Trigonometric/Pythagorean Identities?

[Guest]

In my pre-calculus class, we're working on proving trigonometric and Pythagorean identities as valid solutions for problems given... We are given an unsimplified problem and a solution that we have to prove. I feel totally unsure about how to go about doing this, despite my teacher explaining things to me. Here's a problem I worked on... I have no idea if I'm doing this right or not, so bear with me.

I need to prove the following:

cosx(tanx+cotx) = cscx

Here's what I did:

1. cosx [(sinx/cosx) + (cosx/sinx)] = cscx Conversion to sine and cosine + distribution property.

2. sinx + [(cos²x)/(sinx)] = cscx Multiplied the cosine of x to all in brackets.

3. sin²x + cos²x = cscx Multiplied by denominator sine of x.

4. 1 = cscx Pythagorean identity "sin²x + cos²x = 1."


I feel like something is totally off or missing. Like I didn't actually prove anything and just gave a value to cscx, which we supposedly don't know how to do yet. @_@ I'm always one step ahead, but that always sets me two steps behind. Help?

 

[soroban]

Hello, Aly!

You did very well . . . I just have some advice . . .

\(\displaystyle \text{Prove: }\,\cos x(\tan x\,+\,\cot x)\:=\:\csc x\)

With identities, start with one side of the equation and try to make it equal the other side.
That is, do not take the whole equation. \(\displaystyle \;\) (We don't know if they are equal, do we?)

But your steps were the correct one ... with the correct reasons.
\(\displaystyle \;\;\)I've taken the liberty of polishing it up a bit.

\(\displaystyle 1.\;\cos x\left(\frac{\sin x}{\cos x}\,+\,\frac{\cos x}{\sin x}\right)\;\;\) Conversion to sine and cosine

\(\displaystyle 2.\;\sin x\,+\,\frac{\cos^2x}{\sin x}\;\;\) Distributive property

\(\displaystyle 3.\;\frac{\sin^2x\,+\,\cos^2x}{\sin x}\;\;\) Get a common denominator.

\(\displaystyle 4.\;\frac{1}{\sin x}\;\;\) Pythagorean identity: \(\displaystyle \,\sin^2x\,+\,\cos^2x\:=\:1\)

\(\displaystyle 5.\;\csc x\;\;\) Reciprocal identity: \(\displaystyle \,\frac{1}{\sin x}\,=\,csc x\)

 

[Guest]

Whoa, so you mean my answer is correct? :lol:

Thanks for taking the time to read through my problem and "polish it up."

 

[galactus]

Try it like this:

\(\displaystyle \L\\cos(x)(tan(x)+cot(x))\)

\(\displaystyle \L\\=cos(x)(\frac{sin(x)}{cos(x)}+\frac{cos(x)}{sin(x)})\)

\(\displaystyle =\L\\cos(x)(\frac{1}{sin(x)cos(x)})\)

\(\displaystyle \L\\=\frac{\sout{cos(x)}}{sin(x)\sout{cos(x)}}\)

\(\displaystyle \L\\=\frac{1}{sin(x)}=csc(x)\)

 

[Guest]

Try it like this:

\(\displaystyle \L\\cos(x)(tan(x)+cot(x))\)

\(\displaystyle \L\\=cos(x)(\frac{sin(x)}{cos(x)}+\frac{cos(x)}{sin(x)})\)

\(\displaystyle =\L\\cos(x)(\frac{1}{sin(x)cos(x)})\)

\(\displaystyle \L\\=\frac{\sout{cos(x)}}{sin(x)\sout{cos(x)}}\)

\(\displaystyle \L\\=\frac{1}{sin(x)}=csc(x)\)

Oh... I see what I did. Thanks. That makes a lot of sense. Thank you so much. I was wondering if I needed to do something like that.

 

[Guest]

Another one I'm really stuck on!

Okay, so I got through the "easy" and "intermediate" ones without many problems. However, I'm totally confused how to even APPROACH this final one. Could someone get me started, because I have no clue as to where I need to begin. I tried dividing by cos²x, but that didn't get me very far at all.

Prove:


3sin²x + 4cos²x = 3 + cos²x


Thanks again!

 

[galactus]

You're gonna kick yourself when you see this one:

\(\displaystyle \L\\3sin^{2}(x)+4cos^{2}(x)=3+cos^{2}(x)\)

Let \(\displaystyle \L\\sin^{2}(x)=1-cos^{2}(x)\)

\(\displaystyle \L\\3(1-cos^{2}(x))+4cos^{2}(x)\)

Now, expand out and see what you get.

 

[skeeter]

Re: Another one I'm really stuck on!

Okay, so I got through the "easy" and "intermediate" ones without many problems. However, I'm totally confused how to even APPROACH this final one. Could someone get me started, because I have no clue as to where I need to begin. I tried dividing by cos²x, but that didn't get me very far at all.

Prove:


3sin²x + 4cos²x = 3 + cos²x

3sin²x + 3cos²x + 1cos²x = ?


Thanks again!

 

[Guest]

Re: Another one I'm really stuck on!

3sin²x + 3cos²x + 1cos²x = ?

So, 3 (sin²x + cos²x) + cos²x = 3 + cos²x

=>

3 (1) + cos²x = 3 + cos²x

=>

3 + cos²x = 3 + cos²x


D'oh!!! Thanks a lot to the both of you. I'd never thought to expand it before.