# Math Help - Equation solving trigonomeric

1. ## Equation solving trigonomeric

I have an exercise as follows:
Show that: cos(5*angle)=16*cos^5(angle)-20*cos^3(angle)+cos(angle), and hence show that the roots of: x(16x^4-20x^2+5)=0 are 0; cos(pi/10); cos(3*pi/10); cos(7*pi/10) and cos(9*pi/10)

The first part has me completely confused, but the second part is also confusing.

Could you show me what way to do this?

thx

2. Originally Posted by johnnyboy
I have an exercise as follows:
Show that: cos(5*angle)=16*cos^5(angle)-20*cos^3(angle)+cos(angle), and hence show that the roots of: x(16x^4-20x^2+5)=0 are 0; cos(pi/10); cos(3*pi/10); cos(7*pi/10) and cos(9*pi/10)

The first part has me completely confused, but the second part is also confusing.

Could you show me what way to do this?

thx
cos(5 theta)=Re[e^(5 i theta)]=Re [(e^(i theta))^5]

......=Re[ (cos(theta)+isin(theta))^5]

......=Re[c^5+5 i s c^4-10 s^2 c^3 -10 i s^3 c^2 +5 s^4 c + i s^5]

......=c^5 - 10(1-c^2)c^3 + 5(1-c^2)^2 c

......=16 c^5 -20 c^3 + 5 c

Thence the roots of:

x(16x^4-20x^2+5)=0

are the roots of cos(5 x)=0, and the given roots are distinct roots of this
equation.

RonL

3. Hello, Johnny!

Here's the first part . . .

You're expected to know these identities:

. . cos(A + B) .= .cos(A)·cos(B) - sin(A)·sin(B)

. . sin(2A) .= .2·sin(A)·cos(A)

. . cos(2A) .= .2·cos2(A) - 1

Show that: .cos(5θ) .= .16·cos^5(θ) - 20·cos³(θ) + cos(θ)

cos(5θ) .= .cos(4θ + θ)

. . . . . . = .cos(4θ)cos(θ) - sin(4θ)sin(θ)

. . . . . . = .[2cos²(2θ) - 1]·cos(θ) - [2·sin(2θ)·cos(2θ)]·sin(θ)

. . . . . . = .2·cos²(2θ)·cos(θ) - cos(θ) - 2·[2·sin(θ)·cos(θ)][2·cos²(θ) - 1]·sin(θ)

. . . . . . = .2·[2·cos²(θ) - 1]²·cos(θ) - cos(θ) - 4·sin²(θ)·cos(θ)[2·cos²(θ) - 1]

. . . . . . = .2·[4·cos^4(θ) - 4·cos²(θ) + 1]·cos(θ) - cos(θ) - 4·[1 - cos²(θ)]·cos(θ)·[2·cos²(θ) - 1]

. . . . . . = .8·cos^5(θ) - 8·cos³(θ) + 2·cos(θ) - cos(θ) - 4·cos(θ)&#183l[1 - cos²(θ)][2·cos²(θ) - 1]

. . . . . . = .8·cos^5(θ) - 8·cos³(θ) + cos(θ) - 8·cos³(θ) + 8·cos^5(θ) + 4·cos(θ) - 4·cos³(θ)

. . . . . . = .16·cos^5(θ) - 20·cos³(θ) + 5·cos(θ)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Captain Black's method is much simpler
. . if you're familiar with DeMoivre's Theorem.

Edit: Corrected . . . You're right, Captain, it's a 5.

4. Originally Posted by Soroban
Hello, Johnny!

Here's the first part . . .

You're expected to know these identities:

. . cos(A + B) .= .cos(A)·cos(B) - sin(A)·sin(B)

. . sin(2A) .= .2·sin(A)·cos(A)

. . cos(2A) .= .2·cos2(A) - 1

cos(5θ) .= .cos(4θ + θ)

. . . . . . = .cos(4θ)cos(θ) - sin(4θ)sin(θ)

. . . . . . = .[2cos²(2θ) - 1]·cos(θ) - [2·sin(2θ)·cos(2θ)]·sin(θ)

. . . . . . = .2·cos²(2θ)·cos(θ) - cos(θ) - 2·[2·sin(θ)·cos(θ)][2·cos²(θ) - 1]·sin(θ)

. . . . . . = .2·[2·cos²(θ) - 1]²·cos(θ) - cos(θ) - 4·sin²(θ)·cos(θ)[2·cos²(θ) - 1]

. . . . . . = .2·[4·cos^4(θ) - 4·cos²(θ) - 1]·cos(θ) - cos(θ) - 4·[1 - cos²(θ)]·cos(θ)·[2·cos²(θ) - 1]

. . . . . . = .8·cos^5(θ) - 8·cos³(θ) - 2·cos(θ) - cos(θ) - 4·cos(θ)&#183l[1 - cos²(θ)][2·cos²(θ) - 1]

. . . . . . = .8·cos^5(θ) - 8·cos³(θ) - 3·cos(θ) - 8·cos³(θ) + 8·cos^5(θ) + 4·cos(θ) - 4·cos³(θ)

. . . . . . = .16·cos^5(θ) - 20·cos³(θ) + cos(θ)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Captain Black's method is much simpler
. . if you're familiar with DeMoivre's Theorem.
We seem to disagree about the coefficient of cos. Yours agrees with
that in the question, but I had assumed that was a typo, because of
the second part of the question.

I have just checked with a symbolic maths package and it gives 5 as
this coefficient. There can't be something interesting going on here,
can there?

RonL

5. Thanks for the help guys.