When teaching trigonometry we come across both the difference identities and the half angle identities. Each of these can give what look like very different values for the same thing.
For example by the angle difference identity:
sin(15)=sin(45-30)=sin(45)cos(30)-cos(45)sin(30)
=√2/2*√3/2 - √2/2*1/2
=(√6-√2)/4
Whereas by the half angle identity
sin(15)=sin(30/2)=√((1-cos(30))/2)
=√((1-√3/2)/2)
=√(2-√3)/2
While these are undoubtedly the same, they look very different, especially the second with its radical within a radical. Can you show algebraically that they are indeed equivalent?
Let A = v(6- v2)/4, while:
B = v(2-v3)/2
Let us suppose that :
(P, Q) = (A^2, B^2)
Then,
P = (8 - 4v3)/16
= (2-v3)/4
And,
Q = (v(2-v3)/2)^2
= (2-v3)/4
Accordingly, P = Q........(i)
Now, v6 > v2, so that: A> 0
And, 2> v3, so that: B^2>0, giving |B| > 0
Now, we know that 0< = sin x<=1, for 0 deg<= x<= 90 deg, and consequently B<0 is a contradiction.
Accordingly, B> 0
Thus both A and B respectively corresponds to the positive square roots of P and Q.
But, P = Q in terms of (i), and therefore:
A = B
Consequently, (v6 - v2)/4 = v(2-v3)/2
Edited on June 25, 2007, 10:53 pm
Solution
