Help please, simplify the expression as much as possible:

[calc1]

I'm doing okay on most of this type of problems but on these two I cannot get anywhere.

cos²x + cos²x cot²x

and

(tan x – sec x)²

Thanks for any pointers.

 

[HallsofIvy]

I'm doing okay on most of this type of problems but on these two I cannot get anywhere.

cos²x + cos²x cot²x

An obvious first step is to factor \(\displaystyle cos^2(x)\) out: \(\displaystyle cos^2(x)(1+ cot^2(x))\). Do you know an identity involving \(\displaystyle 1+ cot^2(x)\)?
If not, try writing tan(x) as sin(x)/cos(x) and do a little algebra.

and

(tan x – sec x)²

Thanks for any pointers.

tan(x)= sin(x)/cos(x) and sec(x)= 1/cos(x) so tan(x)- sec(x)= (sin(x)- 1)/cos(x) and so \(\displaystyle (sin(x)- 1)^2/cos^2(x)\).

 

[DrPhil]

Thanks for any pointers.

I believe in two steps:

1) convert all functions to sines and cosines

2) remember THE fundamental identity: \(\displaystyle \sin^2{x} + \cos^2{x} = 1 \)

 

[calc1]

An obvious first step is to factor \(\displaystyle cos^2(x)\) out: \(\displaystyle cos^2(x)(1+ cot^2(x))\). Do you know an identity involving \(\displaystyle 1+ cot^2(x)\)?
If not, try writing tan(x) as sin(x)/cos(x) and do a little algebra.

cos²x + cos²x cot²x = 1

Thank you HallsofIvy. Hey, I was on the right track. My error was that I kept dropping the exponent when converting csc²(x).

(sin(x)- 1)^2/cos^2(x)[/tex].

But, I am still stuck on this one. Your solution is not one of my answer options. I have tried to FOIL, but that turned into a big mess.


I think the correct solution is -1+sin(x)/-1-sin(x). I solved to a similar solution with different signs, but I haven’t been able to do that again. My similar solution was wrong anyway.

My other choices are: 1 + 2tan²(x), which doesn't smell right to me, and tan²(x) + sec²(x), but I don't think the middle sign would change.

Thanks.

 

[calc1]

Thanks DrPhil. I remember that. My course does stress it.

 

[lookagain]

But, I am still stuck on this one. Your solution is not one of my answer options. I have tried to FOIL, but that turned into a big mess.


I think the correct solution is -1+sin(x)/-1-sin(x). . . . calc1, that's not correct anyway, because you're missing grouping symbols
around the numerator and denominator.

I solved to a similar solution with different signs, but I haven’t been able to do that again. My similar solution was wrong anyway.

My other choices are: 1 + 2tan²(x), which doesn't smell right to me, and tan²(x) + sec²(x), but I don't think the middle sign would change.

Thanks.

tan(x)= sin(x)/cos(x) and sec(x)= 1/cos(x) so tan(x)- sec(x)= (sin(x)- 1)/cos(x) and so [FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Math]i[/FONT][FONT=MathJax_Math]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]/[/FONT][FONT=MathJax_Math]c[/FONT][FONT=MathJax_Math]o[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT]

calc1,

you stated that you want it simplified as much as possible. Then, I am looking to get the degrees of the trig functions
cancelled down to the first degree, if possible. Also, I am trying to not have any leading negative signs.


HallofIvy's last expression:

\(\displaystyle \dfrac{(sin(x) - 1)^2}{cos^2(x)}\)


It equals


\(\displaystyle \dfrac{(sin(x) - 1)(sin(x) - 1)}{1 - sin^2(x)}\) ** \(\displaystyle = \)


\(\displaystyle \dfrac{(sin(x) - 1)(sin(x) - 1)}{(1 - sin(x))(1 + sin(x))} \ = \)


\(\displaystyle \dfrac{(-1)(1 - sin(x))(sin(x) - 1)}{(1 - sin(x))(1 + sin(x))} \ = \)


\(\displaystyle \dfrac{(-1)(sin(x) - 1)}{(1)(1 + sin(x))} \ =\)


\(\displaystyle \boxed{\dfrac{ \ 1 - sin(x) \ }{ \ 1 + sin(x) \ }}\)




- - - - - -- - - - - - - - -- - - - - - - - - - - - -



** This stems from the identity \(\displaystyle sin^2(x) + cos^2(x) = 1\)


\(\displaystyle sin^2(x) + cos^2(x) = 1 \ \ \implies \)


\(\displaystyle cos^2(x) = 1 - sin^2(x)\)

 

[calc1]

calc1,

you stated that you want it simplified as much as possible. Then, I am looking to get the degrees of the trig functions
cancelled down to the first degree, if possible. Also, I am trying to not have any leading negative signs.


HallofIvy's last expression:

\(\displaystyle \dfrac{(sin(x) - 1)^2}{cos^2(x)}\)


It equals


\(\displaystyle \dfrac{(sin(x) - 1)(sin(x) - 1)}{1 - sin^2(x)}\) ** \(\displaystyle = \)


\(\displaystyle \dfrac{(sin(x) - 1)(sin(x) - 1)}{(1 - sin(x))(1 + sin(x))} \ = \)

Thanks lookagain. You got me around third base. But, to get to my available option, I had to factor out (-1) from the denominator.

((sin⁡(x)-1)(sin⁡(x)-1))/
((1-sin⁡(x))(1+sin⁡(x)))

((sin(x)-1)(sin(x)-1))/
((-1)(sin⁡(x)-1)(1+sin(x)))

(sin(x)-1)/
((-1)(1+sin(x)))

(-1+sin(x))/
(-1-sin(x))

 

[calc1]

To save time (and keys!), let k = sin(x) ; get my drift?

Thanks. Novice learns new 'duh' concept.
I'll remember to substitute. I will be like icing on the cake.