Simplifying Trigonometric Expressions

[TrueStar]

Homework Statement

I'm not sure if you can ask about two questions in one thread; they are very similar. If it's not allowed I can make a second thread and edit this one.

Simplify the expressions:

1. sin^3 theta + cos^3 theta / sin theta + cos theta

2. 2-tan theta / 2 csc theta-sec theta

Homework Equations

Pythagorean Identities

The Attempt at a Solution

1. if the either of these were squared, they could be equal to 1. I thought that I could get rid of th denominator and make sin^2 and cos^2 that way...and it would simplify to 1. The answer I am given is actually 1-sin theta cos theta. I think this has to factor somehow, but I'm not certain. I'm not used to dealing with problems that look like this yet.

2. I think this has to do with factoring something out as well, but I'm not sure what. I tried changing them all to variations of sin and cos, but nothing clicked for me.

 

[roam]

I wonder how they've got that answer, I don't see anything wrong with what you did.

What happens if we use the identities

sin^2 \theta = \frac{1}{2} (1-cos2 \theta)

cos^2 \theta = \frac{1}{2} (1+cos2 \theta)

 

[rock.freak667]

For

\frac{sin^3 \theta + cos^3 \theta}{sin \theta + cos \theta}

you should notice that sin³θ+cos³θ=0 when sinθ=cosθ, so try factoring the numerator as (sinθ+cosθ)*f(θ)

 

[TrueStar]

For

\frac{sin^3 \theta + cos^3 \theta}{sin \theta + cos \theta}

you should notice that sin³θ+cos³θ=0 when sinθ=cosθ, so try factoring the numerator as (sinθ+cosθ)*f(θ)

I'm sorry, I don't understand. Why would sin³theta+cos³theta=0 when sin theta=cos theta. Maybe it's something we haven't covered yet?

We are expected to get the answer from the even odd properties and the Pythagorean Identities.

I apologize if I'm making this difficult. :(

 

[Elucidus]

My recommendation for the first problem is to ignore momentarily sine and cosine and focus on how one would simplify

\frac{a^3+b^3}{a+b}

and then deal with the resulting expression.

As for the second problem I am assuming you are working with the expression

\frac{2-\tan \theta}{2 \csc \theta - \sec \theta}.

If not, please correct me. If so, then I'd recommend multiplying the numerator and denominator by \sin \theta \cdot \cos \theta and see if that helps simplify the ratio.

--Elucidus

 

[rock.freak667]

I'm sorry, I don't understand. Why would sin³theta+cos³theta=0 when sin theta=cos theta. Maybe it's something we haven't covered yet?

We are expected to get the answer from the even odd properties and the Pythagorean Identities.

I apologize if I'm making this difficult. :(

I meant that you can factor a³+b³ into (a+b)*(something).

 

[TrueStar]

A quick reply for you Elucidus on the first problem. THANK YOU! I get it now. It's the formula (x+y)(x^2-xy+y^2) only with sin and cos. I worked it out on some paper and it fell into place. It's so frustrating when I can't see ways to solve the problem that I've already learned.

As for the second problem, you have the expression correct. I'm going to work on it now.

 

[TrueStar]

OK, I've worked through the second problem. I doubted myself on this one. I changed everything to sin/cos and thought about multiplying everything by sin and cos. I thought that wasn't right and I wasn't sure if I was multiplying correctly anyway. I should just try my ideas next time.

Anyway, it all fell into place and I understand how it works. Now I can sleep well tonight. Thank you all so much for your time and patience. :)