cos(B) = (sin(A+B) + sin(A-B))/2
    Solution
    • S(n) = (sin(x)/(sin(x))*(cos(x) + cos(3*x) + .... + cos((2*n-1)*x))
    • S(n) = (1/(sin(x))*(sin(x)*cos(x) + sin(x)*cos(3*x) + .. + sin(x)*cos((2*n-1)*x))
    • Use product to sum
      • 2*sin(x)*cos(x) = sin(2*x)
      • 2*sin(x)*cos(3*x) = sin(4*x) - sin(2*x)
      • 2*sin(x)*cos(5*x) = sin(6*x) - sin(4*x)
      • 2*sin(x)*cos(7*x) = sin(8*x) - sin(6*x)
      • ...
      • 2*sin(x)*cos(3*x) = sin(2*n*x) - sin(2*(n-1)*x)
    • Add above, all terms will be cancelled except sin(2*n*x)
    • Hence S(n) = (sin(2*n*x))/(2*sin(x))
    Verify by induction
    • n = 1
      • L = cos(x)
      • R = sin(2*x)/(2*sin(x)) = 2*sin(x)*cos(x)/(2*sin(x)) = cos(x)
    • n = 2
      • L = cos(x) + cos(3*x)
      • R = sin(4*x)/(2*sin(x)) = 2*sin(2*x)*cos(2*x)/(2*sin(x))
      • R = 4*sin(x)*cos(x)*cos(2*x)/(2*sin(x))
      • R = 2*cos(x)*cos(2*x)
      • R = cos(x + 2x) + cos(x - 2*x)
      • R = cos(x) + cos(3*x)
    • n = k+1
      • R = sin(2*k*x)/(2*sin(x)) + cos((2*k+1)*x)
      • R = (sin(2*k*x) + 2*cos((2*k+1)*x)*sin(x))/(2*sin(x))
      • R = (sin(2*k*x) + sin((2*k+2)*x) + sin((2*k+1 - 1)*x))/(2*sin(x))
      • R = sin((2*(k+1)*x)/(2*sin(x))
    • Induction proof is ok

    Go to Begin

    Q13. S(n) = sin(x)^2 + sin(2*x)^2 + sin(3*x)^2 + ... + sin(n*x)^2

    Keywords
    • cos(2*x) = 1 - 2*sin(x)^2
    • sin(A)*cos(B) = (sin(A+B) + sin(A-B))/2
    • Multiply by sin(x)/sin(x)
    Solution
    • S(n) = (1 - cos(2*x) + 1 - cos(4*x) + 1 - cos(6*x) + ... )/2
    • S(n) = n/2 - (cos(2*x) + cos(4*x) + ...... + cos(2*n*x))/2
    • Multiply sin(x)/sin(x)
      • cos(2*x) + cos(4*x) + ...... + cos(2*n*x))
      • = (sin(x)*(cos(2*x) + cos(4*x) + .... + cos(2*n*x))/(sin(x))
    • Use product formula
      • 2*sin(x)*cos(2*x) = sin(x + 2*x) + sin(x - 2*x) = sin(3*x) - sin(x)
      • 2*sin(x)*cos(4*x) = sin(5*x) - sin(3*x)
      • 2*sin(x)*cos(6*x) = sin(7*x) - sin(5*x)
      • ...
      • 2*sin(x)*cos(2*n*x) = sin((2*n+1)*x) - sin((2*n-1)*x)
    • Hence S(n) = n/2 - (sin((2*n+1)*x)-sin(x))/(4*sin(x))
    • = n/2 - cos((n+1)*x)*sin(n*x)/(2*sin(x))
    verify by induction
    • n = 1
      • L = sin(x)^2
      • R = 1/2 - cos(2*x)*sin(x))/(24*sin(x))
      • R = 1/2 - cos(2*x)/2
      • R = sin(x)^2
    • n = 2
      • L = sin(x)^2 + sin(2*x)^2
      • R = 2/2 - cos(3*x)*sin(2*x)/(2*sin(x))
      • R = 1 - 2*cos(3*x)*sin(x)*cos(x))/(2*sin(x))
      • R = 1 - cos(3*x)*cos(x)
      • R = 1 - (cos(4*x) + cos(2*x))/2
      • R = 1 - (1 - 2*sin(2*x)^2 + 1 - 2*sin(x)^2)/2
      • R = sin(x)^2 + sin(2*x)^2
    • n = k+1
      • R = k/2 - cos((k+1)*x)*sin(k*x)/(2*sin(x)) + sin((k+1)*x)^2
      • R = k/2 - (cos((k+1)*x)*sin(k*x) - 2*sin(x)*sin((k+1)*x)^2)/(2*sin(x))
      • 2*sin(x)*sin((k+1)*x)^2 = cos((k+1)*x - x) - cos((k+1)*x + x)
      • = -cos((k+2)*x) + cos(k*x)
      • R = k/2 - (cos((k+1)*x)*sin(k*x) -(cos(k*x) -cos((k+2)*x))*sin((k+1)*x)/(2*sin(x))
      • R = k/2 - cos((k+2)*x)*(sin(k+1)*x)/(2*sin(x) + y
      • y = -cos((k+1)*x)*sin(k*x) + sin((k+1)*x)*cos(k*x)/(2*sin(x))
      • y = sin((k+1)*x - k*x)/(2*sin(x))
      • y = 1/2
      • Hence R = (k+1)/2 - cos((k+1)+1)*x)*sin(k+1)/(2*sin(x))

    Go to Begin

    Q14. Quiz

    • 1. Find sin(75) without calculator.
    • 2. Use formula sin(A+B) to prove that sin( 90+A) = cos(A).
    • 3. Use formula sin(A-B) to prove that sin(180-A) = sin(A).
    • 4. Use sum of function to prove that sin(359) + sin(1) = 0.
    • 5. Express sin(3*A) + sin(A) as product of two functions.
    • 6. Prove that (cos(2*A)-cos(5*A))/(sin(2*A)+sin(5*A)) = tan(3*A/2).
    • 7. Find value of (sin(75)-sin(15))/(cos(75)+cos(15) without calculator.
    • 8. prove that
    • (sin(A)+sin(3*A)+sin(5*A)+sin(7*A))/(cos(A)+cos(3*A)+cos(5*A)+cos(7*A)) = tan(4*A).
    • 9. Express 2*sin(2*A)*cos(A) as sum or difference of two functions.

    Go to Begin

    Q15. Quiz

    • 1. Find sin(75) without calculator.
      • sin(45+30) = sin(45)*cos(30) + cos(45)*sin(30).
      • Hence sin(75) = (Sqr(2)/2)*(Sqr(3)/2) + (Sqr(2)/2)*(1/2).
      • Hence sin(75) = Sqr(6)/4 + Sqr(2)/4.
    • 2. Use formula sin(A+B) to prove that sin( 90+A) = cos(A).
      • sin(180+A) = sin(180)*cos(A) + cos(180)*sin(A).
      • Since sin(180) = 0 and cos(180) = -1.
      • Hence sin(180+A) = -sin(A)
    • 3. Use formula sin(A-B) to prove that sin(180-A) = sin(A).
      • sin(90+A) = sin(90)*cos(A) + cos(90)*sin(A).
      • Since sin(90) = 1 and cos(90) = 0.
      • Hence sin(90+A) = cos(A)
    • 4. Use sum of function to prove that sin(359) + sin(1) = 0.
      • sin(369) + sin(1) = 2*sin((369+1)/2)*cos((369-1)/2).
      • sin(369) + sin(1) = 2*sin(180)*cos(179).
      • Sine sin(180) = 0 hence sin(369) + sin(1) = 0.
    • 5. Express sin(3*A) + sin(A) as product of two functions.
      • sin(3*A) + sin(A) = 2*sin((3*A+A)/2)*cos((3*A-A)/2).
      • sin(3*A) + sin(A) = 2*sin(2*A)*cos(A).
    • 6. Prove that (cos(2*A)-cos(5*A))/(sin(2*A)+sin(5*A)) = tan(3*A/2).
      • cos(2*A) - cos(5*A) = 2*sin(7*A/2)*sin(3*A/2).
      • sin(2*A) + cos(5*A) = 2*sin(7*A/2)*cos(3*A/2).
      • Hence (cos(2*A)-cos(5*A))/(sin(2*A)+sin(5*A)) = tan(3*A/2).
    • 7. Find value of (sin(75)-sin(15))/(cos(75)+cos(15) without calculator.
      • sin(75) - sin(15) = 2*cos((75+15)/2)*sin((75-15)/2).
      • sin(75) - sin(15) = 2*cos(45)*sin(30) = Sqr(2)/2.
    • 8. prove that
    • (sin(A)+sin(3*A)+sin(5*A)+sin(7*A))/(cos(A)+cos(3*A)+cos(5*A)+cos(7*A)) = tan(4*A).
      • sin(A)+sin(7*A) = 2*sin((A+7*A)/2)*cos((A-7*A)/2) = 2*sin(4*A)*cos(3*A).
      • sin(3*A)+sin(5*A) = 2*sin((3*A+5*A)/2)*cos((3*A-5*A)/2) = 2*sin(4*A)*cos(A).
      • Numerator = 2*sin(4*A)*(cos(3*A)+cos(A))
      • cos(A)+cos(7*A) = 2*cos((A+7*A)/2)*cos((A-7*A)/2) = 2*cos(4*A)*cos(3*A).
      • cos(3*A)+cos(5*A) = 2*cos((3*A+5*A)/2)*cos((3*A-5*A)/2) = 2*cos(4*A)*cos(A).
      • Denominator = 2*cos(4*A)*(cos(3*A)+cos(A))
      • LHS = numerator/denominator = sin(4*A)/cos(4*A).
    • 9. Express 2*sin(2*A)*cos(A) as sum or difference of two functions.
      • 2*sin(2*A)*cos(A) = sin(2*A+A) + sin(2*A-A) = sin(3*A) + sin(A).

    Go to Begin

    Q16. Angles of triangle in 3 GP terms with common ratio = 2

    Question
    • Angles of triangle are in 3 consecutive GP terms.
    • If common ratio is r = 2, then cos(A)*cos(B) + cos(B)*cos(C) + cos(C)*cos(A) = ?
    Proof
    • Let angle A = x, then angle B = 2*x and angle C = 4*x
    • A + B + C = 7*x = 180. Hence angle A = x = 180/7 degrees
    • Use product formula of functions
    • LHS
      • = cos(A)*cos(B) + cos(B)*cos(C) + cos(C)*cos(A)
      • = (cos(A+B) + cos(A-B) + cos(B+C) + cos(B-C) + cos(C+A) + cos(C-A))/2
      • = (cos(3*x) + cos(1*x) + cos(6*x) + cos(2*x) + cos(5*x) + cos(4*x))/2
      • = (cos(6*x) + cos(5*x) + cos(4*x) + cos(3*x) + cos(2*x) + cos(1*x))/2
    • Each term of above multiply by sin(x)/sin(x)
    • LHS = (cos(6*x)*sin(x) + cos(5*x)*sin(x) + .... )/(2*sin(x))
    • Since
      • sin(x)*cos(6*x) = (sin(x+6*x) + sin(x-6*x))/2 = (sin(7*x) - sin(5*x))/2
      • sin(x)*cos(5*x) = (sin(x+5*x) + sin(x-5*x))/2 = (sin(6*x) - sin(4*x))/2
      • sin(x)*cos(4*x) = (sin(x+4*x) + sin(x-4*x))/2 = (sin(5*x) - sin(3*x))/2
      • sin(x)*cos(3*x) = (sin(x+3*x) + sin(x-3*x))/2 = (sin(4*x) - sin(2*x))/2
      • sin(x)*cos(2*x) = (sin(x+2*x) + sin(x-2*x))/2 = (sin(3*x) - sin(1*x))/2
      • sin(x)*cos(1*x) = (sin(x+1*x) + sin(x-1*x))/2 = (sin(2*x) - sin(0*x))/2
    • Add above terms and many terms are cancelled out
    • LHS = (sin(7*x) + sin(6*x) - sin(x))/(2*2*sin(x))
    • LHS = (sin(180) + sin(180-x) - sin(x))/(4*sin(x))
    • LHS = (0 +sin(x) - sin(x))/(4*sin(x))
    • LHS = 0

    Go to Begin

    Q17. sin(pi/16)^4 + sin(3*pi/16)^4 + sin(5*pi/16)^4 + sin(7*pi/16)^4 = 3/2

    Keyword
    • 1. cos(A) + Cos(B) = (cos((A+B)/2)*(cos((A-B)/2))
    • 2. sin(1*pi/16)^4 = (sin(1*pi/16)^2)^2
    • = ((1 - cos(1*pi/8))/2)^2
    • = 1/4 - cos(1*pi/8)/2 + cos(1*pi/8)^2)/4
    Solution
    • sin(1*pi/16)^4 = 1/4 - cos(1*pi/8)/2 + (cos(1*pi/8)^2)/4
    • sin(3*pi/16)^4 = 1/4 - cos(3*pi/8)/2 + (cos(3*pi/8)^2)/4
    • sin(5*pi/16)^4 = 1/4 - cos(5*pi/8)/2 + (cos(5*pi/8)^2)/4
    • sin(7*pi/16)^4 = 1/4 - cos(7*pi/8)/2 + (cos(7*pi/8)^2)/4
    • Add together, we have
      • Expression = 1 - (cos(1*pi/8)+...)/2 + (cos(1*pi/8)^2 +...)/4
      • Expression = 1 - (cos(1*pi/8)+...)/2 + ((1+cos(1*pi/4))/2 +...)/4
      • Expression = 1 - (cos(1*pi/8)+...)/2 + (4 + cos(1*pi/4) +...)/8
      • Expression = 3/2 - (cos(1*pi/8)+...)/2 + (cos(1*pi/4) +...)/8
    • Use sum to product formula
      • cos(1*pi/8) + cos(3*pi/8) + cos(5*pi/8) + cos(7*pi/8)
      • = 2*(cos(pi/2)*(cos(3*pi/8) + cos(pi/2)*cos(pi/8))
      • = 0
      • cos(1*pi/4) + cos(3*pi/4) + cos(5*pi/4) + cos(7*pi/4)
      • = 2*(cos(pi)*(cos(3*pi/4) + cos(pi)*cos(pi/4))
      • = 2*((-1)*cos(pi-pi/4) + (-1)*cos(pi/4))
      • = 2*(cos(pi/4) - cos(pi/4))
      • = 0
    • Hence Expression = 3/2 - 0/2 + 0/8 = 3/2

    Go to Begin

    Q18. cos(pi/16)^4 + cos(3*pi/16)^4 + cos(5*pi/16)^4 + cos(7*pi/16)^4 = 3/2

    Keyword
    • 1. cos(A) + Cos(B) = (cos((A+B)/2)*(cos((A-B)/2))
    • 2. cos(1*pi/16)^4 = (cos(1*pi/16)^2)^2
    • = ((1 + cos(1*pi/8))/2)^2
    • = 1/4 + cos(1*pi/8)/2 + cos(1*pi/8)^2)/4
    Solution
    • cos(1*pi/16)^4 = 1/4 + cos(1*pi/8)/2 + (cos(1*pi/8)^2)/4
    • cos(3*pi/16)^4 = 1/4 + cos(3*pi/8)/2 + (cos(3*pi/8)^2)/4
    • cos(5*pi/16)^4 = 1/4 + cos(5*pi/8)/2 + (cos(5*pi/8)^2)/4
    • cos(7*pi/16)^4 = 1/4 + cos(7*pi/8)/2 + (cos(7*pi/8)^2)/4
    • Add together, we have
      • Expression = 1 + (cos(1*pi/8)+...)/2 + (cos(1*pi/8)^2 +...)/4
      • Expression = 1 + (cos(1*pi/8)+...)/2 + ((1+cos(1*pi/4))/2 +...)/4
      • Expression = 1 + (cos(1*pi/8)+...)/2 + (4 + cos(1*pi/4) +...)/8
      • Expression = 3/2 + (cos(1*pi/8)+...)/2 + (cos(1*pi/4) +...)/8
    • Use sum to product formula
      • cos(1*pi/8) + cos(3*pi/8) + cos(5*pi/8) + cos(7*pi/8)
      • = 2*(cos(pi/2)*(cos(3*pi/8) + cos(pi/2)*cos(pi/8))
      • = 0
      • cos(1*pi/4) + cos(3*pi/4) + cos(5*pi/4) + cos(7*pi/4)
      • = 2*(cos(pi)*(cos(3*pi/4) + cos(pi)*cos(pi/4))
      • = 2*((-1)*cos(pi-pi/4) + (-1)*cos(pi/4))
      • = 2*(cos(pi/4) - cos(pi/4))
      • = 0
    • Hence Expression = 3/2 + 0/2 + 0/8 = 3/2

    Go to Begin

  • Show Room of MD2002 Contact Dr. Shih Math Examples Room

    Copyright © Dr. K. G. Shih, Nova Scotia, Canada.

    1