Question

verify the identity cos(x+pi/2)=-sinx
Please explain in steps. I give medals!

Answer 1

We need this identity...
\[\Large \cos(A+B) = \cos(A)\cos(B) -\sin(A)\sin(B)\]

And the facts that
\[\Large \sin\frac\pi 2=1\]
\[\Large \cos\frac \pi 2 = 0\]

digest these identities first, then we'll proceed :)

Answer 2

With A being x and B being pi/2?

Answer 3

That is correct :)

Answer 4

Perhaps you could do it from here? :)

Answer 5

so cosx sinpi/2 -sinx cos pi/2 = 1/0?

Answer 6

Okay, a little problem there... (maybe more than a little ;) )

Answer 7

XD alright what do I need to do.

Answer 8

Let's use this identity
\[\Large \cos(A+B) = \cos(A)\cos(B) -\sin(A)\sin(B)\]
and as you said
A = x
B = pi/2

so we now have...
\[\Large \cos\left(x+\frac \pi 2\right) = \cos(x)\cos\left(\frac\pi 2\right) -\sin(x)\sin\left(\frac \pi 2\right)\]

Answer 9

So now, simplify :P

Answer 10

And sinpi/2 = 1 and cospi/2 = 0?

So cosx cos0 - sinx sin1?

Answer 11

How did you get cos 0?

Answer 12

And sin 1 for that matter?

Answer 13

you said sin(pi/2) =1

Answer 14

Yes I did :)
I said sin(pi/2) = 1

I DON'T REMEMBER saying that
sin(pi/2) = sin 1 however :P

Answer 15

Yeah sin1. It's in the second post.

Answer 16

So where do I go from here then.

Answer 17

I don't see any post where I said sin(pi/2) = sin 1

Answer 18

1 and sin 1 are two very different things.

Answer 19

No it equals 1. . . a you're confusing me. Ok what do I do now.

Answer 20

Well, if you're confused, better fix that first.
And since you did it incorrectly, do it again :P
simplify:
\[\Large \cos\left(x+\frac \pi 2\right) = \cos(x)\cos\left(\frac\pi 2\right) -\sin(x)\sin\left(\frac \pi 2\right)\]

Answer 21

I need help with that.

Answer 22

Okay... what is \(\large\cos \frac \pi 2 = \color{red}?\)

Answer 23

0!

Answer 24

Yes. so, what happens to this part:
\[\Large \cos\left(x+\frac \pi 2\right) = \cos(x)\color{red}{\cos\left(\frac\pi 2\right)} -\sin(x)\sin\left(\frac \pi 2\right)\]

Answer 25

Ok so . . .

cosx (0) - sinx(1)?

Answer 26

Much better.
simplify further?

Answer 27

0-sinx which is -sinx! yay ty!

Answer 28

Good job.