Question

Prove that cot (-a) cos (-a) + sin (-a) = -csc a
@zepdrix

Answer 1

Hmm to start, let's change all of the (-a)'s to (a)'s.
To do that, we'll need to remember which trig functions are ODD, and which are EVEN functions.

Answer 2

yes i did that to cos-a and sin-a

Answer 3

which is now cot (-a) cos (a) - sin (a)=-csc (a)

Answer 4

Okay looks good so far :)
Let's apply that to this one as well,\[\large \cot(-a)=\frac{\cos(-a)}{\sin(-a)}\]

Answer 5

okay let me do that will quick

Answer 6

SHould it look like cos (-a)/sin (-a) times cos (a) minus sin (a)= -csc (a)

Answer 7

No, we want to change the cot(-a) to cot(a) somehow. So we're converting cotangent to sines and cosines first,

now apply the rule you just did to sine and cosine above, to this fraction, changing both of the (-a)'s

Answer 8

so would Cot (-a) be the same as Cos (a)/-Sin (a)

Answer 9

Yes very good.

Answer 10

okay so i have this so far...let me send pic

Answer 11

here it is

Answer 12

\[\large \left(-\frac{\cos a}{\sin a}\right)\cos a -\sin a\]
Ok looks good so far :)

Answer 13

i dont know what to do from there haha

Answer 14

do i get common denominators for all 3 of those factors

Answer 15

So let's get a common denominator, looks like the right term needs another Sine.

Answer 16

what do i multiply sin(a)/sin (a) to?

Answer 17

to the sin(a) term.

Answer 18

just that term?

Answer 19

Yes, the other one already has a sine on the bottom.

Answer 20

yeah i know csc is 1/sin

Answer 21

but how come I am not multiplying sina/sins to both cosa and sina

Answer 22

sina/sina* not sins

Answer 23

\[\large -\frac{\cos a}{\sin a} \cdot \cos a\quad =\quad -\frac{\cos^2 a}{\sin a}\]

\[\large -\frac{\cos^2a}{\sin a}-\sin a\]
We would like to be able to combine these fractions. But they don't have the same denominator.

Answer 24

We need to get a sin(a) Under the right term somehow, so we can combine them.

Answer 25

Are you just confused because you feel that if we're going to multiply, we need to multiply the entire side of the equation by something?

Answer 26

no i know what you mean! let me send you what i just did!

Answer 27

here is what i have so far from what you said

Answer 28

Ok good :)

Answer 29

so next, should we factor out that negative sign from the cos?

Answer 30

From both top terms, yes.

Answer 31

okay so how would that look like if you worked it out from both top terms

Answer 32

\[\large \frac{-\cos^2a-\sin^2a}{\sin a}\qquad \rightarrow \qquad \frac{-(\cos^2a+\sin^2a)}{\sin a}\]
Understand how the sign on sine changed to PLUS?

Answer 33

No i dont know how that changed to a plus. I just know it should be a plus to make it 1 but i dont know the logic of how it is a plus

Answer 34

Let's revisit factoring a moment, so we can understand what's going on :)

Answer 35

thanks teacher:)

Answer 36

When you factor, you're doing 2 things... You're dividing and you're multiplying.
You're DIVIDING, ... but you're saving the multiplication for later.

Example:\[\large 2a+4b\]If we want to factor out a 2 from each term, we'll divide both terms by 2, and leave the multiplication for later,
We'll show that our multiplication is being applied to BOTH terms by putting brackets around them.\[\large 2a+4b \qquad \rightarrow \qquad 2\left(\frac{2a}{2}+\frac{4b}{2}\right) \qquad \rightarrow \qquad 2(a+2b)\]Understand how that works?

Answer 37

yes

Answer 38

Let's say we had,\[\large -2a-4b\]This is an important thing to understand, it's going to sound crazy though,
THERE IS NO SUCH THING AS SUBTRACTION! :O
It's all just a different way of writing ADDITION.

We can rewrite our example as,\[\large -2a+(-4b)\]It's really just addition, with a negative 4b.

Let's factor a -2 out of each term,\[\large -2a+(-4b) \qquad \rightarrow \qquad -2\left(\frac{-2a}{-2}+\frac{-4b}{-2}\right) \qquad \rightarrow \qquad -2(a+2b)\]

Answer 39

Sorry the last part got cut off.\[\large \rightarrow \qquad -2(a+2b)\]

Answer 40

okay i get it thank you very much

Answer 41

So understand how our -1 came out of each term in the top? :) Able to simplify it down from there?

Answer 42

yes! i did it already:)

Answer 43

Yay !

Answer 44

thank you you are so helpful.