How do you prove these trig identities?

Question

I'm very bad with trig, and we never learned how to prove these types of trig identities...

cos (x/2) = sqrt [(1+cosx) / 2]

and

sin (x/2) = sqrt [(1-cosx) / 2]

sqrt = square root...

Help please?

Answer 1

Start with cos 2a = 2(cos a)^2 - 1
Now call x = 2a
cos x = 2(cos 1/2x)^2 -1
1 + cos x = 2(cos 1/2x)^2
(1 + cos x)/2 = (cos 1/2x)^2
±sqrt(1 + cos x)/2 =cos 1/2x

For sin 1/2x, start with cos 2x = 1 - 2(sin x)^2, and do exactly the same steps.

Answer 2

I will do the second part first and start with the formula for the cosine of a double angle.

cos 2θ = 1 - 2 sin 2θ

Now, if we let

θ =x/2

our formula becomes:

cos x = 1 − 2 sin² (x/2)

We now solve for

sin² (x/2)

That is, we get sin² (x/2) on the left of the equation and everything else on the right:

2sin² (x/2) = 1 − cos x
sin² (x/2) = (1 − cos x)/2

Solving gives us the following sine of a half-angle identity:

sin (x/2) = ±√ [(1-cosx) / 2]

now the first part is the same, but you start with the cos 2x formula:

cos 2θ = 2cos² θ − 1

Now, let:

θ = x/2

our formula becomes:

cos 2(x/2) = 2 cos² x/2 − 1

We now solve for

cos² (x/2)

That is, we get cos² (x/2) on the left of the equation and everything else on the right:

2cos² (x/2) = 1 + cos x
cos² (x/2) = (1 + cos x)/2

Solving gives us the following sine of a half-angle identity:

cos (x/2) = ±√[(1+ cosx) / 2]

for proof of a double angle formula I started with, check the website below:

Answer 3

cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
cos(2y)=cos(y)cos(y)-sin(y)sin(y)
=[cos(y)]^2-[sin(y)]^2

[cos(y)]^2+[sin(y)]^2=1
-->[cos(y)]^2=1-[sin(y)]^2
-->[sin(y)]^2=1-[cos(y)]^2

Plugging those 2 into the formula for cos(2y),

cos(2y)=1-2[sin(y)]^2
cos(2y)=1+2[cos(y)]^2

Let y = x/2 -->

cos(x)=1-2[sin(x/2)]^2
cos(x)=1+2[cos(x/2)]^2

Solve for sin(x/2) and cos(x/2) yields

sin(x/2)=sqrt([1-cos(x)]/2)
cos(x/2)=sqrt([1+cos(x)]/2)