cos(x/2) = +/- √((1 + cosx) / 2)
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To prove the given identity:
We can use the half-angle formula for cosine, which states that cos(x/2) = √((1+cosx)/2) ± √((1-cosx)/2). We want to show that one of the square roots is always equal to zero.
Starting with the right-hand side of the given identity:
+/-√((1+cosx)/2)
We want to simplify this expression to something involving only cos(x/2). To do this, we can use the double-angle formula for cosine, which states that:
cos2θ = 2cos^2θ – 1
Rearranging this formula, we can solve for cosθ in terms of cos2θ:
cosθ = +/-√((1+cos2θ)/2)
Notice that this looks very similar to the right-hand side of the given identity. We can use this formula to simplify our expression:
+/-√((1+cosx)/2) = +/- sqrt((1 + cos(2(x/2)))/2)
= +/- sqrt((1 + cosx)/2)
Notice that we can drop the factor of 2 from the argument of the cosine function, since cos2(x/2) = cos(x) + 1. This simplifies the expression to:
+/- sqrt((1 + cosx)/2)
which is exactly what we started with!
Therefore, we have shown that cos(x/2) = +/- √((1 + cosx) / 2), which proves the given identity.
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