(a) Copy and complete the following table of values for \(y = 9 \cos x + 5 \sin x\) to one decimal place. x 0° 30° 60° 90° 120° 150° 180° 210° y 10.3 -0.2 -...

(a) Completing the table of values

For \(y = 9\cos x + 5\sin x\), evaluate at the missing angles (to one decimal place):

\(x = 0^\circ:\; 9(1)+5(0)=9.0\)
\(x = 60^\circ:\; 9(0.5)+5(0.866)=4.5+4.33=8.8\)
\(x = 90^\circ:\; 9(0)+5(1)=5.0\)
\(x = 180^\circ:\; 9(-1)+5(0)=-9.0\)

(b) Graph of \(y = 9\cos x + 5\sin x\)

Using a scale of 2 cm to 30° on the x-axis and 2 cm to 1 unit on the y-axis, plot the eight points from the table and join them with a smooth curve for \(0^\circ \le x \le 210^\circ\).

(c) Solving from the graph

(i) \(9\cos x + 5\sin x = 0\). This is where the curve cuts the x-axis (\(y = 0\)). Reading the graph, the curve crosses between \(x=110^\circ\) and \(x=120^\circ\), at:

(Check: \(\tan x = -\dfrac{9}{5}=-1.8\Rightarrow x = 180^\circ-61^\circ = 119^\circ.\))

(ii) \(9\cos x + 5\sin x = 3.5\). Draw the horizontal line \(y = 3.5\) and read where it meets the curve. It meets the descending part of the curve at:

(Check, using \(9\cos x+5\sin x = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}=10.3\) and \(\alpha = \tan^{-1}\frac{5}{9}=29^\circ\): \(10.3\cos(x-29^\circ)=3.5\Rightarrow \cos(x-29^\circ)=0.340\Rightarrow x-29^\circ=70^\circ\Rightarrow x\approx 99^\circ.\))

(d) Maximum value of y

The highest point of the curve occurs near \(x = 30^\circ\). Since \(y = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}\):

The maximum value of \(y\) is 10.3 (occurring at \(x \approx 29^\circ\)).