Question

Find sin 105 degrees by using 105 degrees = 60 degrees + 45 degrees

Answer 1

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 $\frac{ \sqrt{3} }{2}$ and the cos of 60 is 1/2.  Both the sin and cos of 45 is $\frac{1}{ \sqrt{2} }$.  So your formula is filled in like this: $( \frac{ \sqrt{3} }{2} * \frac{1}{ \sqrt{2} })+( \frac{1}{2} * \frac{1}{ \sqrt{2} })$.  When you do the multiplication on those sets of parenthesis, you get $\frac{ \sqrt{3} }{2 \sqrt{2} } + \frac{1}{2 \sqrt{2} }$.  When you add those your solution is $\frac{1+ \sqrt{3} }{2 \sqrt{2} }$.  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!