sin780.sin480+cos240.cos300
P pappus Junior Member Joined Feb 13, 2012 Messages 220 Jul 1, 2012 #2 sjnaureen2012 said: sin780.sin480+cos240.cos300 Click to expand... I assume that the arguments of the trgonometric functions are in degree(?). If so: You are supposed to know that \(\displaystyle \displaystyle{\sin(0^\circ) = \sin(360^\circ)}\) and therefore \(\displaystyle \displaystyle{\sin(780^\circ) = \sin(60^\circ) = \frac12 \sqrt{3}}\) Re-write the arguments of the other values. You should come out with \(\displaystyle \displaystyle{\frac12}\) as the total result.
sjnaureen2012 said: sin780.sin480+cos240.cos300 Click to expand... I assume that the arguments of the trgonometric functions are in degree(?). If so: You are supposed to know that \(\displaystyle \displaystyle{\sin(0^\circ) = \sin(360^\circ)}\) and therefore \(\displaystyle \displaystyle{\sin(780^\circ) = \sin(60^\circ) = \frac12 \sqrt{3}}\) Re-write the arguments of the other values. You should come out with \(\displaystyle \displaystyle{\frac12}\) as the total result.
