This is the rule of sums of trig identities. It says that sin(a + b) = sina*cosb+sinb*cosa. Set up your right triangles. In a 30-60-90 triangle, the side across from the 60 angle is the square root of 3, the side across from the 30 is 1, and the hypotenuse is 2. In a 45-45-90 the sides across from the 45 degree angles are 1 and the hypotenuse is square root of 2. With that being said, the sin of 60 is [tex] \frac{ \sqrt{3} }{2} [/tex] and the cos of 60 is 1/2. Both the sin and cos of 45 is [tex] \frac{1}{ \sqrt{2} } [/tex]. So your formula is filled in like this: [tex]( \frac{ \sqrt{3} }{2} * \frac{1}{ \sqrt{2} })+( \frac{1}{2} * \frac{1}{ \sqrt{2} }) [/tex]. When you do the multiplication on those sets of parenthesis, you get [tex] \frac{ \sqrt{3} }{2 \sqrt{2} } + \frac{1}{2 \sqrt{2} } [/tex]. When you add those your solution is [tex] \frac{1+ \sqrt{3} }{2 \sqrt{2} } [/tex]. Depending upon your instructor you may have to rationalize that denominator to get that radical out from under there, but if not, that's pretty much it!
