Question

Simplify the expression.

sine of x to the second power minus one divided by cosine of negative x

Answer 1

@campbell_st @ParthKohli @IMStuck @SithsAndGiggles

Answer 2

I can try...let me take a look!

Answer 3

Thanks! This is the last one I have to do today :D

Answer 4

Is this your expression?\[\sin ^{2}x-\frac{ 1 }{ \cos ^{-x} }\]?

Answer 5

Got it. Let me study that for a tiny minute.

Answer 6

Ok, looking at the numerator, we have an identity from which we can derive the simplification. Do you know your Pythgorean identities in trig?

Answer 7

sin^2 θ + cos^2 θ = 1

tan^2 θ + 1 = sec^2 θ

cot^2 θ + 1 = csc^2 θ

Answer 8

\[
\cos(-x) = \cos(x)
\]

Answer 9

Ok use the first one. Get it into that same format as your numerator. Do you know what I mean? Looking at your identity, what does sin^2(x) - 1 equal?

Answer 10

\[
\sin^2(x) = 1-\cos^2(x)
\]

Answer 11

Please let her figure it out for herself. She needs to learn.

Answer 12

-cos?

Answer 13

-cos^2 (x). \[\sin ^{2}x-1=-\cos ^{2}x\]That was good. Now let's replace that numerator with its identity.

Answer 14

\[-\frac{ \cos ^{2}x }{ \cos(-x) }\]As wio stated up above, cos(-x) = cos(x). So we can make that replacement as well.

Answer 15

\[-\frac{ \cos ^{2}x }{ cosx }\]How do you reduce/simplify that now?

Answer 16

-cos/1?

Answer 17

The denominator will completely cancel. Left up on top, when you factor one cos x out what are you left with?

Answer 18

cos or -cos?

Answer 19

Oh, I see what you did. You're right! Sorry! The 1 in the denominator threw me for a sec, as I always leave it out when it's a 1. But you're right!

Answer 20

It's -cos(x)

Answer 21

so that's it, right? just factor out cos and I'm left with -cosx?

Answer 22

Yep!! Trig uses identities ALL the TIME! So get used to looking for a way to use one in most problems.

Answer 23

Thanks again! :D

Answer 24

You're most welcome!