Trig Identities - Pre-calculus in a Nutshell - Section 4 Question 1

In summary, the conversation is about solving an identity involving trigonometric functions. The conversation starts with a user asking for help in establishing the identity, and another user providing a direct approach to solving it. The conversation also includes the user sharing their progress and thanking another user for their help. Eventually, the user is able to solve the identity using fractions and the
  • #1
Homework Statement
Morning, I hope someone could help me. I've tackled this one a few times and keep getting stuck.
Essentially its to establish the following identity
Relevant Equations
$$\frac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} = \sin \theta \tan \theta$$
My latest attempt
\begin{align*}
\frac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} = \\
\frac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} \cdot \frac{\csc \theta - \cot \theta}{\csc \theta - \cot \theta} =\\
\frac{\sin \theta \csc \theta + \tan\theta \csc \theta - \sin \theta \cot \theta - \tan \theta \cot \theta}{\csc^2 \theta - \cot^2 \theta} = \\
\frac{1 + \tan\theta \csc \theta - \sin \theta \cot \theta - 1}{\csc^2 \theta - \cot^2 \theta} = \\
\frac{\tan\theta \csc \theta - \sin \theta \cot \theta }{\csc^2 \theta - \cot^2 \theta} = \\
\frac{\tan\theta \csc \theta - \sin \theta \cot \theta }{1} = \\
\frac{\tan\theta}{\sin \theta} - \frac{\sin \theta} {\tan \theta} = \\
\end{align*}

I would even be grateful for any resources on solving problems like this, I have checked out few youtube videos which are helping, but the problems explained tend to have less terms.
Many thanks in advance
 
Physics news on Phys.org
  • #2
I'd convert everything to [itex] \sin\theta[/itex] and [itex] \cos\theta[/itex] first then simplify the expression.
 
  • Like
Likes DaveE, MichaelRocke and PeroK
  • #3
MichaelRocke said:
Homework Statement:: Morning, I hope someone could help me. I've tackled this one a few times and keep getting stuck.
Essentially its to establish the following identity
Relevant Equations:: $$\frac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} = \sin \theta \tan \theta$$

The first thing I saw was a factor of ##\sin \theta## on the LHS:
$$\sin \theta + \tan \theta = \sin \theta(1 + \frac{ 1}{ \cos \theta})$$
 
  • Like
Likes MichaelRocke
  • #4
FWIW I simply cleared the fraction by multipying by denominator. Everything cancels apart from the numerator.

$$ \frac{sin+tan}{csc+cot}=sin.tan \ \ \ \ \text {[multiply bs by } (csc+cot) \text{ ]} $$
$$⇒ sin+tan = csc.sin.tan + cot.sin.tan$$
$$⇒sin+tan=tan + sin \ \ \ \ \ W^5 $$
 
  • Like
Likes PeroK
  • #5
Thank you for your replies!

I have gotten so far\begin{align*}
\frac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} = \\
\frac{\sin \theta + \frac{\sin \theta} {\cos \theta} }{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}} = \\
\frac{\sin \theta (1 + \frac{1} {\cos \theta}) }{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}} = \\
\frac{\sin \theta (1 + \frac{1} {\cos \theta}) }{\frac{1}{\sin \theta}(1 + \cos \theta)} = \\
\frac{\sin^2 \theta (1 + \frac{1} {\cos \theta}) }{(1 + \cos \theta)} = \\
\frac{\sin^2 + \frac{\sin^2 \theta} {\cos \theta} }{(1 + \cos \theta)} = \\
...\\
\frac{\sin^2 \theta} {\cos \theta} = \\
\sin \theta \tan\theta
\end{align*}

Where I think I want to get it to the $$\frac{\sin^2 \theta} {\cos \theta}$$ bit of but keep running into brick walls.
Gonna have a little break and crack on again later tonight. I feel such an idiot for not getting this, but that won't stop me from trying!
 
  • #6
MichaelRocke said:
Thank you for your replies!

I have gotten so far\begin{align*}
\frac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} = \\
\frac{\sin \theta + \frac{\sin \theta} {\cos \theta} }{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}} = \\
\frac{\sin \theta (1 + \frac{1} {\cos \theta}) }{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}} = \\
\frac{\sin \theta (1 + \frac{1} {\cos \theta}) }{\frac{1}{\sin \theta}(1 + \cos \theta)} = \\
\frac{\sin^2 \theta (1 + \frac{1} {\cos \theta}) }{(1 + \cos \theta)} = \\
\end{align*}
Just take a factor of ##\cos \theta## out the denominator and you are done.
 
  • Like
Likes scottdave and MichaelRocke
  • #7
There was a direct approach:
$$\frac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} = \sin \theta (\frac{1 + \frac{1}{\cos \theta}}{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}}) = (\sin \theta \tan \theta)(\frac{\cos \theta}{\sin \theta})( \frac{1 + \frac{1}{\cos \theta}}{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}})$$
The strategy is to take out the answer and reduce what's left to identically ##1##.
 
  • Like
Likes MichaelRocke
  • #8
Thank you!
I literally stuck at it for a few minutes more and learned how to fiddle with the fractions so it cancels
\begin{align*}
\frac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} = \\
\frac{\sin \theta + \frac{\sin \theta} {\cos \theta} }{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}} = \\
\frac{\sin \theta (1 + \frac{1} {\cos \theta}) }{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}} = \\
\frac{\sin \theta (1 + \frac{1} {\cos \theta}) }{\frac{1}{\sin \theta}(1 + \cos \theta)} = \\
\frac{\sin^2 \theta (1 + \frac{1} {\cos \theta}) }{(1 + \cos \theta)} = \\
\frac{\sin^2 \theta (\frac{\cos \theta + 1}{\cos \theta}) }{(1 + \cos \theta)} = \\
\frac{\sin^2 \theta} {\cos \theta} = \\
\sin \theta \tan\theta
\end{align*}

Shout out to PeroK for your help! Cheers!
 
  • Like
Likes scottdave and PeroK
  • #9
I think you'd have found it much easier to keep the cot, csc and sec. The whole idea (I thought) of using these auxiliary functions was to avoid all the fractions.
Then start by clearing the given fraction and work with sums and products.

Once you get a term containing a pair of reciprocals, like cot and tan, they just disappear.

I'm not sure this would help if the result were a complicated fraction, but here you knew the result was simple sin and tan, so all the other functions must cancel.
 
  • #10
Merlin3189 said:
FWIW I simply cleared the fraction by multipying by denominator. Everything cancels apart from the numerator.

$$ \frac{sin+tan}{csc+cot}=sin.tan \ \ \ \ \text {[multiply bs by } (csc+cot) \text{ ]} $$
$$⇒ sin+tan = csc.sin.tan + cot.sin.tan$$
$$⇒sin+tan=tan + sin \ \ \ \ \ W^5 $$

This is a very neat solution, but I think it would be better like this:
$$\sin \theta \tan \theta (\csc \theta + \cot \theta) = \tan \theta + \sin \theta$$
Since:
$$\sin \theta \csc \theta = 1,\ \tan \theta \cot \theta = 1$$
And the result follows.
 
  • Like
Likes neilparker62
  • #11
Fair enough. I'm not very math-sophisticated.
I did wonder whether I should have done it by some formal procedure, but it seemed so obvious once you simplified the equation.
 
  • #12
Thank you Merlin3189 for your solution too.
I like the idea, very elegant too.

For my approach, I tried to reach the end of $$\sin\theta\tan\theta$$ without using it in any steps (if that makes sense), I don't know if I was being overly strict on myself?
 
  • #13
As I say, not sophisticated at math. If I needed to do it as you suggest, following my simplification, I then have the clue to do it:

## \frac{sin+tan}{csc+cot}= \frac {sin.tan}{sin.tan} \frac { (sin+tan)}{ (csc+cot)} ##

## \frac{sin+tan}{csc+cot} = \frac {sin.tan (sin+tan) } { sin.tan.csc + sin.tan.cot } ##

## \frac{sin+tan}{csc+cot} = \frac { sin.tan (sin+tan) } { tan.(sin.csc) + sin.(tan.cot) } ##

## \frac{sin+tan}{csc+cot} = \frac {sin.tan (sin+tan) } { (tan + sin) } ##

## \frac{sin+tan}{csc+cot} = sin.tan ##

Which would be shorter if I didn't try to explain.

PS. Sorry it's taken so long - I've been doing battle with Latex to try to show cancellation (and as you see, losing!)
 
  • #14
MichaelRocke said:
Homework Statement:: Morning, I hope someone could help me. I've tackled this one a few times and keep getting stuck.
Essentially its to establish the following identity
Relevant Equations:: ## \dfrac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} = \sin \theta \tan \theta##

My latest attempt:
...

##\dfrac{\tan\theta \csc \theta - \sin \theta \cot \theta }{1} ##
I see you have completed this, but looking at your Original Post, I see that you nearly had the result there.

After that you have: ##\dfrac{\sin \theta}{\cos \theta}\cdot \dfrac{1}{\sin \theta} - \dfrac{\sin \theta}{1} \cdot \dfrac{\cos \theta}{\sin \theta}##

##\ =\dfrac{1}{\cos \theta}-\dfrac{\cos \theta}{1}##​
##\ =\dfrac{1-\cos^2 \theta}{\cos \theta}##​

etc.
 
  • Like
Likes MichaelRocke
  • #15
Good work!
For the life of me, I can't figure out why anyone every invented or used sec, csc, and cot functions. I thought they were pointless when I first saw them and 40 years later, I still think they are.
Of course it's just a personal preference, but I will always prefer a little simple algebra in place of creating a whole family of redundant functions.
 
  • Like
Likes MichaelRocke, ehild and Dr Transport
  • #16
DaveE said:
Good work!
For the life of me, I can't figure out why anyone every invented or used sec, csc, and cot functions.
I met sec and csc here, in PF first in my life, and I always have to check which is 1/sin and 1/cos. We did not learn them here, in Hungary.
 
  • Like
Likes MichaelRocke
  • #17
DaveE said:
Good work!
For the life of me, I can't figure out why anyone every invented or used sec, csc, and cot functions. I thought they were pointless when I first saw them and 40 years later, I still think they are.
Of course it's just a personal preference, but I will always prefer a little simple algebra in place of creating a whole family of redundant functions.

I hardly ever use ##\csc## or ##\cot##, but I find ##\sec## more useful, because of its relation to ##\tan##:
$$\sec^2 \theta = 1 + \tan^2 \theta; \ \ \ \frac{d }{d\theta}\tan \theta = \sec^2 \theta$$
That said, I'd be happy to use ##\frac 1 {\cos \theta}##, if ##\sec## were abolished with the others.
 
  • #18
MichaelRocke said:
Thank you Merlin3189 for your solution too.
I like the idea, very elegant too.

For my approach, I tried to reach the end of $$\sin\theta\tan\theta$$ without using it in any steps (if that makes sense), I don't know if I was being overly strict on myself?

Here's what I would take from this exercise. There are several different strategies for proving an identity, both for trig identities and more generally:

1) Express everything in common terms. In this case, ##\cos \theta## and ##\sin \theta##. I would say this is a reliable generic method.

2) Try to extract the answer and then simplify what's left to ##1## or ##0##, as appropriate. In this case, that was to take out a factor of ##\sin \theta \tan \theta##.

3) Try to re-express the identity. In this case it was easier to prove the equivalent identity: ##\sin \theta \tan \theta (\csc \theta + \cot \theta) = \tan \theta + \sin \theta##.

I use all these techniques frequently. If things don't obviously simplify one way, it's amazing the way things just fall out in 1-2 lines when you try a different approach.
 
  • Like
Likes MichaelRocke
  • #19
PeroK said:
I'd be happy to use 1cosθ\frac 1 {\cos \theta}, if sec\sec were abolished with the others.

The exact reason I said to convert everything to [itex] \sin[/itex] and [itex] \cos[/itex] initially.
 
  • #20
I wonder how much easier the problem would be if it were simply to show that:
$$\frac{a + \frac a b}{\frac 1 a + \frac b a} = \frac{a^2}{b}$$
 
  • Like
Likes DaveE
  • #21
PeroK said:
Here's what I would take from this exercise. There are several different strategies for proving an identity, both for trig identities and more generally:

1) Express everything in common terms. In this case, ##\cos \theta## and ##\sin \theta##. I would say this is a reliable generic method.

2) Try to extract the answer and then simplify what's left to ##1## or ##0##, as appropriate. In this case, that was to take out a factor of ##\sin \theta \tan \theta##.

3) Try to re-express the identity. In this case it was easier to prove the equivalent identity: ##\sin \theta \tan \theta (\csc \theta + \cot \theta) = \tan \theta + \sin \theta##.

I use all these techniques frequently. If things don't obviously simplify one way, it's amazing the way things just fall out in 1-2 lines when you try a different approach.

Brilliant, thank you for your advice here. I have plenty of more exercises to go and starting to tear through them at a much faster pace.
 
  • #22
##
\begin{align}
\frac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} &\stackrel{?}{=} \sin \theta \tan \theta\\
\frac{\sin \theta + \tan \theta}{\frac{1}{\sin \theta }+ \frac{1}{\tan \theta}} &\stackrel{?}{=} \sin \theta \tan \theta\\
\frac{S + T}{\left(\frac{1}{S }+ \frac{1}{T}\right)} &\stackrel{?}{=} S\ T\\
\frac{S + T}{S\ T} &\stackrel{\checkmark}{=} \left(\frac{1}{S }+ \frac{1}{T}\right)
\end{align}
##
 
  • Like
Likes MichaelRocke
  • #23
Possibly enlightening...
1582097399873.png

https://commons.wikimedia.org/wiki/File:Circle-trig6.svg
https://blogs.scientificamerican.co...unctions-your-math-teachers-never-taught-you/

https://www.theonion.com/nation-s-math-teachers-introduce-27-new-trig-functions-1819575558
 
  • Like
Likes SammyS and MichaelRocke
  • #24
MichaelRocke said:
Homework Statement:: Morning, I hope someone could help me. I've tackled this one a few times and keep getting stuck.
Essentially its to establish the following identity
Relevant Equations:: $$\frac{\sin \theta + \tan \theta}{\csc \theta + \cot \theta} = \sin \theta \tan \theta$$
The expression has denominators of ##\sin\theta## and ##\cos\theta## so I would multiply top and bottom of the fraction by ##\sin\theta \cos\theta## to clear.
 
  • #25
PeroK said:
I wonder how much easier the problem would be if it were simply to show that:
$$\frac{a + \frac a b}{\frac 1 a + \frac b a} = \frac{a^2}{b}$$
Or more simply [tex]
\frac{a + b}{a^{-1} + b^{-1}} = ab.[/tex]
 
  • Like
Likes neilparker62

Suggested for: Trig Identities - Pre-calculus in a Nutshell - Section 4 Question 1

Replies
1
Views
852
Replies
4
Views
477
Replies
9
Views
2K
Replies
10
Views
1K
Replies
9
Views
1K
Replies
3
Views
988
Replies
18
Views
2K
Back
Top