Question

Simplify:
(csc x)(sin ^2)(pi/2 -x)(tan x)

Answer 1

I keep trying to apply the trig identities but it just gets me nowhere.

Answer 2

:) hi again.

Answer 3

is that \[\csc x*\sin^2(\pi/2-x)*\tan x\]?

Answer 4

yup.

Answer 5

What does the graph of \[y=\sin(\pi/2-x)\] look like compared to the graph of \[y=\sin(-x)\]

Answer 6

I dont know lemme graph it...

Answer 7

sin(-x) is shifted to the left

Answer 8

No i guess it wouldbe the first one thats been shifted.

Answer 9

:( Am I missinig something obvious again?

Answer 10

Does it look like the graph of any other function?

Answer 11

y=sin x

Answer 12

\[\sin(\pi/2-x) = \cos(x)\]

x pi/2-x sin pi/2-x cos x
0 pi/2 1 1
pi/2 0 0 0
pi -pi/2 -1 -1
etc.

Answer 13

If we make that substitution, what do we have?

Answer 14

but its a sin squared. not just a plain ol' sin.

Answer 15

that's fine. if sin(pi/2-x) = cos(x) then sin^2(pi/2-x) = cos^x

Answer 16

it's not a different function when squared, it's the same function, squared.

Answer 17

okay... so then it'd be csc x cos^2 x tan x

Answer 18

\[\sin^2x = (\sin x)*(\sin x)\]

Answer 19

oops, didn't get the exponent in for cos^2x a few posts ago

Answer 20

I figured it was supposed to be there :)

Answer 21

Ugh i hate these problems I cant even work through them! I just sit here staring at it.

Answer 22

What was the definition of csc again?

Answer 23

1/sin x

Answer 24

Okay I get to about \[\frac{ (\cos ^{2} x)(\tan x) }{ \sin x }\]

Answer 25

Then I'm stuck and I have no idea if thats even right.

Answer 26

Good. Do you know the definition of tan x?

Answer 27

tan=sin/cos

Answer 28

Okay, why don't you plug that in as well

Answer 29

is it seriously that freakin simple? I'm going to shoot myself.

Answer 30

the real stumbling block is realizing that sin(pi/2-x) = cos(x)

Answer 31

\[\frac{ \cos ^{2} x}{ \sin x }\times \frac{ \sin x }{ \cos x }\]

Answer 32

so then it =cos x

Answer 33

yep, cos x is the final answer. Forehead, look out, incoming palm! :-)

Answer 34

lol no joke...Do you wanna help with the next one too? Maybe if we stick together we'll just get super fast.

Answer 35

Give it a shot first, then post it, there's someone else pleading for help...

Answer 36

lol okay. :) us poor retarded people.