1. ## Simplifying trigonometric expressions

Two questions for now:

1. Simplify sin$\displaystyle \theta$/1+cos$\displaystyle \theta$ * 1-sin^2$\displaystyle \theta$/cot$\displaystyle \theta$.

I got as far as sin^2$\displaystyle \theta$cos$\displaystyle \theta$/1 + cos$\displaystyle \theta$. Is there any further i can go??

2. Simplify sin^3$\displaystyle \theta$-cos^3$\displaystyle \theta$/sin^2$\displaystyle \theta$-cos^2$\displaystyle \theta$.

Can i use difference and sum of cubes here to factorise or is this not possible?

Thanks!
Steven

2. Hello,

wHat exactly is in the denominator ?

3. can you post what have you done so far?

4. lol sorry i really struggle with the math tags

Ok for the first one:

(sin$\displaystyle \theta$) (1-sin^2$\displaystyle \theta$)
--------------------------
(1 + cos$\displaystyle \theta$) (cot$\displaystyle \theta$)

So far all i've been able to do is simplify it somehow to:

(sin^2$\displaystyle \theta$ * cos$\displaystyle \theta$)
----------------------
(1 + cos$\displaystyle \theta$)

I don't even know if this is right though. We are supposed to be using trigonometric identities but im a bit rusty with the algebra (which i know should be quite easy)

5. Hmmm, for the latex, use $$\frac{numerator}{denominator}$$ for a fraction

You can notice that $\displaystyle \sin^2 \theta=1-\cos^2 \theta=(1-\cos \theta)(1+\cos \theta)$

Can i use difference and sum of cubes here to factorise or is this not possible?
Yes you can ! (it's a difference here )
And factorise the denominator too : $\displaystyle a^2-b^2=(a-b)(a+b)$

6. Thanks!

But for the first one doesn't that mean that

sin^2$\displaystyle \theta$*cos$\displaystyle \theta$
-----------
1 + cos$\displaystyle \theta$

becomes

(1-cos^2$\displaystyle \theta$)(cos$\displaystyle \theta$)
-------------
1 + cos$\displaystyle \theta$

???

7. Yes, it is the idea ! This is due to the identity $\displaystyle \cos^2 \theta+\sin^2 \theta=1$

Then it will yield $\displaystyle \frac{(1-\cos \theta)(1+ \cos \theta)(\cos \theta)}{1+\cos \theta}$

8. That's much better!

So the back of my book tells me the answer is

cos$\displaystyle \theta$
----------
1 + cos$\displaystyle \theta$

Is this the same thing (obviously a more simplified version)? If so, how do they get to the final stage?

P.S. Thanks for alerting me to the mistake! I was panicking for a moment there

9. Originally Posted by Stevo_Evo_22
P.S. Thanks for alerting me to the mistake! I was panicking for a moment there
Yes, I'm really sorry for that

cos$\displaystyle \theta$
----------
1 + cos$\displaystyle \theta$

Is this the same thing (obviously a more simplified version)? If so, how do they get to the final stage?
This is weird

It's not the same result at all...

I'll start from scratch, ok ?
--------------------------------

$\displaystyle \frac{\sin \theta (1-\sin^2 \theta)}{(1+\cos \theta) \cot \theta}$

$\displaystyle =\frac{\sin \theta \cos^2 \theta}{(1+\cos \theta) \cdot \frac{\cos \theta}{\sin \theta}}$

$\displaystyle =\frac{\sin \theta \cos \theta}{(1+\cos \theta)} \cdot \sin \theta$

$\displaystyle =\frac{\sin^2 \theta \cos \theta}{1+\cos \theta} \neq \frac{\cos \theta}{1+\cos \theta}$

Is there a possible mistake in the book ?

10. It is extremely possible that there is a mistake in the book (it is full of them). So i will take your word for it (particularly because yours makes sense and i got almost the same answer). That help was invaluable to me, I was so close, yet so far Just one more thing, in the end can you cancel the bottom cos$\displaystyle \theta$ with the top one? or is that one the final answer?

Ok, so for the second one, i am at this point:

Sin^2$\displaystyle \theta$ + cos$\displaystyle \theta$sin$\displaystyle \theta$ + cos^2$\displaystyle \theta$
-------------------
sin$\displaystyle \theta$ + cos$\displaystyle \theta$

This is after I have factorised it, but this par confuses me, especially the top bit. Am i on the right track or way off?

11. Originally Posted by Stevo_Evo_22
It is extremely possible that there is a mistake in the book (it is full of them). So i will take your word for it (particularly because yours makes sense and i got almost the same answer). That help was invaluable to me, I was so close, yet so far Just one more thing, in the end can you cancel the bottom cos$\displaystyle \theta$ with the top one? or is that one the final answer?
How do you want to simplify the cosines ??

Isn't the final answer $\displaystyle (1-\cos \theta) \cos \theta$ ?

Well, it's correct you were on the right track Now you just need some practice ^^

Ok, so for the second one, i am at this point:

Sin^2$\displaystyle \theta$ + cos$\displaystyle \theta$sin$\displaystyle \theta$ + cos^2$\displaystyle \theta$
-------------------
sin$\displaystyle \theta$ + cos$\displaystyle \theta$

This is after I have factorised it, but this par confuses me, especially the top bit. Am i on the right track or way off?
Right track
This is the good factorisation !

Now you can play with it...

$\displaystyle \sin^2 \theta+\cos \theta \sin \theta+\cos^2 \theta=\underbrace{\sin^2 \theta+{\color{red}2} \cos \theta \sin \theta+\cos^2 \theta}_{a^2+2ab+b^2=\dots ?} {\color{red}-\cos \theta \sin \theta}$

and then $\displaystyle \frac{\cos \theta \sin \theta}{\cos \theta+\sin \theta}=\frac{1}{\frac{1}{\cos \theta}+\frac{1}{\sin \theta}}=\frac{1}{\sec \theta+\csc \theta}$

I'm not used to csc and sec, so if you know some identities, use them ^^

---------------------------------
Or you can write $\displaystyle \cos^2+\sin^2=1 \implies \sin^2 \theta+\cos \theta \sin \theta+\cos^2 \theta=1+\cos \theta \sin \theta$

So you can have at least 2 displays of the final answer

12. Originally Posted by Moo
How do you want to simplify the cosines ??

Isn't the final answer $\displaystyle (1-\cos \theta) \cos \theta$ ?

Well, it's correct you were on the right track Now you just need some practice ^^
Yes definitely ( and that is what i meant by simplifying it, thanks for clearing that up for me)

Originally Posted by ;Moo164767
Or you can write $\displaystyle \cos^2+\sin^2=1 \implies \sin^2 \theta+\cos \theta \sin \theta+\cos^2 \theta=1+\cos \theta \sin \theta$

So you can have at least 2 displays of the final answer
Yes this makes it much better, i guess i just needed to rearrange it a bit better. So is this the final answer?

13. Originally Posted by Stevo_Evo_22
Yes this makes it much better, i guess i just needed to rearrange it a bit better. So is this the final answer?
These were only the numerator, so after that, simplify with the denominator.

For the method you quoted, you can simplify cos sin/(cos+sin) like i showed you above

14. Thanks, the answer given is
sin$\displaystyle \theta$cos$\displaystyle \theta$
---------
sin$\displaystyle \theta$ + cos$\displaystyle \theta$

I got the same, except aat the top im left with 1- cos$\displaystyle \theta$sin$\displaystyle \theta$

What happens to the 1 at the top?

15. Hmm, it should be 1+cos(theta)sin(theta)

Can you show what you have done ?

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