Solving Trig Identities

**mike_302** · May 24th 2008, 03:24 PM

So my homework is to solve these trig identities. I'm in grade 11, so we have only begun by learning the Pythagorean Identity, the Quotient identity and its inverse, as well as the 3 (or 6... depending how you look at it) that include sin=1/csc, cos=1/sec, and tan=1/cot (and the inverses for the last 3). So Our task was to solve thse trig identities. I've scanned my work in for 5 that I cannot seem to solve. Mind you guys, I know there's a couple in there that will end up having the same method of solutions, so if you find a couple like that in the 5, just do the one that seems easiest, please. I'm sure I could then figure out the other one. But I really do need some tips in the right direction on these. Also mind you, I have attempted a few steps, as you can see, but I may have gone about solving these in the improper method, so don't take the steps that I have included as correct.

Thanks in advance!

http://i160.photobucket.com/albums/t...identities.jpg (On this page, I need help on (e) AND (f) .
http://i160.photobucket.com/albums/t...dentities2.jpg

**Jonboy** · May 24th 2008, 04:02 PM

i find it best to get things in terms of sin and cosine.

So we have: $\frac{1 + sin\,\theta}{1 + csc\,\theta} = sin\theta$

Change $csc\,\theta$ in terms of Sine:

$\frac{1 + sin\,\theta}{1 + \frac{1}{sin\,\theta}}$

Get one fraction on the bottom: $\frac{1 + sin\,\theta}{\frac{sin\,\theta + 1}{sin\,\theta}}$

When dividing, you may multiply the numerator by the reciprocal (flipping the fraction) of the denominator.

So: $\frac{1 + sin\,\theta}{1}\cdot\frac{sin\,\theta}{sin\,\theta + 1}$

The $sin\,\theta + 1$ cross cancel, and you have the right side.

**mike_302** · May 24th 2008, 04:07 PM

wow, definate thanks. I completed that one as you said and I understand the idea behind it (personally, I hate these... they're like puzzles but there's no way to tell if the route you're taking is going to make your life easy or difficult..) . I'll see if that helps me with any of the others. In the mean time, if anyone else notices any of the others that are completely different, I coul still use a hand with those.

thanks again!

**Jonboy** · May 24th 2008, 04:12 PM

yeah that's how i felt about these as well. but once you do so many, you just know what to do. i'll show you a way for f. it's a little more tricky.

**Jonboy** · May 24th 2008, 04:24 PM

that simple fundamental, change things into sin and cosine, will help us here.

We have: $\frac{sin\,\theta + tan\,\theta}{cos\,\theta\,+\,1} = tan\,\theta$

Recall: $tan\,\theta\,=\,\frac{sin\,\theta}{cos\,\theta}$

So apply that: $\frac{sin\,\theta + \frac{sin\,\theta}{cos\,\theta}}{cos\,\theta\,+\,1 }$

Get one fraction on the top:

$\frac{\frac{cos\,\theta\cdot sin\,\theta + sin\,\theta}{cos\,\theta}}{cos\,\theta + 1}$

Factor the very top:

$\frac{\frac{sin\,\theta(cos\,\theta + 1)}{cos\,\theta}}{cos\,\theta + 1}$

Multiply by the reciprocal: $\frac{sin\,\theta(cos\,\theta + 1)}{cos\,\theta}\cdot\frac{1}{cos\,\theta + 1}$

The $cos\,\theta + 1$ cancels, and you're left with:

$\frac{sin\,\theta}{cos\,\theta}$

Which is $tan\,\theta$ . There you go!

**mike_302** · May 24th 2008, 04:35 PM

absolutly, purely genius. I love it. DEFINATELY a different way of looking at it. Thanks ! I mean, not an easier way, but the different way of looking at the questions and solving makes life that much more interesting. Thanks VERY much for that.

On the other hand, I tried using that method (of switching all to sine/cos) on the other page of questions and had luck on only 4. c) in the sense that i got it all down to sin^2=cos/cos (Sorry, I'm not very knowledgeable with writing with LaTeX... If I had some sort of latex generator maybe that would help us)

Edit: Found one, google. Will retype that equation

**mike_302** · May 24th 2008, 04:50 PM

woow, sorry, that latex generator works as in it renders the latex image, but I cannot seem to put it into here properly. Sorry bout that. But definately 4. c), by changing the tan to sin over cos (or sin^2 over cos^2 in the one case), I was able to work out : sin^2=cos/cos ... or sin^2=1

**Soroban** · May 24th 2008, 04:55 PM

Hello, Mike!

You seem to have the basics that are needed.
You should learn to "clear" the complex fractions . . .

$(e)\;\;\frac{1 + \sin\theta}{1 + \csc\theta} \:=\:\sin\theta$

The left side is: . $\frac{1 + \sin\theta}{1 + \frac{1}{\sin\theta}}$

Multiply by $\frac{\sin\theta}{\sin\theta}\!:\;\;\frac{\sin\the ta(1 + \sin\theta)}{\sin\theta\left(1 + \frac{1}{\sin\theta}\right)} \;=\;\frac{\sin(1 + \sin\theta)}{\sin\theta + 1} \;=\;\sin\theta$

$(f)\;\;\frac{\sin\theta + \tan\theta}{\cos\theta + 1} \;=\;\tan\theta$

The left side is: . $\frac{\sin\theta + \frac{\sin\theta}{\cos\theta}}{\cos\theta + 1} \;=\;\frac{\sin\theta\left(1 + \frac{1}{\cos\theta}\right)}{\cos\theta + 1}$

Multiply by $\frac{\cos\theta}{\cos\theta}\!:\;\;\frac{\cos\the ta\cdot\sin\theta\left(1 + \frac{1}{\cos\theta}\right)}{\cos\theta(\cos\theta + 1)} \;=\;\frac{\sin\theta(\cos\theta + 1)}{\cos\theta(\cos\theta + 1)}$

. . . . . . $= \;\frac{\sin\theta}{\cos\theta} \;=\;\tan\theta$

$4(b)\;\;\csc^2\!\theta - 1 \;=\;\csc^2\!\theta\cos^2\!\theta$

The left side is a basic identity . $\csc^2\!\theta-1 \;=\;\cot^2\!\theta$

Then: . $\cot^2\!\theta \;=\;\frac{\cos^2\!\theta}{\sin^2\!\theta} \;=\;\frac{1}{\sin^2\!\theta}\cdot\cos^2\!\theta \;=\;\csc^2\!\theta\cos^2\!\theta$

$(c)\;\;\sin^2\!\theta \;=\;\frac{\tan^2\!\theta}{1 + \tan^2\!\theta}$

The right side is: . $\frac{\frac{\sin^2\!\theta}{\cos^2\!\theta}}{1 + \frac{\sin^2\!\theta}{\cos^2\!\theta}}$

Multiply by $\frac{\cos^2\!\theta}{\cos^2\!\theta}\!:\;\;\frac{ \cos^2\!\theta\left(\frac{\sin^2\!\theta}{\cos^2\! \theta}\right)}{\cos^2\!\theta\left(1 + \frac{\sin^2\!\theta}{\cos^2\!\theta}\right)}$

. . . . . $= \;\;\frac{\sin^2\!\theta}{\underbrace{\cos^2\!\the ta + \sin^2\!\theta}_{\text{This is 1}}} \;\;=\;\;\sin^2\!\theta$

Edit: While I was typing up this long disseration, Jonboy did a great job!
.

**mike_302** · May 24th 2008, 05:03 PM

Thanks for the help there. My only query is with 4 b and c, since any other basic identity is not supposed to be used until grade 12 math, and so, from what I can guess, we are not supposed to use those, but figure out our own ways around that in order to prove these identities. You have managed to solve them, but with identities of which we are not supposed to know until grade 12.

Thanks though! I can appreciate the amount of time that took you.

So, for b and c, I still require methods to prove the identity using the identities that I listed in my first post . (Sorry though! I really do appreciate the post)

**Jonboy** · May 24th 2008, 05:12 PM

thanks for the post and compliment Soroban! it's always a pleasure to read your amazing post. i'm out, as i'm starving.

**mike_302** · May 24th 2008, 05:13 PM

I must thank you again Jonboy, and same Soroban

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