Link to the slides post.

Link to the slides page.

Hey guys, its Paul here posting my scribe for our class on Double Identities on Tuesday (just in time amirite?).

So we started off the class with a review of the previous class on slide 2. The basic concept here is that the distance between Q and P is equal to cos(a-b).

In slide 3, we used this idea to find the Sum Identity with Even and Odd functions since we already have the Difference Identity.

Sum Identity: cos(a+b) = cosacosb - sinasinb

Difference Identity: cos(a-b) = cosacosb + sinasinb

Cosine is an even function, sine and tangent are odd.

We use the same concept for finding the Sum and Difference identities of the distance between R and Q on slides 4 and 5.

Then we take a break from our "regularly scheduled programming" (I wonder if that counts as one of Mr.K's catchphrases that we should put on the board) with a short and simple quiz on slides 7, 8 and 9.

Slide 7: Simplify the expressions. Pretty self explanatory, use identities to find the simplest expression.

Slide 8: Prove the identities. Again, stuff we've done before.

Slide 9: Find the exact value of sin(5pi/12). This one took a little more work, but it was actually easy because the values a = pi/6 and b = pi/4 were given to us. As Mr.K explained, on tests we will simply be given a value to find with a formula, which means we'll have to find a and b on our own. Since they're given to us, we can just plug the values into the formula and get our answer.

As we return to our regularly scheduled programming on slide 10, we are given and identity to prove. The Sum Identity for tangent, to be precise.

The solution is pretty long and looks complicated but it basically follows these steps:

tan(a+b) = sin(a+b)/cos(a+b)

And since we already know what sin(a+b) and cos(a+b) equal to, we get a really long equation (one I won't bother to type since its right there in the slides). Once you have your really long equation, most of it simplifies or reduces to the proper solution of tana+tanb/1-tanatanb.

And since Math is the science of patterns, you'll probably know what we're going to do next. Next, we find the Difference Identity of tan (tan(a-b)) since we have the Sum Identity. We do this on Slide 11 in an equally long but rewarding process.

In summary:

tan(a+b) = tana+tanb/1-tanatanb

tan(a-b) = tana-tanb/1+tanatanb

On Slides 12 and 13 we learn about Double Angle Identities (which is different from a double identity). Here we apply some old stuff to a new problem, and turn something like:

sin(2theta)

Into:

sin(theta + theta)

Which would look more familiar as say... sin(a+b)?

Since its exact same thing (so long as b = a), we can rewrite sin(2theta) as:

(sintheta)(costheta) + (costheta)(sintheta)

And then simplify it so it looks like

2sintheta(costheta)

We then apply this concept to find the Double Angle Identities of cos and tan.

sin(2theta) = 2sintheta(costheta)

cos(2theta) = cos^2(theta) - sin^2(theta), 1-2sin^2(theta), 2cos^2(theta) -1

tan(2theta) = 2tantheta/1-tan^2(theta)

And that basically sums up what we did that day in class.

Now I'd like to add a small note to the sine dance to make provisions for the tan identities, which is obviously a pain to figure out via sin(a+b) and cos(a+b). Therefore, I propose a tan dance to help us remember the tan identities.

Since this is all in terms of tangent, we do the dance in variable order, something like this:

Alpha, Beta, Divide, 1, Bust-a-move, Alphabeta

or in stick man form...

Im just putting it out there. Seems easier than solving sin(a+b)/cos(a+b) every time you want to do a tan identity to me.

And that concludes my painfully brief scribe post about Double Identities. Hope you guys get a good nights sleep and enjoy our last day before SPRING BREAK.

Yeah man.

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