# Mathematics - Trigonometric Identities - Reiny

Mathematics - Trigonometric Identities - Reiny, Friday, November 9, 2007 at 10:30pm

(sinx - 1 -cos^2x) (sinx + 1 - cos^2x)

should have been

(sinx - 1 + cos^2x) (sinx + 1 - cos^2x) and then the next line should be

sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2 x + cos^2xsinx - cos^4x
= cos^2 x - cos^4 x
= RS

an easier way would have been

LS = sin^2x(1-sin^2x) by common factor
=sin^2x(cos^2x)

RS = cos^2x(1-cos^2x) also by common factor
=cos^2x(sin^2x)
= LS

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

What happened to the negative in the first line?

You wrote:

an easier way would have been

LS = sin^2x(1-sin^2x) by common factor
=sin^2x(cos^2x)

--

Shouldn't it be LS = sin^2x (-sin^2x) ?

And for my way ...

You wrote:

(sinx - 1 -cos^2x) (sinx + 1 - cos^2x)

should have been

(sinx - 1 + cos^2x) (sinx + 1 - cos^2x) and then the next line should be

sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2 x + cos^2xsinx - cos^4x
= cos^2 x - cos^4 x
= RS

--

and I'm asking why is the "cos^2x" a negative from "sinx - 1 + cos^2x"?

This line, "sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2 x + cos^2xsinx - cos^4x" ----> I get "sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2x + cos^2xsinx + cos^2x - cos^4x" ----> I get an extra "cos^2x" ...

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1. sin^2x - sin^4x = cos^2x - cos^4x

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2. Ok, let's start from the beginning

sin^2x - sin^4x = cos^2x - cos^4x

LS = (sinx - sin^2 x)(sinx + sin^2 x)

(sinx - 1 -cos^2x) (sinx + 1 - cos^2x)

I will put in the in-between step

=(sinx - (1 -cos^2x)) (sinx + 1 - cos^2x)

= (sinx - 1 + cos^2x) (sinx + 1 - cos^2x)
= sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2 x + cos^2xsinx + cos^2 x - cos^4x
(and yes, you are right, I left out a +cos^2 x , so.......

= sin^2x+cos^2x + sinx-sinx - 1 + cos^2xsinx-cos^2xsinx + cos^2 x - cos^4 x
= 1 + 0 - 1 + 0 + cos^2 x - cos^4 x
= cos^2 x - cos^4 x
= RS

For my second solution I worked on both the LS and the RS, reaching the same expression at both ends.

Surely your teacher must have shown you that as an acceptable method.

Here is what you wrote:

"hat happened to the negative in the first line?"

regarding:

LS = sin^2x(1-sin^2x) by common factor
=sin^2x(cos^2x)

didn't you recognize the replacement of
1 - sin^2 x with cos^2 x ???

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posted by Reiny
3. I see it now, thanks

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