Q2 POW #3 -Trig Identities

Question

1) given the double angle identity for cosine: cos2θ=cos²θ-sin²θ

     Prove: cos3θ = cos³θ - 3cosθsin²θ

        and: cos4θ = cos⁴θ - 6cos²θsin²θ + sin²θ

2) Continue this process, finding an identity for cos5θ and cos6θ.

3) By examining the above identities (and a few more if needed), try to identify a pattern in order to generalize an identity for cosnθ.

(HINT: You will need to look at Pascal's Triangle.)

Answer 1

1) cos(3θ)

= cos(2θ+θ)

= cos2θ cosθ - sin2θ sinθ

= (cos^2 θ - sin^2θ)cosθ - 2sin^2θ cosθ

= cos^3θ - 3cosθ sin^2θ

 

cos4θ

= cos^2(2θ) - sin^2(2θ)

= (cos^2θ-sin^2θ)^2 - 4sin^2θ cos^2θ

= cos^4θ - 6cos^2θ sin^2θ + sin^4θ <==You had a typo here.)

 

You can follow this way to get it done. But it is lot simpler if you use De Moivre's theorem.

cosnθ + i sinnθ = (cosθ + i sinθ)^n

Expand and compare real parts:

cosnθ

= cos^nθ - nC2cos^(n-2)θsin^2θ + nC4cos^(n-4)θsin^4θ -...+

= ∑{i = 0, n}nCi cos^(n-i) θ sin^i θ cos(0.5(n-i)pi)

Answer 2

1) given: cos(2θ) = cos^2(θ) - sin^2(θ)
prove: cos(3θ) = cos^3(θ) - 3 cos(θ) sin^2(θ)
and: cos(4θ) = cos^4(θ) - 6 cos^2(θ) sin^2(θ) + sin^2(θ)
2) Continue this process, finding an identity for cos(5θ) and cos(6θ).
3) By examining the above identities (and a few more if needed), try to identify a pattern in order to generalize an identity for cos^n(θ).

1) cos(3θ) = cos(2θ + θ) = cos(2θ)cos(θ) - sin(2θ)sin(θ)
= (cos^2(θ) - sin^2(θ)) cos(θ) - (2 sin(θ) cos(θ)) sin(θ)
= cos^3(θ) - sin^2(θ) cos(θ) - 2 sin^2(θ) cos(θ)
= cos^3(θ) - 3 sin^2(θ) cos(θ)
= cos^3(θ) - 3 cos(θ) sin^2(θ)

cos(4θ) = cos(2θ + 2θ) = cos(2θ)cos(2θ) - sin(2θ)sin(2θ)
= (cos^2(θ) - sin^2(θ)) (cos^2(θ) - sin^2(θ))
- (2 sin(θ) cos(θ)) (2 sin(θ) cos(θ))
= cos^4(θ) - 2 cos^2(θ) sin^2(θ) + sin^4(θ) - 4 cos^2(θ) sin^2(θ)
= cos^4(θ) - 6 cos^2(θ) sin^2(θ) + sin^4(θ)

Note that last term is sin^4(θ), NOT sin^2(θ).
GeoGebra graphs confirm that the correct term IS sin^4(θ).

2) cos(5θ) = cos(3θ + 2θ) = cos(3θ)cos(2θ) - sin(3θ)sin(2θ)

sin(3θ) = sin(2θ + θ) = sin(2θ)cos(θ) + cos(2θ)sin(θ)
= (2 sin(θ) cos(θ)) cos(θ) + (cos^2(θ) - sin^2(θ)) sin(θ)
= 3 sin(θ) cos^2(θ) - sin^3(θ)

cos(5θ) = (cos^3(θ) - 3 cos(θ) sin^2(θ)) (cos^2(θ) - sin^2(θ))
- (3 sin(θ) cos^2(θ) - sin^3(θ)) (2 sin(θ) cos(θ))
= cos^5(θ) - cos^3(θ) sin^2(θ) - 3 cos^3(θ) sin^2(θ)
+ 3 cos(θ) sin^4(θ) - 6 sin^2(θ) cos^3(θ) + 2 sin^4(θ) cos(θ)
= cos^5(θ) - 10 cos^3(θ) sin^2(θ) + 5 cos(θ) sin^4(θ)

cos(6θ) = cos(3θ + 3θ) = cos(3θ)cos(3θ) - sin(3θ)sin(3θ)
= (cos^3(θ) - 3 cos(θ) sin^2(θ))^2 - (3 sin(θ) cos^2(θ) - sin^3(θ))^2
= cos^6(θ) - 6 cos^4(θ) sin^2(θ) + 9 cos^2(θ) sin^4(θ)
- 9 sin^2(θ) cos^4(θ) + 6 sin^4(θ) cos^2(θ) - sin^6(θ)
= cos^6(θ) - 15 cos^4(θ) sin^2(θ) + 15 cos^2(θ) sin^4(θ) - sin^6(θ)

3) cos(2θ) = cos^2(θ) - sin^2(θ)
cos(3θ) = cos^3(θ) - 3 cos(θ) sin^2(θ)
cos(4θ) = cos^4(θ) - 6 cos^2(θ) sin^2(θ) + sin^4(θ)
cos(5θ) = cos^5(θ) - 10 cos^3(θ) sin^2(θ) + 5 cos(θ) sin^4(θ)
cos(6θ) = cos^6(θ) - 15 cos^4(θ) sin^2(θ) + 15 cos^2(θ) sin^4(θ) - sin^6(θ)

Pascal’s Triangle:

1 0   0   0  0  0 0
1 1   0   0  0  0 0
1 2   1   0  0  0 0
1 3   3   1  0  0 0
1 4   6   4  1  0 0
1 5 10 10  5  1 0
1 6 15 20 15 6 1

Let P(r,c) be the element in row r and column c of Pascal’s Triangle. The second column contains the row number.

Then cos(n θ) = cos^n(θ) - P(n,3) (cos(θ))^(n-2) sin^2(θ) + P(n,5) (cos(θ))^(n-4) sin^4(θ) - P(n,7) (cos(θ))^(n-6) sin^6(θ) + . . .