How Do You Simplify Trigonometric Expressions Using Basic Identities?

[Jen23]

Homework Statement

Express (1+cot^2 x) / (cot^2 x) in terms of sinx and/or cosx

Homework Equations

cot(x) = 1/tan(x)
sin^2(x) + cos^2(x) = 1

The Attempt at a Solution

I do not know if I am solving this problem correctly. Is there an easier route than the way I have solved it, if it is solved correctly?

= (1+cot^2x) / (cot^2x)
= 1+ [ (cos^2x) / (sin^2x) ] ÷ [ (cos^2x) / (sin^2x) ]
= 1 + [ (cos^2x) / (sin^2x) ] x [ (sin^2x / cos^2x) ]
= [ (sin^2x / sin^2x) + (cos^2x / sin^2x) ] x [ (sin^2x) / (cos^2x) ]
= [ (sin^2x + cos^2x) / (sin^2x) ] x [ (sin^2x )/ (cos^2x) ]
= [ 1 / sin^2x ] x [ sin^2x / cos^2x]
= sin^2x / (sin^2x)(cos^2x)
= 1 / cos^2x

 

[Dick]

Homework Statement

Express (1+cot^2 x) / (cot^2 x) in terms of sinx and/or cosx

Homework Equations

cot(x) = 1/tan(x)
sin^2(x) + cos^2(x) = 1

The Attempt at a Solution

I do not know if I am solving this problem correctly. Is there an easier route than the way I have solved it, if it is solved correctly?

= (1+cot^2x) / (cot^2x)
= 1+ [ (cos^2x) / (sin^2x) ] ÷ [ (cos^2x) / (sin^2x) ]
= 1 + [ (cos^2x) / (sin^2x) ] x [ (sin^2x / cos^2x) ]
= [ (sin^2x / sin^2x) + (cos^2x / sin^2x) ] x [ (sin^2x) / (cos^2x) ]
= [ (sin^2x + cos^2x) / (sin^2x) ] x [ (sin^2x )/ (cos^2x) ]
= [ 1 / sin^2x ] x [ sin^2x / cos^2x]
= sin^2x / (sin^2x)(cos^2x)
= 1 / cos^2x

Looks just fine to me.

 

[Ray Vickson]

Homework Statement

Express (1+cot^2 x) / (cot^2 x) in terms of sinx and/or cosx

Homework Equations

cot(x) = 1/tan(x)
sin^2(x) + cos^2(x) = 1

The Attempt at a Solution

I do not know if I am solving this problem correctly. Is there an easier route than the way I have solved it, if it is solved correctly?

= (1+cot^2x) / (cot^2x)
= 1+ [ (cos^2x) / (sin^2x) ] ÷ [ (cos^2x) / (sin^2x) ]
= 1 + [ (cos^2x) / (sin^2x) ] x [ (sin^2x / cos^2x) ]
= [ (sin^2x / sin^2x) + (cos^2x / sin^2x) ] x [ (sin^2x) / (cos^2x) ]
= [ (sin^2x + cos^2x) / (sin^2x) ] x [ (sin^2x )/ (cos^2x) ]
= [ 1 / sin^2x ] x [ sin^2x / cos^2x]
= sin^2x / (sin^2x)(cos^2x)
= 1 / cos^2x

Why don't you use the fact that
$$
\frac{1 +\cot^2 x}{\cot^2 x} = \frac{1}{\cot^2 x} + 1 ?
$$
Then you can finish off the whole thing in one more line of simple algebra (plus the identity $\cos^2 x + \sin^2 x = 1$).

 

[Jen23]

Why don't you use the fact that
$$
\frac{1 +\cot^2 x}{\cot^2 x} = \frac{1}{\cot^2 x} + 1 ?
$$
Then you can finish off the whole thing in one more line of simple algebra (plus the identity $\cos^2 x + \sin^2 x = 1$).

so if 1 / cot^2x = tan^2x = sin^2x / cos^2x

Then all we have to do is: = (sin^2x/cos^2x ) + (cos^2x/cos^2x)
=( sin^2x + cos^2x)/ cos^2x
= 1 / cos^2x

Thanks so much haha, I didn't even notice that. Also another question for proofs. We know that sin^2x + cos^2x = 1 (pythagorean identity). Can we say the same for sinx + cosx= 1? I am pretty sure not but just double checking.

 

[Dick]

so if 1 / cot^2x = tan^2x = sin^2x / cos^2x

Then all we have to do is: = (sin^2x/cos^2x ) + (cos^2x/cos^2x)
=( sin^2x + cos^2x)/ cos^2x
= 1 / cos^2x

Thanks so much haha, I didn't even notice that. Also another question for proofs. We know that sin^2x + cos^2x = 1 (pythagorean identity). Can we say the same for sinx + cosx= 1? I am pretty sure not but just double checking.

Absolutely not. Try it if $x=\pi$.

 

[jimkris69]

The brackets [ ] on your second line are incorrect