To express sin5θ in terms of sine and cosine of multiple angles, we use Euler's formula and the binomial theorem.
Step 1: Express sinθ in terms of z and z−1.
Given z=cosθ+isinθ, we know that z=eiθ.
Then z−1=cosθ−isinθ=e−iθ.
Subtracting z−1 from z:
z−z−1=(cosθ+isinθ)−(cosθ−isinθ)=2isinθ
Therefore, sinθ=2iz−z−1.
Step 2: Raise sinθ to the power of 5.
sin5θ=(2iz−z−1)5=(2i)5(z−z−1)5
Step 3: Expand (z−z−1)5 using the binomial theorem.
The binomial expansion for (a−b)5 is a5−5a4b+10a3b2−10a2b3+5ab4−b5.
Substitute a=z and b=z−1:
(z−z−1)5=z5−5z4(z−1)+10z3(z−1)2−10z2(z−1)3+5z(z−1)4−(z−1)5
=z5−5z3+10z−10z−1+5z−3−z−5
Step 4: Group terms and convert back to sine of multiple angles.
Group the terms as (zn−z−n):
(z−z−1)5=(z5−z−5)−5(z3−z−3)+10(z−z−1)
Using the identity zn−z−n=einθ−e−inθ=(cos(nθ)+isin(nθ))−(cos(nθ)−isin(nθ))=2isin(nθ):
z5−z−5=2isin(5θ)
z3−z−3=2isin(3θ)
z−z−1=2isinθ
Substitute these into the expanded expression:
(z−z−1)5=2isin(5θ)−5(2isin(3θ))+10(2isinθ)
=2i(sin(5θ)−5sin(3θ)+10sinθ)
Step 5: Calculate (2i)5 and substitute back into the expression for sin5θ.
(2i)5=25⋅i5=32⋅(i4⋅i)=32⋅1⋅i=32i
Now, substitute this and the expanded binomial back into the expression for sin5θ:
sin5θ=32i2i(sin(5θ)−5sin(3θ)+10sinθ)
sin5θ=161(sin(5θ)−5sin(3θ)+10sinθ)
The final expression for sin5θ in terms of sine of multiple angles is:
\frac{1{16} (\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta)}
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