Your question is unclear but I guess you want to find the maximal value of the function, which argument is the smallest non-negative among arguments of all maximums (trigonometric functions, being periodic in general, have infinitely many maximal and minimal values). You need neither calculus nor algebra to solve your problem (let alone graphing, which never gives you a mathematical solution, only a hint); all you need is definitions and basic properties of sin and cos - plus some simple geometry and arithmetic. Well, you need just a little algebra, or, rather, precalculus. Notice that the argument of the cos term in your expression is greater than the argument of the sin term by Pi/2 (arithmetic). Cos(x+Pi/2)=-sin(x) (can be seen immediately from the definitions on the unit circle - definitions plus a little geometry). As a result, your expression reduces to P=-(sin(t-Pi/6)^2 (^ means raising to a degree). Raising to the second degree leaves all the maxima and minima in place since this is a monotonic transformation. The negative sign in front of the expression turns all maxima into minima (the plot gets reversed - this is a hint, not a proof!). Therefore, your solution will be given by the coordinates of the first _minimum_ of the function P=sin(t-P/6), whose plot is the same as that of P=sin(t), only shifted to the right along the X axis by Pi/6.