Three Trigonometry Identities - Please help ASAP, thank you!

Question

I have three identities I cannot seem to figure out.  I'm hoping someone can figure them out for me ASAP, I have to hand the problems in tomorrow.  Please show all work.  I thank you VERY, VERY much!

COS(4x) = COS^4x - 6SIN^2x COS^2x + SIN^4x

COS^4x = 1/8 (3 + 4 COS(2x) + COS(4x))

SEC^2 x/2 = 2 SEC x / SEC x + 1

Answer 1

cos(4x)=cos²(2x)-sin²(2x);

sin(2x)=2sin(x)cos(x), so sin²(2x)=4sin²(x)cos²(x);

cos(2x)=cos²(x)-sin²(x), so cos²(2x)=(cos²(x)-sin²(x))²;

cos(4x)=(cos²(x)-sin²(x))²-4sin²(x)cos²(x),

cos(4x)=cos⁴(x)-2sin²(x)cos²(x)+sin⁴(x)-4sin²(x)cos²(x),

cos(4x)=cos⁴(x)-6sin²(x)cos²(x)+sin⁴(x) QED

cos⁴(x)=cos(4x)+6sin²(x)cos²(x)-sin⁴(x) (from the line above)

cos⁴(x)=cos(4x)+6(1-cos²(x))cos²(x)-(1-cos²(x))²,

cos⁴(x)=cos(4x)+6cos²(x)-6cos⁴(x)-1+2cos²(x)-cos⁴(x),

8cos⁴(x)=cos(4x)+8cos²(x)-1=cos(4x) +(8cos²(x)-4)+3,

8cos⁴(x)=cos(4x)+4(2cos²(x)-1)+3=cos(4x)+4cos(2x)+3,

cos⁴(x)=⅛(3+4cos(2x)+cos(4x)) QED

sec²(x/2)=1/cos²(x/2); cos(x)=2cos²(x/2)-1, cos²(x/2)=½(1+cos(x)),

sec²(x/2)=2/(1+cos(x)).

Multiply top and bottom by sec(x): sec²(x/2)=2sec(x)/(sec(x)+1) QED