Express sin7 There Interms of sin Theta

Question

Trigonometry

Answer 1

Hello Prakash.

 

I did this both ways and even checked my answer...  I used trig double angle and triple angle formulas.  Then I did it with De Moivers Theorem like you are required.  It takes much less paper!!!

 

 

De Moivers Theorem states that (cos(theta) + i sin(theta))ⁿ = cos(n theta) + i sin(n theta)

 

Since we want sin(7theta) for us DeMoivers will look like:

 

[cos(theta) + i sin(theta)]⁷ = cos(7theta) + i sin(7theta)

 

Since we are looking for the sin(7theta), if we do a binomial expansion on [cos(theta) + i sin(theta)]⁷ and just keep the imaginary part, we will have the equivalent of i sin(7theta).

 

Binomial expansion for [cos(theta)+i sin(theta)]⁷ =  (i am going to skip all of the "theta"s, but will put them in for the final answer just to save time and s6pace...)

 

cos⁷ +7 cos⁶ (i sin) + 21 cos⁵ (i sin)² + 35 cos⁴ (i sin)³ + 35 cos³ (i sin)⁴ + 21 cos² (i sin)⁵ + 7 cos (i sin)⁶ + (i sin)⁷

 

 

We are only keeping the imaginary parts, so any "i" with an even exponent is real and we will not use it.  

 

So i sin(7theta) = i[7 cos⁶ sin   -  35 cos⁴ sin³   +   21 cos² sin⁵   -  sin⁷]

 

Divide both sides by "i" and get

 

sin(7θ) = 7 cos⁶(θ) sin(θ)  - 35 cos⁴(θ) sin³(θ)   +   21 cos²(θ) sin⁵(θ)   -   sin⁷(θ)

 

 

Now, just change every even power of cos into multiple "cos²(θ)"s.  Then replace each cos²(θ) with "1 - sin²(θ)"  and multiply out all of those factors, collect like terms  and you will have your answer.

 

 

sin(7θ) = 7 cos²(θ) cos²(θ) cos²(θ) sin(θ)  -  35 cos²(θ)cos²(θ)sin³(θ)   +  21 cos²(θ) sin⁵(θ)   -    sin²(θ)

 

 

Now replace every cos²(θ)  with   "1 - sin²(θ)"  and  carefully multiply every thing that needs multiplying and gather like terms.

 

I ended up with

 

7 sin(θ) - 56 sin³(θ) + 112 sin⁵(θ) - 64 sin⁷(θ)