WAEC Mathematics 2016 Theory — Question 12

Model answer

(a) The table is completed by calculating y=3sinx+2cosx at each value of x from 0° to 360° in steps of 30°, giving values that range from y=2.0 at x=0° up to a maximum around y=3.6 near x=56°, back down through zero, to a minimum around y=-3.6 near x=236°, and back up to y=2.0 at x=360°. (b) The graph is a sine-like curve (sinusoidal curve) with amplitude √(3²+2²)=√13≈3.6, oscillating between approximately -3.6 and 3.6, plotted using the given scales. (c)(i) To solve 3sinx+2cosx=0, read the graph where the curve crosses the x-axis (y=0). The solutions are x=147° and x=327°. (ii) To solve 2+2cosx+3sinx=0, rearrange to 3sinx+2cosx=-2, i.e. y=-2. This is solved by finding where the graph y=3sinx+2cosx cuts the horizontal line y=-2. The solutions are x=180° and x=291°.