The expression [sin(θ/2)-cos(θ/2)]^2 does not equal 1 for all values of θ. The confusion arises from the incorrect application of the algebraic identity (A - B)², which expands to A² - 2AB + B², rather than A² - B². This misunderstanding leads to the false conclusion that the expression simplifies to 1, as demonstrated through the unit circle context and the correct trigonometric identities, specifically that cos²(θ/2) - sin²(θ/2) equals cos(θ), not 1.
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The above is wrong. (A - B)2 ##\neq## A2 - B2.NanoChrisK said: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)
NanoChrisK said:=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 said:The above is wrong. (A - B)2 ≠\neq A2 - B2.
Mark44 said:The above is wrong. (A - B)2 ##\neq## A2 - B2.
No, this part is wrong: A2 + AB - AB + (-B)2, namely the +AB term.NanoChrisK said: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
Mark44 said:No, this part is wrong: A2 + AB - AB + (-B)2, namely the +AB term.
Follows from ##cos {2θ}=cos^2 θ - sin^2θ##...HallsofIvy said:It is fairly easy to show that cos^2(\theta/2)- sin^2(\theta/2)= cos(\theta), not 1.