Question

verify sin^4(x)-1 / cos^2(x) = cos^2(x)-2

Answer 1

I don't understand how to get rid of the ^4

Answer 2

Is it

\[\frac{\sin^4x - 1}{\cos^2x} = \cos^2x - 2\]

Answer 3

yes

Answer 4

Okay so according to the difference of squares rule \(a^2 - b^2 = (a + b)(a - b)\)

\[\frac{\sin^4x - 1}{\cos^2x} = \frac{(\sin^2x + 1)(\sin^2x - 1)}{\cos^2x}\]

Right?

Answer 5

yes

Answer 6

right

Answer 7

Correction:

By Pythagorean Identity we can write

\[\frac{(\sin^2x + 1)(\sin^2x - 1)}{\cos^2x} = \frac{(\sin^2x + 1)(\sin^2x - 1)}{1 -\sin^2x }\]


Right?

Answer 8

okay yes

Answer 9

i agree

Answer 10

Actually, if we take out the negative from the denominator, we will get:

\[\frac{(\sin^2x + 1)(\sin^2x - 1)}{1 -\sin^2x } = -\frac{(\sin^2x + 1)(\sin^2x - 1)}{\sin^2x - 1 }\] Okay?

Answer 11

okay

Answer 12

Which means \(\sin^2x - 1\) cancels to get:

\[-(\sin^2x + 1)\]

Answer 13

Now of course \(\sin^2x = 1 - \cos^2x\)

so \(-(\sin^2x + 1) = -((1- \cos^2x) + 1)\)

Do you agree?

Answer 14

yes

Answer 15

And we can remove the inner set of parentheses to simplify it to:

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

Do you see how we get the RHS from there?

Answer 16

yes

Answer 17

Okay great :)

Any questions?

Answer 18

nope! thank you SO much