Posted by Priya on Thursday, May 1, 2008 at 4:46pm.
<<I hope Im not overburdening you with question:/>>
lol, I love trig and do these kind of questions for recreation.
for this one you will need to know
1. sin2A = 2sinAcosA, and
2. cos2A = cos^2 A - sin^2 A
or 2cos^2 A -1
or 1 - 2sin^2 A
I will do the first, you do the second
cos2A = 1-2sin^2A
1/3 = 1 - 2sin^2 A
2 sin^2 A = 1 - 1/3
sin^2 A = 1/3
sinA = ±1/√3 (I used A instead of theta)
Can we use both ± ?
2theta is in either I or IV
then theta is in either I or II, but the sine in both of these is positive, so
sinA = +1/√3 only.
you try the second one.
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