(a) Copy and complete the table for the relation: (y = 2cos x + 3sin x) for (0° leq x leq 360°).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 2.00 | 3.23 | 1.60 | -3.23 |
(b) Using a scale of 2 cm to 60° on the x- axis and 2 cm to one unit on the y- axis, draw the graph of (y = 2cos x + 3sin x) for (0° leq x leq 360°).
(c) From the graph, find the : (i) maximum value of y, correct to two decimal places ; (ii) solution of the equation (frac{2}{3} cos x + sin x = frac{5}{6}).
Explanation
(a)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 2.00 | 3.23 | 3.60 | 3.00 | 1.60 | -0.23 | -2.00 | -3.23 |
(b) Scale : 2 cm to 60° on the x- axis
2 cm to 2 units on y- axis.
(c)(i) Maximum value of y is 3.60.
(ii) Graph : (frac{2}{3} cos x + sin x = frac{5}{6})
Equation : (2 cos x + 3 sin x = y)
i.e. (frac{2}{3} cos x + sin x = frac{y}{3} …. (2))
(2) – (1) : (frac{y}{3} – frac{5}{6} = 0)
(frac{y}{3} = frac{5}{6} implies y = 2.50)
We draw the line y = 2.5 and find the x- values at the points of intersection of graph with y = 2.5.
x = 12° and 102°.