Emergency help on an Algebra II exam!
Prove: sin θ - sin θ•cos2 θ = sin3 θ. You must show all work.

- anonymous

Emergency help on an Algebra II exam!
Prove: sin θ - sin θ•cos2 θ = sin3 θ. You must show all work.

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- schrodinger

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- nincompoop

it looks like this is an exam

- anonymous

It's a practice exam, but it's worth a good bit of points and while I can understand the proofs written out in my lessons, this particular problem is stumping me. I've been working on these tests for hours :(

- anonymous

So I know that identity of sin^2theta + cos^2theta = 1, or...something pretty close to that...

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## More answers

- nincompoop

show me the attempts you have made.

- anonymous

Okay. So the sin theta should cancel, right?
cos^2 theta = sin^3 theta ****(that 3 is a cube, not sin(3))****
I don't know what to do with that!

- nincompoop

how does it cancel?

- anonymous

Bc it's just subtracted. Like 1 - 1 = 0.

- anonymous

sin theta - sin theta = 0.

- nincompoop

no

- nincompoop

sin is multiplied to cos
sin θ - (sin θ•cos2 θ) = sin3 θ
ORDERS of operation

- anonymous

Parentheses first, then

- nincompoop

btw, is that \(cos^2 \theta \) or \(cos (2 \theta) \)

- anonymous

The first, it's squared.

- anonymous

Okay, so...then do we distribute the negative into the parentheses??

- anonymous

I'm thinking maybe we subtract sin theta from both sides to give us (sin theta * cos^2 theta) = sin^2 theta

- nincompoop

is that \(sin^3 \theta \) or \(sin (3 \theta) \)

- anonymous

Cubed, the first one.

- nincompoop

you need to start typing these things appropriately because they mean different things

- anonymous

I'm sorry, I didn't realize it before I posted bc I just copy and pasted it to preserve the theta symbols :(

- nincompoop

like this?
PROVE
\(sin ~\theta - (sin~ \theta ~cos^2 \theta) = sin^3 \theta \)

- anonymous

There weren't any parentheses, but otherwise, yes.

- nincompoop

so tell me what should be the best identity we can use?

- anonymous

The only one I know of is the one I typed above, the sin^2 theta + cos^2 theta = 1.

- anonymous

We didn't focus a lot on identities, so I suppose I may be forgetting some, but they aren't in my notes =/

- nincompoop

what did you focus on?

- nincompoop

let us use trigonometric expansion and make use of some algebra techniques like factoring, yes?

- nincompoop

do you think you can factor the left-hand-side (LHS) of the equation for me?

- anonymous

Sorry! Lost connection!

- anonymous

I can tryyy, hold on

- nincompoop

|dw:1440734256689:dw|

- anonymous

sin theta(1 - cos^2 theta) = sin^3 theta

- nincompoop

we can skip writing the theta for now and just add it in the end so it is easier, agreed?

- anonymous

Okay!
sin(1 - cos^2) = sin^3

- nincompoop

do you have a note for your pythagorean identities? if not it is very simple
\(\large sin^2 + cos^2 = 1 \)
keep in mind for future reference that \(sin = y\) and \(cos = x \)

- nincompoop

|dw:1440734748047:dw|

- anonymous

Yep, I remember that part from the unit circle section we just did.

- nincompoop

you probably already learned the Pythagorean Theorem which states that the length of the longest leg \(\large r \) is equal to the square root of the sum of the two shorter legs
|dw:1440734959343:dw|

- nincompoop

do you think we can manipulate our identity algebraically so it would look familiar as the problem you have?

- anonymous

Let me see...

- anonymous

y(1 - x^2) = y^3 when you say cos = x and sin = y...

- nincompoop

|dw:1440735114209:dw|

- anonymous

I'm sorry this is taking me so long to understand ._.

- nincompoop

|dw:1440735291067:dw|

- nincompoop

|dw:1440735446837:dw|

- nincompoop

what part did I lose you?

- nincompoop

|dw:1440735591245:dw|

- anonymous

Where did the sin^3 go?

- nincompoop

that's just the left-hand-side (LHS) since the board is too small for me to fit everything in

- anonymous

So it's 1^2 + sin^3 on the right hand?

- nincompoop

no, the right-hand-side stays the same since it is what we're trying to accomplish

- anonymous

...so then where is the sin^3 if you said it's not on the left hand? D:

- nincompoop

what we are basically doing is trying to manipulate and re-arrange the left-hand-side of your problem so it would EQUAL to the right-hand side of the problem

- nincompoop

|dw:1440735790511:dw|

- nincompoop

|dw:1440735817436:dw|

- anonymous

Okay, I'm mostly following

- nincompoop

##### 1 Attachment

- nincompoop

from the pythagorean identity, we derived that
\(1-cos^2 = sin^2 \)
because from our original identity (derived from the unit circle)
\(sin^2 + cos^2 = 1 \)
we can re-arrange it to
\(sin^2 + cos^2 \color{red}{-cos^2} =1 \color{red}{-cos^2}\)
simplifying to:
\(sin^2 = 1-cos^2 \)

- anonymous

I don't think any of this is getting through anymore, it's really late here and I'm thinking my exam has probably timed out by now ._. I'm tracking but I don't see where this is actually leading, we keep ending up back at this identity and I'm not sure what to do with it

- nincompoop

then it would suffice to say that
\(sin - sin~cos^2 = sin^3 \)
because
\(sin (1-cos^2) = sin^3 \)
because the left-hand side is equivalent to
\(sin (sin^2) \)
therefore
\(sin^3 = sin^3 \)

- anonymous

I'll definitely reference this when I retake. Thank you so so so much for all of your time and help <3

- anonymous

You were very patient, I hope it comes back to you tenfold :x

- nincompoop

you're not even telling me what part you are not following

- nincompoop

up to what part you are able to grasp and what part seems a little fuzzy still

- anonymous

I'm sorry, it's just because I don't really know where I'm lost, I'm just really tired >.<

- nincompoop

come back when your brain is fully rested. there is no point in studying when your brain is not working to its potential.

- nincompoop

go to sleep

- anonymous

Exactly :( I hope you have a good night, thanks so much for the guidance, I'm sure it'l help a lot more in the morning

- anonymous

You, too!

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