Trig Identities -- example problem confusion

[opus]

Homework Statement

Back with more trig identities.

Verify that the following is an identity

$-tan\frac{a}{2} = cot\left(a\right)-csc\left(a\right)$

Homework Equations

All pythagorean identities, double angle formulas, half angle formulas

The Attempt at a Solution

In the picture that I've attached, I am confused about where I annotated with the blue arrow. If I have $cos\left(a\right)-1$ and multiply by -1, I get $1-cos\left(a\right)$. Then I'm told by the example to square root and square this expression. However, it still keeps a negative out in front of the fraction. So how can I multiply by -1, to change the expression, and also keep a negative out front of the fraction that wasn't there before? Shouldn't it be one or the other?

 

[tnich]

Homework Statement

Back with more trig identities.

Verify that the following is an identity

$-tan\frac{a}{2} = cot\left(a\right)-csc\left(a\right)$

Homework Equations

All pythagorean identities, double angle formulas, half angle formulas

The Attempt at a Solution

In the picture that I've attached, I am confused about where I annotated with the blue arrow. If I have $cos\left(a\right)-1$ and multiply by -1, I get $1-cos\left(a\right)$. Then I'm told by the example to square root and square this expression. However, it still keeps a negative out in front of the fraction. So how can I multiply by -1, to change the expression, and also keep a negative out front of the fraction that wasn't there before? Shouldn't it be one or the other?

Aside from the fact that the example may be poorly worded, do you agree that
$\frac{cos α - 1}{\sqrt{1-cos^2α}}=-\frac{\sqrt{(1-cos α)^2}}{\sqrt{1-cos^2α}}$

 

[Charles Link]

To keep the expression unchanged, you need to multiply it by $(-1)(-1)=+1$. $\\$ Alternatively, you can convert $\cos(\alpha)-1$ to $1-\cos(\alpha)$ if you factor out $-1$. You don't simply multiply by $-1$. It is poorly worded.

 

[opus]

Aside from the fact that the example may be poorly worded, do you agree that
$\frac{cos α - 1}{\sqrt{1-cos^2α}}=-\frac{\sqrt{(1-cos α)^2}}{\sqrt{1-cos^2α}}$

To keep the expression unchanged, you need to multiply it by $(-1)(-1)=+1$. $\\$ Alternatively, you can convert $\cos(\alpha)-1$ to $1-\cos(\alpha)$ if you factor out $-1$. You don't simply multiply by $-1$. It is poorly worded.

So if I were to multiply by (-1) solely, I would say that there should not be a negative preceding the fraction.
So are we saying that we multiplied the numerator and denominator by (-1) because you have to do the same to both? And in doing so, the numerator expresson changed, but there was still a negation in the denominator and we just pulled it out front?
Or did we just multiply the numerator by (-1) to change the expression, then multiply it again by (-1) to keep the original value?

 

[opus]

To keep the expression unchanged, you need to multiply it by $(-1)(-1)=+1$. $\\$ Alternatively, you can convert $\cos(\alpha)-1$ to $1-\cos(\alpha)$ if you factor out $-1$. You don't simply multiply by $-1$. It is poorly worded.

So we can multiply by (-1) to get the expression in the form that we want, but that changes the value of the expression. So we can keep the new form, but we have to multiply by (-1) again to keep its original value. And this is okay because (-1)(-1)=1, and multiplying by 1 doesn't change the value. Is this correct?

 

[Mark44]

Verify that the following is an identity
$-tan\frac{a}{2} = cot\left(a\right)-csc\left(a\right)$

A much simpler approach than the one shown in the screen shot:
Let $x = \frac a 2$
Then the identity above becomes $-\tan(x) = \cot(2x) - \csc(2x)$
In just a few steps you can go from the right side to the left, after converting the expressions on the right to expressions involving only sine and cosine, and some double angle identies.

 

[Charles Link]

The way shown by @Mark44 is also much more mathematically solid, in the sense that when you start taking square roots in the half angle formulas, depending on the quadrant of the angle of interest, you can't always choose the positive square root. This one happened to work for every quadrant, and every possible choice of angle $a$, but there really was no guarantee without a more solid proof.

 

[opus]

Mark did you simply double all the angles there? This is allowed? I've been having serious issues with these identities. They take me over an hour each usually. Been putting in a lot of practice time with them but its coming very slowly. I think I have a misunderstanding of what is allowed and what isnt.

 

[Mark44]

Mark did you simply double all the angles there?

Basically, I just did a substitution.
My work looks like this:
Prove that $-\tan(\frac a 2) = \cot(a) - \csc(a)$
Let $x = \frac a 2$
The first identity then becomes, "Prove that $-\tan(x) = \cot(2x) - \csc(2x)$
I then proved that the right side can be simplified to the expression on the left, using only the most basic identities (e.g, $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$ and similar), and double angle identities.
The only thing remaining is to "undo" the substitution, by replacing x with a/2. This proves that $-\tan(\frac a 2) = \cot(a) - \csc(a)$.
Clear?

I think I have a misunderstanding of what is allowed and what isnt.

You're always allowed to make simple substitutions, as long as you're consistent in what you change.

For example, your textbook probably proved the addition formula $\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)$ first, then replaced b with a to get the double angle formula for cosine: $\cos(a + a) = \cos(2a) = \cos^2(a) - \sin^2(a)$. It might have done something similar to get the half-angle formula.

 

[opus]

Ohhh! That makes sense. I'm glad you said that because I had my midterm today and there was a similar question where I did this same thing after reading this. Very helpful, thank you!

 

[YoshiMoshi]

OP,

I would like to add something. I remember memorizing all these trigonometric identities was a pain in the butt. Plus you'll probably forget them in less than a year (talking from experience) just remember this

and you should be good. All these verifying trigonometric identities can be solved using these formulas. You can solve your problem using these formulas, much easier IMO. Otherwise I have to go digging through trigonometric formulas I have forgotten.