Question

prove
sin(pi/2-x)=sin(pi/2+x)

Answer 1

@campbell_st

Answer 2

I'd do it this way, using the sum and difference of sin expansions

\[\sin(\frac{\pi}{2}- x) = \sin(\frac{\pi}{2})\cos(x) - \cos(\frac{\pi}{2})\sin(x)\]

and

\[\sin(\frac{\pi}{2} + x) = \sin(\frac{\pi}{2})\cos(x) + \cos(\frac{\pi}{2})\sin(x)\]

simplify each... and see what you get...?

you need to remember

\[\sin(\frac{\pi}{2}) = 1 ~~~~~~~\cos(\frac{\pi}{2}) = 0\]

hope it helps

Answer 3

@ganeshie8

Answer 4

@campbell_st has given you what you need just simplify
sin(pi/2) is what?
cos(pi/2) is what?

Answer 5

That was how i did, but it turns out its not the way to prove according to my professor

Answer 6

and i think he is right because if u solve for the other part ull get a cosx

Answer 7

so what does you teachers wants? did he indicate what you should use?
for example we can use cofunction of sin
sin(pi/2-x)=cos(x)
sin(pi/2+x)=sin(pi/2-(-x))=cos(-x)

Answer 8

both side most be equal to each other

Answer 9

and since cosine is an even function
so what would cos(-x) be

Answer 10

of course they should equal to each other

Answer 11

cos(x)

Answer 12

yes but but professor said not the way to solve this

Answer 13

i have tried all possible methods that i know

Answer 14

to find a way to prove that one side is equal the other but the answer isnt cosx

Answer 15

the hall class did the way we just did and he said not the way to prove it

Answer 16

whole

Answer 17

well thanks

Answer 18

you deserve a medal :D

Answer 19

idk what your prof want, there is a geometric proof but
it is also using cos

Answer 20

so you want to take the expression on left hand side and deduce the right hand side, eh ?

Answer 21

well if that will work would you mind walking me through it

Answer 22

without doing that cosx business

Answer 23

i see what ur professor wants, it looks tricky though

Answer 24

@Kainui

Answer 25

for the past 2 days its the only thing in my mind lol

Answer 26

Well since sine is an odd function, \[\LARGE \sin(\theta)=-\sin(-\theta)\] and since sine is periodic, after half a period it's its own negative. \[\LARGE \sin(\theta)=-\sin(\theta + \pi)\]

This should be all you need to find the answer.

Answer 27

wow awesome

Answer 28

i could never think of this lol just saying

Answer 29

I didn't invent it either, don't worry I just read about it and see that it makes sense and understand it and now it's one of my tools in solving problems =)

You just have to look at the graph to see that it's true. \[\LARGE \sin(x)=-\sin(-x)\] Look at the graph of this, we can see that if we start putting in 1, 2, 3, etc... on the left side we get sine wiggling to the right. But if we put in 1, 2, 3, etc... on the right equation -sin(-x) then we get the values of sine as if you were moving to the left. Since those ones are upside down, we have to flip them with another negative sign.

Answer 30

well thank you very much

Answer 31

can you also send me link of where you read it.. plz

Answer 32

I don't have links to the inside of my brain. I don't know, I read and learned about this a long time ago, but you are free to ask me questions for the next 30 minutes and I'll help you before I go out partying for halloween =P

Answer 33

hahaha :D i guess that will be it for today.. ill keep in touch :D thanks to you and @ganeshie8 and @xapproachesinfinity and @campbell_st

Answer 34

you guys are the best 4real

Answer 35

Anytime!

Answer 36

watch first 4 minutes of this video to find out why your professor thinks ur proof is not good

http://www.youtube.com/watch?v=NuGDkmwEObM

Answer 37

Eh I agree with the professor!