Proving Trigonometric Identities problem

[acen_gr]

Homework Statement

Verify that \frac{cosθ}{1-tanθ} + \frac{sinθ}{1-cotθ} = sinθ + cosθ is an identity.

Homework Equations

- Reciprocal Trigonometric Identities
- Pythagorean Trig Identities

The Attempt at a Solution

Every time I try to manipulate the LHS of the equation I always get -1 and as far as I know, the sum of sine and cosine always has to be both squared to get 1. But in this case, the RHS has sinθ + cosθ. Is it possible for sinθ + cosθ to equal 1? Or I just made the wrong way of manipulating in LHS? Thank you!

PS: A hint for the problem will do. I would love to show my work from what I will have to learn from your replies.

 

[Simon Bridge]

It would help if you showed how you kept getting -1 for the LHS.

I'd have started from \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)} and \cot(\theta)=\tan(\frac{\pi}{2}-\theta) remembering how sine and cosine transform when you shift the angle by π/2.

 

[acen_gr]

It would help if you showed how you kept getting -1 for the LHS.

I'd have started from \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)} and \cot(\theta)=\tan(\frac{\pi}{2}-\theta) remembering how sine and cosine transform when you shift the angle by π/2.

Here it is:

\frac{cosθ}{1-tanθ} + \frac{sinθ}{1-cotθ}

\frac{cosθ}{1-\frac{sinθ}{cosθ}} + \frac{sinθ}{1-\frac{cosθ}{sinθ}}

\frac{cosθ}{\frac{cosθ-sinθ}{cosθ}} + \frac{sinθ}{\frac{sinθ-cosθ}{sinθ}}

\frac{cos^2\theta}{cosθ-sinθ} + \frac{sin^2\theta}{sin\theta-cos\theta}

\frac{cos^2\theta}{-(sin\theta-cos\theta} + \frac{sin^2\theta}{sin\theta-cos\theta}

-(cos² + sin²) = -1

 

[CAF123]

I get to \frac{cos^2θ}{cosθ-sinθ} + \frac{sin^2θ}{sinθ-cosθ} in your line of working.
Then you could take out -1 as a factor:
\frac{cos^2θ}{cosθ-sinθ} + \frac{sin^2θ}{-(cosθ-sinθ)}. Use the rules for adding fractions and subsequently simplify.

 

[acen_gr]

I get to \frac{cos^2θ}{cosθ-sinθ} + \frac{sin^2θ}{sinθ-cosθ} in your line of working.
Then you could take out -1 as a factor:
\frac{cos^2θ}{cosθ-sinθ} + \frac{sin^2θ}{-(cosθ-sinθ)}. Use the rules for adding fractions and subsequently simplify.

Hi, CAF123. Thank you for your reply! Yes sir. I did that. And from that I could get the simplified form -1 which is wrong because the RHS of the equation has sinθ + cos θ and there couldn't be any way for sinθ + cosθ to be equal to -1. Right?

 

[CAF123]

To go a little further, you get something like:
\frac{-cos^2θ + sin^2θ}{-(cosθ - sinθ)}, yes?

The reason you get -1 is because you must be manipulating the fractions wrong. Do you see that the above is not equal to -1?

 

[acen_gr]

To go a little further, you get something like:
\frac{-cos^2θ + sin^2θ}{-(cosθ - sinθ)}, yes?

The reason you get -1 is because you must be manipulating the fractions wrong. Do you see that the above is not equal to -1?

If you have two fractions with the same denominator (like the fractions above), you could cancel the denominator and proceed to addition right? That's what I did sir. Is it not right?

 

[acen_gr]

To go a little further, you get something like:
\frac{-cos^2θ + sin^2θ}{-(cosθ - sinθ)}, yes?

The reason you get -1 is because you must be manipulating the fractions wrong. Do you see that the above is not equal to -1?

This is now my work based on what you said:

\frac{cosθ}{1-tanθ} + \frac{sinθ}{1-cotθ}

\frac{cosθ}{1-\frac{sinθ}{cosθ}} + \frac{sinθ}{1-\frac{cosθ}{sinθ}}

\frac{cosθ}{\frac{cosθ-sinθ}{cosθ}} + \frac{sinθ}{\frac{sinθ-cosθ}{sinθ}}

\frac{cos^2}{cos\theta-sin\theta} + \frac{sin^2}{sin\theta-cos\theta}

\frac{-cos^2\theta+sin^2\theta}{-(cos\theta-sin\theta)}

\frac{-(cos^2\theta-sin^2\theta)}{-(cos\theta-sin\theta)}

\frac{cos^2\theta-sin^2\theta}{(cos\theta-sin\theta)}

\frac{(cos\theta-sin\theta)(cos\theta+sin\theta)}{(cos\theta-sin\theta)}

cosθ + sinθ

I would like to confirm my answer from you sir. Is this right now? Thank you so much! :)

 

[CAF123]

If I understand what you mean, then no, this is wrong.
If I understand you correctly, what you have done is said \frac{cos^2θ}{-(sinθ -cosθ)}+ \frac{sin^2θ}{sinθ-cosθ} = -(cos^2θ + sin^2θ) = -1 ?

This is not correct. To see why, consider \frac{1}{-(2)} +\frac{1}{2}. The above approach gives -(1+1) = -2 and clearly -1/2 + 1/2 =0.

EDIT: Your answer is now correct!

 

[Simon Bridge]

\frac{cos²}{cosθ-sinθ} + \frac{sin²}{sinθ-cosθ}

\frac{cos²}{-(sinθ-cosθ} + \frac{sin²}{(sinθ-cosθ)}

... these last two didn't render because you are mixing bv codes with LaTeX too much. It's usually better to write it out by hand.

Some tips:

θ = \theta
you raise to a power with the ^ character
sin and cos and tan are \sin and \cos and \tan
you can get fractions to work better with the tex tag instead of the itex tag.

So you got:
\Leftrightarrow \frac{\cos^2\theta}{\cos\theta - \sin\theta} + \frac{\sin^2\theta}{\sin\theta - \cos\theta}... put LHS under a common denominator gives you:\Leftrightarrow \frac{\cos^2\theta}{\cos\theta - \sin\theta} - \frac{\sin^2\theta}{\cos\theta - \sin\theta}... that's the LHS: multiply both sides by the common denominator ;)

[edit] Ah: more happened while I was typing.

 

[acen_gr]

If I understand what you mean, then no, this is wrong.
If I understand you correctly, what you have done is said \frac{cos^2θ}{-(sinθ -cosθ)}+ \frac{sin^2θ}{sinθ-cosθ} = -(cos^2θ + sin^2θ) = -1 ?

This is not correct. To see why, consider \frac{1}{-(2)} +\frac{1}{2}. The above approach gives -(1+1) = -2 and clearly -1/2 + 1/2 =0.

EDIT: Your answer is now correct!

Yay! Thank you! I should have considered that it's not always true for every situation. I could claim such for that was what we are taught about doing it by our teachers. Now I feel terrible knowing we must be doing wrong ways all the time in math because that was what we are taught of how it must be done :( But I'm glad that you helped me cease this wrong process now that I used to know.

... these last two didn't render because you are mixing bv codes with LaTeX too much. It's usually better to write it out by hand.

Some tips:

θ = \theta
you raise to a power with the ^ character
sin and cos and tan are \sin and \cos and \tan
you can get fractions to work better with the tex tag instead of the itex tag.

So you got:
\Leftrightarrow \frac{\cos^2\theta}{\cos\theta - \sin\theta} + \frac{\sin^2\theta}{\sin\theta - \cos\theta}... put LHS under a common denominator gives you:\Leftrightarrow \frac{\cos^2\theta}{\cos\theta - \sin\theta} - \frac{\sin^2\theta}{\cos\theta - \sin\theta}... that's the LHS: multiply both sides by the common denominator ;)

[edit] Ah: more happened while I was typing.

I should thank you a lot for this. I was dying trying to edit over and over my post because it seems to appear not right. I just figured it out now with your help and was able to make it right now. Thanks!

Thanks for all who shared their ideas and tips. I'm really having fun learning here from you guys!

 

[acen_gr]

I'd have started from \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)} and \cot(\theta)=\tan(\frac{\pi}{2}-\theta) remembering how sine and cosine transform when you shift the angle by π/2.

This seems to be a different method from mine. I actually got the problem from the exercises for basic trigonometric identities in the book. So I wasn't able to use this method. But I would like to know this method. I believe your method involves other group of identities like cofunctions identities and half angle identities?

 

[CAF123]

Consider \tan(\frac{π}{2} - θ).
This is equal to \frac{\sin(\frac{π}{2} - θ)}{\cos(\frac{π}{2} - θ)} = \frac{\cosθ}{\sinθ} = \cotθ.

 

[Simon Bridge]

If you look at a right angled triangle - you normally label one angle \theta, and the sides would be H, O, and A as normal ... since the sum of the angles in a triangle is \pi radiens, the other angle, the one that is not theta, is \frac{\pi}{2}-\theta. That "other angle" is called the "complimentary angle" or "co-angle" for short.

The sine of \theta would be O/H
The cosine is, by definition, the complimentary sine: sine of the complimentary angle: i.e. \sin(\frac{\pi}{2}-\theta) ... you work it out you'll see it comes to A/H for your triangle ... which is the usual definition of the cosine you are used to.

Similarly, the cotangent is the tangent of the complimentary angle.
Since \tan(\theta)=O/A it follows that \cot(\theta)=A/O.

You should have these angle shifts in your list of trig identities. It may look like: \sin(90-\theta)=\cos(\theta)

You notice I didn't immediately go "cot = cos/sin"? I don't actually memorize these things: I just use an understanding of what the trig functions actually mean.

The trig functions are pretty much just the names for specially drawn line segments on the unit circle.

http://en.wikipedia.org/wiki/Unit_circle#Trigonometric_functions_on_the_unit_circle

[Aside:
... see how the trig functions in CAF123's replies are in normal text and in yours they are in italics? That is what the backslash does. "\sin" etc is a special function that tells LaTeX to write out the letters "sin" in a special way. It also affects the spacing.
... you can see the LaTeX markup i someone's reply by hitting the "quote" button at the bottom of the reply ... the code will be displayed in the quoted text.]