Question

find the points of intersection of the curves R= 2sina and R= 2sin2a a:angel

Answer 1

Well, we know that
\[ sin(2x) = 2sin(x)cos(x) \]
So now to find intersections we can just:
\[ 2sin(a) = 2sin(2a) \\
2sin(a) = 2 * 2sin(a)cos(a) \]
Now, we can divide both sides by 2sin(a) as long as a =/= 180K (because sin(180K) = 0 and we can't divide by zero..)
\[ \frac{2sin(a)}{2sin(a)} = \frac{2 \cdot 2sin(a)cos(a)}{2sin(a)} \\
1 = 2cos(a)\\
cos(a) = 0.5\\
a = +-60^\circ + 360^\circ K \]
If a = 180K It's easy to see that the following is true:
\[ 2sin(a) = 2sin(2a) \\
2sin(0) = 2sin(0) \\
0 = 0 \]
So, those are the values we got:
\[ a = 180^\circ K \\
\text{OR} \\
a = \pm60^\circ + 360^\circ K \]
Hope that helps =]