Proving a Trig Identity

math_knight

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so the book is talking about proving an identity as follows:

(sorry I wish I knew how to use LaTex font)


sin^2(-t) - cos^2(-t) / sin(-t)-cos(-t) = cos (t) - sin (t)

using even/odd identities, and looking at only the left side and the numerator of the equation it becomes

Numerator : -sin^2(t) - cos^2(t)

or

[-sin t]^2 - [cos t]^2 (difference of squares)

but the book writes it as

[sin t]^2 - [cos t]^2

Where did the negative sign go?

OK, wait, I thought about it some, but just want to make sure I did it correctly. I guess they factored out (-1) TWICE?

so that

(-sin t - cos t) (-sin t + cost t)

= (-1)
(sin t + cost t) (-1)(sin t - cos t)

=(-1)(-1)
(sin t + cost t)(-1)(sin t - cos t)

=
(1)(sin t + cost t)(sin t - cos t)

=(sin t + cost t)(sin t - cos t)


Is that legal in all 50 states? :p I guess they are factoring out (1) so the sign has been accounted for. I have seen (-1) get factored out before, just never twice like that, if that's in fact what they did.

***It's still hard to see how they got [sin t]^2 - [cos t]^2 from [-sin t]^2 - [cos t]^2 though??

I would think you would have to wait to factor out (-1) twice only after you factored the difference of squares, which is what I did in the orange equations. How did they factor out (-) sign where they did?

Any help would be really appreciated.
 
so the book is talking about proving an identity as follows:

(sorry I wish I knew how to use LaTex font)

What you've done here is fine; everything is perfectly readable. Just be sure to use parentheses around the numerator and denominator of a big fraction, so that what you write is what you mean.

***It's still hard to see how they got [sin t]^2 - [cos t]^2 from [-sin t]^2 - [cos t]^2 though??

I would think you would have to wait to factor out (-1) twice only after you factored the difference of squares, which is what I did in the orange equations. How did they factor out (-) sign where they did?

There are several ways you could go through this, but they are doing what I would probably do.

When you square -sin(x), you get (-1)^2 (sin(x))^2 = sin^2(x), because (-1)^2 = 1. It's often a good idea to get rid of negative signs as quickly as you can, to avoid tripping over them.

With experience, you just know that when you square something negative, you can ignore the sign. Without experience, you wish they'd shown every little step!
 
[-sin t]^2 - [cos t]^2

but the book writes it as

[sin t]^2 - [cos t]^2

Where did the negative sign go?
Think of the negation sign as a factor of -1.

-sin(t) = (-1) ∙ sin(t)

When we square -sin(t), the factor -1 gets squared to become +1.

[ -sin(t) ]^2

[ (-1) ∙ sin(t) ]^2

(-1) ∙ sin(t) ∙ (-1) ∙ sin(t)

(-1)(-1) ∙ sin(t)^2

sin(t)^2
 
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