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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
Prove the following Trig Identity using De Moivre's Theorem
Really not sure where to start with this
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#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
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
You could usefully start by expanding the fourth powers of those two expressions.
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#3
De Moivre:
Note that
By taking real part in both sides and using De Moivre,
Im too lazy to type the rest, but use double angle formula twice on
and rearrange the terms to get the answer.
Note that
By taking real part in both sides and using De Moivre,
Im too lazy to type the rest, but use double angle formula twice on
and rearrange the terms to get the answer.
Last edited by cflau_; 1 month ago
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(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.
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.
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
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#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!
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
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#6
Last line should have been: Expand (z+1/z)^k using the binomial theorem. (Stupid TSR not letting me edit my post...)
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#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
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
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?
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(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?
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?
(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
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#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
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
Continue with the other two terms in each bracket.
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(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.
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.
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