Herbie0209
Badges: 8
Rep:
?
#1
Report Thread starter 1 month ago
#1
Cos^4 + sin^4 = 1/4(cos4(theta)+3)

Prove the following Trig Identity using De Moivre's Theorem

Really not sure where to start with this
0
reply
old_engineer
Badges: 11
Rep:
?
#2
Report 1 month ago
#2
(Original post by Herbie0209)
Cos^4 + sin^4 = 1/4(cos4(theta)+3)

Prove the following Trig Identity using De Moivre's Theorem

Really not sure where to start with this
Let z = e^ia. You should already know that cos(a) = (1/2)(z + 1/z) and sin(a) = (1/2i)(z - 1/z).

You could usefully start by expanding the fourth powers of those two expressions.
0
reply
cflau_
Badges: 14
Rep:
?
#3
Report 1 month ago
#3
De Moivre:
 (\cos \theta + i \sin \theta)^n=\cos(n\theta)+i \sin(n \theta)
Note that
 (\cos \theta + i \sin \theta)^4=\cos ^4 \theta + \sin ^4 \theta + 4i \sin \theta \cos ^3 \theta - 6 \sin ^2 \theta \cos ^2 \theta - 4i \sin ^3 \theta \cos \theta
By taking real part in both sides and using De Moivre,
 \cos (4\theta) = \cos ^4 \theta + \sin ^4 \theta - 6\sin ^2 \theta \cos ^2 \theta
Im too lazy to type the rest, but use double angle formula twice on
- 6\sin ^2 \theta \cos ^2 \theta
and rearrange the terms to get the answer.
Last edited by cflau_; 1 month ago
0
reply
Herbie0209
Badges: 8
Rep:
?
#4
Report Thread starter 1 month ago
#4
(Original post by old_engineer)
Let z = e^ia. You should already know that cos(a) = (1/2)(z + 1/z) and sin(a) = (1/2i)(z - 1/z).

You could usefully start by expanding the fourth powers of those two expressions.
thanks for the help, any chance you could show me the first line so i understand what you mean. I have little experience with this method.

I know the formula will be:

1+j(theta)+(j(theta)^2/2!+(j(theta)^3/3!+(j(theta)^4/4!

Im unsure how I will implement the z+1/z within the equation

Thanks
0
reply
DFranklin
Badges: 18
Rep:
?
#5
Report 1 month ago
#5
(Original post by Herbie0209)
thanks for the help, any chance you could show me the first line so i understand what you mean. I have little experience with this method.

I know the formula will be:

1+j(theta)+(j(theta)^2/2!+(j(theta)^3/3!+(j(theta)^4/4!
Why would you think that's the formula?

Im unsure how I will implement the z+1/z within the equation
If cos theta = (z+1/z)/2, then cos^k theta = (z+1/z)^k / 2^k. Expand (z+1/z) using the binomial theorem.
0
reply
DFranklin
Badges: 18
Rep:
?
#6
Report 1 month ago
#6
Last line should have been: Expand (z+1/z)^k using the binomial theorem. (Stupid TSR not letting me edit my post...)
0
reply
old_engineer
Badges: 11
Rep:
?
#7
Report 1 month ago
#7
(Original post by Herbie0209)
thanks for the help, any chance you could show me the first line so i understand what you mean. I have little experience with this method.

I know the formula will be:

1+j(theta)+(j(theta)^2/2!+(j(theta)^3/3!+(j(theta)^4/4!

Im unsure how I will implement the z+1/z within the equation

Thanks
OK if you're happy that cos(a) = (1/2)(z + 1/z), that means that cos^4(a) = (1/16)(z^4 + 4z^2 + 6 + 4/(z^2) + 1/(z^4))
You need to do the same for sin^4(a) and add to the expansion of cos^4(a)
Can you see the way this is going now?
0
reply
Herbie0209
Badges: 8
Rep:
?
#8
Report Thread starter 1 month ago
#8
(Original post by old_engineer)
OK if you're happy that cos(a) = (1/2)(z + 1/z), that means that cos^4(a) = (1/16)(z^4 + 4z^2 + 6 + 4/(z^2) + 1/(z^4))
You need to do the same for sin^4(a) and add to the expansion of cos^4(a)
Can you see the way this is going now?
Ive got:

(1/8)(cos4(theta)-4cos2(theta)+3)+(1/8)(cos4(theta)+4cos2(theta)+3)

Im sure they are both correct I just dont know how to add them together as stupid as it sounds
0
reply
old_engineer
Badges: 11
Rep:
?
#9
Report 1 month ago
#9
(Original post by Herbie0209)
Ive got:

(1/8)(cos4(theta)-4cos2(theta)+3)+(1/8)(cos4(theta)+4cos2(theta)+3)

Im sure they are both correct I just dont know how to add them together as stupid as it sounds
Well, (1/8)cos([email protected]) + (1/8)cos([email protected]) = (1/4)cos([email protected])
Continue with the other two terms in each bracket.
0
reply
Herbie0209
Badges: 8
Rep:
?
#10
Report Thread starter 1 month ago
#10
(Original post by old_engineer)
Well, (1/8)cos([email protected]) + (1/8)cos([email protected]) = (1/4)cos([email protected])
Continue with the other two terms in each bracket.
Thanks for your help, appreciate it. Completed the question
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (149)
62.61%
I don't have everything I need (89)
37.39%

Watched Threads

View All