proving identities

banfill_89

1. The problem statement, all variables and given/known data

prove that cos ((pi/2)-x) = sinx

2. Relevant equations



3. The attempt at a solution

i extended it to: (cos pi/2) (cos -x) + (sin pi/2) (sin -x)
=1-sinx

Dick

i) cos(a-b)=cos(a)cos(b)+sin(a)sin(b). ii) cos(pi/2)=0. Where did that 1 come from?

banfill_89

i got the 1 from the sin of pi/2.....isnt that 1?

rocomath

You cannot expand trig identities like that.

It's not like [tex]x^2+x=x(x+1)[/tex]

[tex]\sin{(x+2)}\neq\sin x+\sin2[/tex]

Have you learned the Sum and Differences formula?

You can also prove this through triangles.

banfill_89

yea we have the sum and difference identities

Dick

Ok, so 1-sinx actually means 1*(-sin(x))?? That isn't the clearest way to write it, wouldn't you agree?? You still have a sign error.

banfill_89

yea ur right...i forgot the brackets...but it still come sout at -sin(x)......

banfill_89

oh wait....do i need to include the - on the x?

banfill_89

cause the subtraction formula is cos ( x - y), and the part of the formula im using is sinxsiny, so do i just need the y number?

Dick

Look at the second post. You have a sign error in cosine sum rule.

rocomath

Are you familiar with even and odd functions? It's the same with trig functions.

even: f(x)=f(-x)

odd: f(-x)=-f(x)

Dick

Yes. You just need the 'y number'.

banfill_89

ah ****in eh....thanks guys

banfill_89

and rocomath, i tried it with the odd even funtions and i got :

-sinx, because its an odd number infront of the pi/2, and feta=-x....am i missing something?