Solution - Episode 3; The Identity

Hey guys, it's Lawrence here, and I'm gonna give you the solution to the second question in Episode two that had to do with the Identities.

So, because we're solving for identities, we draw the "Great Wall of China" perpendicular to the equal sign and NEVER cross over it.

So to solve it, we'll work with the left side because it'll be easier to "decode".

So FIRST, we'll expand Cos(x+y) * Cos(x-y) so we can see a better picture. Using the infamous "Sine Dance" Cos(x+y) becomes (cosXcosY - sinXsinY) and expand Cos(x-y) to become (cosXcosY + sinXsinY).

When we complete that, the identity now looks like;
(cosXcosY - sinXsinY) * (cosXcosY + sinXsinY) = Cos²X - Sin²Y

From there, we'll proceed algebraically and do the multiplication on the left side of the "Great Wall". We will multiply so that we get:
cos²Xcos²Y + cosXcosYsinXsinY - cosXcosYsinXsinY - sin²Xsin²Y

If you haven't already noticed, there is some canceling going on here.

cos²Xcos²Y + cosXcosYsinXsinY - cosXcosYsinXsinY - sin²Xsin²Y

and we then get...

cos²Xcos²Y - sin²Xsin²Y = Cos²X - Sin²Y

So, from here, we observe the equation and look at what we need to get to solve the identity. Look at the right side and you'll notice it's Cos²X - Sin²Y. On the left side, we already have the Cos²X and the Sin²Y, it's just "with" another. So because we want those ones, we'll use the trig identities to switch cos²Y into (1 - Sin²Y) and sin²X into (1 - Cos²X). In which we then get:
cos²X(1 - Sin²Y) - (1 - Cos²X)sin²Y = Cos²X - Sin²Y

And we do the multiplication on the left side again.

cos²X - cos²XSin²Y - sin²Y - Cos²Xsin²Y = Cos2X - Sin2Y

We now have more cancelling we can do.

cos²X - cos²XSin²Y - sin²Y - Cos²Xsin²Y = Cos2X - Sin2Y

And the identity is now complete:

Cos²X - Sin²Y = Cos2X - Sin2Y






Remember your trig identities as it can help to solve identities like these. Also remember there are more than one way to solve identities so try to experiment sometimes.

Rence ~ Out