Mathematics Dictionary
Dr. K. G. Shih
Trigonometric Functions of Compound Angles
Subjects
Symbol Defintion
Example : Sqr(x) = square root of x
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
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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
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