Hmm if I can remember what I did of complex numbers
I can do 1)
I will use @ for theta
cos4@+i sin4@=(cos@+isin@)<sup>4</sup>
Let cos@=c
sin@=s
(cos@+isin@)<sup>4</sup>=c<sup>4</sup>+4c<sup>3</sup>is+6c<sup>2</sup>i<sup>2</sup>s<sup>2</sup>+4ci<sup>3</sup>s<sup>3</sup>+i<sup>4</sup>s<sup>4</sup>
Since i<sup>2</sup>=-1
c<sup>4</sup>+4c<sup>3</sup>is+6c<sup>2</sup>i<sup>2</sup>s<sup>2</sup>+4ci<sup>3</sup>s<sup>3</sup>+i<sup>4</sup>s<sup>4</sup>=c<sup>4</sup>-6c<sup>2</sup>s<sup>2</sup>+s<sup>4</sup>+4c<sup>3</sup>is-4cis<sup>3</sup>
Equating real and imaginary
cos4@=c<sup>4</sup>-6c<sup>2</sup>s<sup>2</sup>+s<sup>4</sup>
.:cos4@=cos<sup>4</sup>@-6cos<sup>2</sup>@sin<sup>2</sup>@+sin<sup>4</sup>@
Now s<sup>2</sup>=1-c<sup>2</sup>
=c<sup>4</sup>-6c<sup>2</sup>(1-c<sup>2</sup>)+(1-c<sup>2</sup>)<sup>2</sup>
=8c<sup>4</sup>-8c<sup>2</sup>+1
.:
cos4@=8cos<sup>4</sup>@-8cos<sup>2</sup>@+1
The second one I will post up as soon as I can type it
cos<sup>5</sup>@=(1/2[e<sup>i@</sup>+e<sup>-i@</sup>]<sup>5</sup>
=1/2<sup>5</sup>(e<sup>5i@</sup>+5e<sup>4i@</sup>e<sup>-i@</sup>+10e<sup>3i@</sup>e<sup>-2i@</sup>+10e<sup>2i@</sup>e<sup>-3i@</sup>+5e<sup>i@</sup>e<sup>-4i@</sup>+e<sup>-5i@</sup>)
=1/2<sup>5</sup>(e<sup>5i@</sup>+5e<sup>3i@</sup>+10e<sup>i@</sup>+10e<sup>-i@</sup>+5e<sup>-3i@</sup>+e<sup>-5i@</sup>)
=(1/16)cos5@+(5/16)cos3@+(10/16)cos@
I hope thats right because I havent touched complex numbers since the start of term 4 last year.