Why does [sin(θ/2)-cos(θ/2)]^2 not always equal 1?

[NanoChrisK]

Homework Statement

Does [sin(θ/2)-cos(θ/2)]^2 equal 1 for all values of θ?

I need to figure this out to solve a physics problem.

Homework Equations

sin^2(A)+cos^2(A)=1

The Attempt at a Solution

[sin(θ/2)-cos(θ/2)]^2
=sin^2(θ/2)+cos^2(θ/2)
=1

But this isn't what I get when I put f(x)=[sin(θ/2)-cos(θ/2)]^2 in my graphing calculator. What's going on?

 

[Mark44]

Homework Statement

Does [sin(θ/2)-cos(θ/2)]^2 equal 1 for all values of θ?

I need to figure this out to solve a physics problem.

Homework Equations

sin^2(A)+cos^2(A)=1

The Attempt at a Solution

[sin(θ/2)-cos(θ/2)]^2
=sin^2(θ/2)+cos^2(θ/2)

The above is wrong. (A - B)² $\neq$ A² - B².

=1

But this isn't what I get when I put f(x)=[sin(θ/2)-cos(θ/2)]^2 in my graphing calculator. What's going on?

 

[phinds]

Draw a unit circle with a radius at an arbitrary angle and looks at the equation in terms of what it means in that context. This should show you pretty quickly whether it's true or not.

 

[NanoChrisK]

The above is wrong. (A - B)2 ≠\neq A2 - B2.

Hmm I may be wrong, but I think you meant to type "(A -B)² ≠ A²+A²" (with a + in the second half of the equation) This is the form I gave it in, unless I'm missing something. Is this not correct? I'm getting this from:

(A-B)² = (A-B)(A-B) = A² + AB - AB + (-B)² = A² + B²

 

[NanoChrisK]

The above is wrong. (A - B)² $\neq$ A² - B².

Ok I see what I said previously was incorrect! Because

(A-B)² = (A-B)(A-B) = A² - 2AB + (-B)² = A² - 2AB + B²

NOT what I had earlier. I guess I just need to re-take basic algebra! Thanks Mark!

 

[Mark44]

Hmm I may be wrong, but I think you meant to type "(A -B)² ≠ A²+A²" (with a + in the second half of the equation) This is the form I gave it in, unless I'm missing something. Is this not correct? I'm getting this from:

(A-B)² = (A-B)(A-B) = A² + AB - AB + (-B)² = A² + B²

No, this part is wrong: A² + AB - AB + (-B)², namely the +AB term.

 

[NanoChrisK]

No, this part is wrong: A² + AB - AB + (-B)², namely the +AB term.

Yes, I understand now :D There should have been a -2AB in the middle there. Thanks for making that clear to me!

 

[HallsofIvy]

It is fairly easy to show that $cos^2(\theta/2)- sin^2(\theta/2)= cos(\theta)$, not 1.

 

[chwala]

It is fairly easy to show that $cos^2(\theta/2)- sin^2(\theta/2)= cos(\theta)$, not 1.

Follows from $cos {2θ}=cos^2 θ - sin^2θ$...
⇒$\cos θ = cos^2 \frac {1}{2} θ - sin^2 \frac {1}{2} θ$