# Applying the cosine even identity to the cosine difference identity

I'm slightly confused over what happens when you're applying cosine's "even identities" to the difference identity. Here's how I go about, please tell correct me as I feel i'm going wrong somewhere.

Applying the identity $\cos ( - \theta ) = \cos \theta$ to the cosine difference identity:

$\cos (\alpha - \beta ) = \cos \alpha \cos\beta + \sin\alpha \sin\beta$

Okay, so if I applied the identity $\cos ( - \theta ) = \cos \theta$ I should get:

$\cos (-(\alpha - \beta) ) = (\cos \alpha \cos\beta + \sin\alpha \sin\beta)$

$\cos (\beta - \alpha) = \cos \alpha \cos\beta - \sin\alpha \sin\beta$

However I'm confused here, we just multiplied the angles $(\alpha - \beta)$ that we're going to input by -1 essentially; I know that the angles being negative does not effect the values of cosine in the identity, but what about the sine values? Originally we had:

$\cos (\alpha - \beta ) = \cos \alpha \cos\beta + \sin\alpha \sin\beta$

So after applying the identity and multiplying the angle $(\alpha - \beta)$ by $-1$ shouldn't we have:

$\cos (\beta -\alpha) = \cos (-\alpha) \cos\beta + \sin(-\alpha) \sin\beta$

So that would give:

$\cos (\beta -\alpha) = \cos (\alpha) \cos\beta - \sin(\alpha) \sin\beta$

Can someone tell me where my thinking is erroneous?

• You replaced both $\alpha$ and $\beta$ by their opposites, so both sines change sign, with no net effect since they are multiplied, right? – MPW Jul 14 '14 at 12:04
• Yes you're right! Thanks! – seeker Jul 14 '14 at 12:16

You write:

$\cos (\beta -\alpha) = \cos (-\alpha) \cos\beta + \sin(-\alpha) \sin\beta$.

This is not correct. You should either use the sum of $(\beta)$ and $(-\alpha)$. Or you should use the difference of $(\beta)$ and $(\alpha)$.

1st case: $\cos (\beta +(-\alpha)) = \cos (-\alpha) \cos(\beta) - \sin(-\alpha) \sin(\beta)= \cos (\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$.

2nd case: $\cos (\beta -\alpha) = \cos (\beta) \cos(\alpha) + \sin(\alpha) \sin(\beta)= \cos (\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$.

• Just what I was looking for! – seeker Jul 14 '14 at 12:31

We know that $\cos(A+B)=\cos A\cos B-\sin A\sin B$.

Now:

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