Show that
Sin(A/2)=root(1/2(1-cosA)??
remember your double-angle formula?
cos(2x) = 1 - 2sin^2(x)
That means that
cos(x) = 1 - 2sin^2(x/2)
That should help.
To show that sin(A/2) = √[1/2(1-cosA)], we can use the half-angle identity for sine.
The half-angle identity for sine states that sin(A/2) = √[(1-cosA)/2].
Let's start by proving this identity:
Step 1: Start with the double-angle identity for sine.
The double-angle identity for sine is: sin(2θ) = 2sinθcosθ.
Step 2: Divide both sides of the double-angle identity by 2.
sin(2θ)/2 = sinθcosθ.
Step 3: Replace θ with A/2.
sin(A)/2 = sin(A/2)cos(A/2).
Step 4: Rearrange the equation to isolate sin(A/2).
sin(A/2) = sin(A)/2cos(A/2).
Step 5: Recall the identity sin(A) = 2sin(A/2)cos(A/2).
Substituting this identity into the equation from Step 4 gives us:
sin(A/2) = (2sin(A/2)cos(A/2))/(2cos(A/2)).
Step 6: Cancel out the common factor of 2cos(A/2) in the numerator and denominator.
sin(A/2) = sin(A/2).
So, we have proven that sin(A/2) = sin(A/2).
Now, let's use this half-angle identity to prove the desired equation sin(A/2) = √[1/2(1-cosA)].
Step 1: Substitute the half-angle identity for sine.
sin(A/2) = √[(1-cosA)/2].
Step 2: Simplify the expression on the right.
sin(A/2) = √[1/2(1-cosA)].
Therefore, we have shown that sin(A/2) = √[1/2(1-cosA)].