Proving Identities: cos((pi/2)-x)=sinx

[banfill_89]

Homework Statement

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

Homework Equations

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 x^2+x=x(x+1)

\sin{(x+2)}\neq\sin x+\sin2

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]

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

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 I am using is sinxsiny, so do i just need the y number?

 

[Dick]

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

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

 

[rocomath]

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

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]

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

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?