Prove

Prove

Similar to the previous one, you need to make the angle the same as on the right.

How do you change compound angles to single angles?

Yah similar

So? Remember that:

\cos(A+B) = \cos A \cos B - \sin A\sin B

\cos(A-B) = \cos A \cos B + \sin A\sin B

Yes

So... if you expand the left side, what do you get?

(cos pi)(cos theta) - (sin pi)(sin theta)(cospi/2)(costhet) + (sin pi/2)(sin theta)

Simplify it. There are terms you can, such as \cos(\pi) =-1

-1cos(theta)-sin(pi)sin(theta)cos(pi/2)cos(theta)+sin(pi/2)sin(theta)

-1cos(theta)-sin(pi)sin(theta)cos(pi/2)cos(theta)+sin(pi/2)sin(theta)

You wrongly did the substitution...

Erm... don't you know the others?

\Large{\cos\(\frac{\pi}{2}\) =?}

\Large{\sin\(\frac{\pi}{2}\) =?}

\Large{\sin\(\pi\) =?}

-1cos(theta)-0sin(theta)0cos(theta)+1sin(theta)

You can simplify this.

Don't forget the brackets you had. It's supposed to be:

(-1cos\theta-0\sin\theta)(0\cos\theta+1\sin\theta)

Okay know what ?

I told you that you can simplify this...

is going to be 0-2costheta -0-0 =2cos( theta )

I don't understand how you could get that...

I'll take it bit by bit.

(-1cos\theta-0\sin\theta)(0\cos\theta+1\sin\theta)

Can be broken down into:

-1cos\theta\ =\ ?

-0\sin\theta\ =\ ?

0\cos\theta\ =\ ?

1\sin\theta\ =\ ?

-costheta-costheta sinthta -sintheta costheta- sine^2 theta like that

:confused:

Okay, what is: -1\ \times\ \cos\theta