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

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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" ...

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

Ok, let's start from the beginning

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

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

then you had:
(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 ???

I see it now, thanks

In the given expression, (sinx - 1 + cos^2x) (sinx + 1 - cos^2x), let's break it down step by step:

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

Here, there seems to be a typo in the expression in your question. It should have been (sinx - 1 + cos^2x) instead of (sinx - 1 - cos^2x).

2. Simplifying the expression:

Expanding the expression by using the distributive property:

(sin^2x + sinx - cos^2xsinx - sinx - 1 + cos^2 x + cos^2xsinx - cos^4x)

Here, we combine like terms to simplify:

-1 and -sinx cancel each other, and the same happens for sinx and -sinx terms:

(sin^2x - cos^2xsinx - cos^4x + cos^2 x)

3. Simplifying further:

(sin^2x - cos^2xsinx + cos^2 x - cos^4x)

4. Rearranging and combining terms:

(sin^2x + cos^2x - cos^4x - cos^2xsinx)

5. Simplifying:

Using the identity sin^2x + cos^2x = 1, we have:

(1 - cos^4x - cos^2xsinx)

6. Simplifying further:

Rearranging the terms:

(1 - cos^4x) - cos^2xsinx

And using the identity 1 - cos^2x = sin^2x, we have:

sin^2x - cos^2xsinx

Now, let's address your questions:

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

No, it shouldn't be LS = sin^2x (-sin^2x). In the easier way you mentioned, it is correct to factor out sin^2x from (1 - sin^2x). It results in LS = sin^2x(cos^2x), rather than sin^2x(-sin^2x).

2. "Why is the "cos^2x" a negative from "sinx - 1 + cos^2x"?"

There seems to be a typo in the original question. The expression should be (sinx - 1 + cos^2x) instead of (sinx - 1 - cos^2x). So, the "cos^2x" is not negative in the original expression.