Question

5. Expand sin 4theta in ascending powers of theta. ​

Answer 1

question is not proper repost please

Answer 2

Answer:

Apply De Moivre's Theorem

2) Use Pascals Triangle (Proves quicker for me than the method of Binomial Expansion)

3) Know your Trig Identities because this is where you're headed

My working so far:

(cosθ+isinθ)3(cos⁡θ+isin⁡θ)3 = (cos3θ+isin3θ)(cos⁡3θ+isin⁡3θ) By De Moivre's Theorem

For this specific problem I am using the 1 3 3 1 tier of Pascal's Triangle

Now I apply this with my powers incremented by 1 for cos and decremented by 1 for sin. (Hope this makes sense?)

1cos3θi0sin0θ+3cos2θi1sin1θ+3cos1θi2sin2θ+cos0θi3sin3θ1cos3⁡θi0sin0⁡θ+3cos2⁡θi1sin1⁡θ+3cos1⁡θi2sin2⁡θ+cos0⁡θi3sin3⁡θ

Now I know i2=(−1)i2=(−1) so I proceed as follows:

cos3θ+3cos2θisinθ−3cosθsin2θ−isin3θcos3⁡θ+3cos2⁡θisin⁡θ−3cos⁡θsin2⁡θ−isin3⁡θ

Now I would equate the real and imaginary parts as follows:

cos3θ=cos3θ+3cosθsin2θcos⁡3θ=cos3⁡θ+3cos⁡θsin2⁡θ

sin3θ=3cos2θisinθ+sin3θsin⁡3θ=3cos2⁡θisin⁡θ+sin3⁡θ

I know that I am supposed to derive the cube trig identities here namely:

cos3θ=4cos3θ−3cosθcos⁡3θ=4cos3⁡θ−3cos⁡θ

sin3θ=3sinθ−4sin3θsin⁡3θ=3sin⁡θ−4sin3⁡θ

Perhaps I've missed something here. My workings of cos4θcos⁡4θ and cos2θcos⁡2θ are spot on though.

Thanks for taking the time.