Question

*Trigonometric Graphs question*
Simplify the expression: cos^2(pi/2)-x)/square root1-sin^2(x). *Please help will give medal for this*

Answer 1

Simplify this?\[\large\rm \frac{\cos^2\left(\frac{\pi}{2}-x\right)}{\sqrt{1-\sin^2(x)}}\]

Answer 2

The mathway calculator won't let me square the cos or the sin

Answer 3

Ahhh sorry I gotta get food out of the oven a sec >.<
I can help in a few minutes, unless one of these fellas gets to it first hehe

Answer 4

lol okay save me some though

Answer 5

XD

Answer 6

Recall that sine and cosine are `cofunctions`.
Example of this concept:\[\large\rm \sin(30^o)=\cos(60^o)\]\[\large\rm \sin(70^o)=\cos(20^o)\]We can generalize this by saying:\[\large\rm \sin(x)=\cos(90^o-x)\]Or so it's more useful for us, let's see this relationship in radians,\[\large\rm \sin(x)=\cos\left(\frac{\pi}{2}-x\right)\]

Answer 7

Do you see how we can apply this relationship to the numerator?\[\large\rm \frac{\color{orangered}{\cos^2\left(\frac{\pi}{2}-x\right)}}{\sqrt{1-\sin^2(x)}}\]

Answer 8

Yeah I see

Answer 9

So what's the top become? :)

Answer 10

sin(x)?

Answer 11

Well, I was showing you this relationship,\[\large\rm \sin(x)=\cos\left(\frac{\pi}{2}-x\right)\]If we square each side,\[\large\rm \sin^2(x)=\cos^2\left(\frac{\pi}{2}-x\right)\]this right side is what we have in our numerator.
So I guess we end up with sin^2(x), ya? :)

Answer 12

\[\large\rm \frac{\color{orangered}{\sin^2(x)}}{\sqrt{1-\sin^2(x)}}\]So that takes care of the top, cool.

Answer 13

Alright

Answer 14

Recall your Pythagorean Identity which involves sine and cosine:\[\large\rm \sin^2(x)+\cos^2(x)=1\]What happens if we subtract sin^2(x) from each side?
What do we get? :d
Anything useful?

Answer 15

is the bottom tan(x)?

Answer 16

Hmm no :o

Answer 17

My answer choices are weird, they go:A) tan(x), B) cos(x) tan(x), C) cos(x) cot(x), D) sin(x) tan(x).

Answer 18

We're not quite there yet :)
we're working on the denominator.

When you subtract sin^2(x) from each side,
you get something like this:\[\large\rm \cos^2(x)=\color{orangered}{1-\sin^2(x)}\]

Answer 19

oh okay

Answer 20

Do you see how this might be useful for our problem?\[\large\rm \frac{\sin^2(x)}{\sqrt{\color{orangered}{1-\sin^2(x)}}}\]What can we do with the denominator here?

Answer 21

subtract cos^2(x)?

Answer 22

subtract...? 0_o

Answer 23

add? sorry my brain is fried been working on this math stuff all day I'm homeschooled.

Answer 24

We've shown, using our Pythagorean Identity, that 1-sin^2(x) is the same as cos^2(x).
So we can simply `replace` our 1-sin^2(x) with cos^2(x), ya?

Answer 25

alright, I see

Answer 26

That leaves us here,\[\large\rm \frac{\sin^2(x)}{\sqrt{\color{orangered}{\cos^2(x)}}}\]

Answer 27

I guess we can proceed by taking the square root of that square,\[\large\rm =\frac{\sin^2(x)}{\cos(x)}\]And we'll have to do a little clever work at this point.

Answer 28

Is it tan^2(x)

Answer 29

simplified.

Answer 30

Hmm, no.
It's close to tan^2(x), but notice we don't have enough cosines in the bottom! :)

Answer 31

Do you remember how tangent is defined in terms of sines and cosines?
Bah I'll just type it out :)\[\large\rm \tan(x)=\frac{\sin(x)}{\cos(x)}\]

Answer 32

sin(x) tan(x)?

Answer 33

So for our problem, maybe we can do something clever with the sines,\[\large\rm =\frac{\sin^2(x)}{\cos(x)}\quad=\frac{\sin(x)\cdot \sin(x)}{\cos(x)}\quad=\frac{\sin(x)}{\cos(x)}\cdot \sin(x)\]Replace the sin(x)/cos(x) by tan(x) to get
sin(x)tan(x)?

Yay good job \c:/

Answer 34

Yay thanks for putting up with my slow self..lol I appreciate the help a lot.

Answer 35

np :3