Question

2 part question

Answer 1

Afternoon @zarkam21, how far have you gotten with this one?

Answer 2

I would isolate sin right

Answer 3

Strange that it says to complete the left hand side of the column. The left hand side of the column contains the expression
\(\dfrac{1}{2}\left[\cos(a - b)-\cos(a + b)\right]\)

So we have to manipulate that until we get the expression of the left side of the equation. Hope that's not confusing for you.

Answer 4

In other words it is expecting you to use sum and difference formulas to simplify the expression. Do you know the sum formula for \(\cos(a + b)\)?

Answer 5

Okay, would you mind posting your steps here since you have the formulas there right in front of you?

Answer 6

YEs I am doing that, its just taking me a min

Answer 7

Okay, take your time.

Answer 8

Sin a cos b - cos a sin b

Answer 9

Im having a harf time

Answer 10

this is what i got so far

Answer 11

I'd prefer if you started with the original expression, then post up to the point where you are now. I want to see how you are manipulating the expression.

Answer 12

and then:

2sin(a) sin(b) = cos (a-b) - cos (a+b)

Answer 13

? But remember, we're only working with the expression on the right side of the equation.

Answer 14

oh beginning with the 1/2?

Answer 15

Yes and you're trying to manipulate that until you get the left side of the equation. But you don't do anything to the left side. You leave that as is.

Answer 16

Yes, manipulate that until you get the left hand side of the equation. Post each of your steps here. What would be the first step?

Answer 17

that is expansion

Answer 18

Here, let's do it this way. I'll just walk you through the whole thing. I'm going to write the expression. Then tell you the steps. You will just have to perform the steps afterwards. Use \(\LaTeX\) though.

Answer 19

Okay, so we start with:
\(\dfrac{1}{2}\left[\cos(a - b)-\cos(a + b)\right]\)

The first step is to replace the expression \(\cos(a - b)\) with \(\cos(a)\cos(b)+\sin(a)\sin(b)\)

Answer 20

@zarkam21, go ahead and try performing that step.

Answer 21

\[1/2 [\cos(a) \cos(b)+\sin(a) \sin(b) - \cos(a+b)]\]

Answer 22

Okay, that's great. Next replace \(\cos(a + b)\) with \(\cos(a)\cos(b) - \sin(a)\sin(b)\). Make sure to put that entire expression in parentheses due to the minus on the outside.

Answer 23

\[1/2[\cos(a)\cos(b)+\sin(a)\sin(b)−(\cos(a) \cos(b)-\sin(a)\sin(b))]\]

Answer 24

Very good, now distribute the negative to expression in the parentheses, then remove the parentheses. Basically you will flip the signs of each term in the parentheses. Let's see if you can perform this step correctly.

Answer 25

would it be each like (b)

Answer 26

or the entire thing like sin (b)

Answer 27

It would be like \(-(a - b) = -a + b\) for example.

Answer 28

\[1/2[\cos -a \cos -b +\sin -a \sin -b − \cos-a \cos-b−\sin-a \sin(-b))]\]

Answer 29

ugh :/

Answer 30

Yep, I see that I confused you. It doesn't work that way. The variables are in parentheses. You can't do anything with them in this instance.

Answer 31

But don't get frustrated. This is how you learn. The correct distribution is this:

Answer 32

At this point, we have:
\(1/2[\cos(a)\cos(b)+\sin(a)\sin(b)−(\cos(a) \cos(b)-\sin(a)\sin(b))]\)

After distributing the negative and removing the parentheses we end up with:
\(1/2[\cos(a)\cos(b)+\sin(a)\sin(b)−\cos(a) \cos(b)+\sin(a)\sin(b))]\)

Notice how the signs have flipped now after distributing the negative.

Answer 33

Next step is to combine like terms. Do you think you can perform this next step? Yea or Nea?

Answer 34

Yes I think I can

Answer 35

Okay, good luck :D

Answer 36

sin(a)sin(b)
1/2 (cos(a-b)-cos(a+b))

Answer 37

@zarkam21 I don't understand what you have written here. I asked you to combine like terms.

Answer 38

well cos cancels out doesn't it

Answer 39

and we are left with sin

Answer 40

Yes but you still must write out the expression correctly. Otherwise, you can make a critical mistake and get it wrong.
When you combine like terms you should have ended up with

\(1/2[\cos(a)\cos(b) - \cos(a)\cos(b) + \sin(a)\sin(b) + \sin(a)\sin(b)]\)

Answer 41

1/2[sin(a)sin(b)+sin(a)sin(b)]

Answer 42

Yes correct.

Answer 43

And then you would combine the expressions involving sine to get what?

Answer 44

well there are 2 a's and 2 b's so maybe a^2

Answer 45

By the way, we should write the fraction properly at this point to get:
\(\dfrac{2\sin(a)\sin(b)}{2}\)

Answer 46

At this point it is clear that 2 cancels at top and bottom to get just
\(\sin(a)\sin(b)\)

Answer 47

yay

Answer 48

thank you so much i know thay took a lot of patience

Answer 49

@zarkam21, you're welcome. Hopefully you understand the process a little better now.