Graphs of Trigonometric Functions: math test solutions, Grade 12 – download the PDF

1 Analyzing a sinusoid / 4 pts

(a) \(3x-\pi=3\left(x-\dfrac{\pi}{3}\right)\), so \(y=-2\sin\left(3\left(x-\dfrac{\pi}{3}\right)\right)+1\): amplitude 2, period \(\dfrac{2\pi}{3}\). (1 pt)

(b) Phase shift \(\dfrac{\pi}{3}\) to the right. (1 pt)

(c) Midline \(y=1\), maximum \(3\), minimum \(-1\). (1 pt)

(d) At \(x=\dfrac{\pi}{3}\): \(\sin0=0\), so \(y=1\). At \(x=\dfrac{\pi}{2}\): \(3x-\pi=\dfrac{\pi}{2}\), \(\sin\dfrac{\pi}{2}=1\), so \(y=-2+1=-1\). (1 pt)

2 Writing an equation / 4 pts

(a) Amplitude \(\dfrac{12-2}{2}=5\); midline \(\dfrac{12+2}{2}=7\); half a period is 3, so the period is 6. (1.5 pt)

(b) \(B=\dfrac{2\pi}{6}=\dfrac{\pi}{3}\), so \(y=5\cos\left(\dfrac{\pi x}{3}\right)+7\). (1 pt)

(c) \(y(1)=5\cos\dfrac{\pi}{3}+7=9.5\). For \(y=9.5\): \(\cos\dfrac{\pi x}{3}=\dfrac12\Rightarrow\dfrac{\pi x}{3}=\dfrac{\pi}{3}\) or \(\dfrac{5\pi}{3}\), so \(x=1\) or \(x=5\). (1.5 pt)

3 Tangent and cotangent / 3 pts

(a) Period \(\dfrac{\pi}{3}\); asymptotes \(3x=\pm\dfrac{\pi}{2}\), i.e. \(x=\pm\dfrac{\pi}{6}\); \(y\left(\dfrac{\pi}{12}\right)=\tan\dfrac{\pi}{4}=1\). (1.5 pt)

(b) Asymptotes \(x=0\) and \(x=\pi\); zero at \(x=\dfrac{\pi}{2}\); \(\cot\dfrac{\pi}{4}=1\) and \(\cot\dfrac{3\pi}{4}=-1\). (1.5 pt)

4 Cosecant / 3 pts

(a) Asymptotes \(x=k\pi\) (zeros of sine). Range \((-\infty,-3]\cup[3,\infty)\). (1 pt)

(b) \(y\left(\dfrac{\pi}{6}\right)=\dfrac{3}{1/2}=6\); \(y\left(\dfrac{7\pi}{6}\right)=\dfrac{3}{-1/2}=-6\). (1 pt)

(c) Local minimum \(3\) at \(x=\dfrac{\pi}{2}\); local maximum \(-3\) at \(x=\dfrac{3\pi}{2}\). (1 pt)

5 Inverse trigonometric values / 3 pts

(a) \(\dfrac{\pi}{3}\) (0.5 pt)

(b) \(\dfrac{\pi}{4}\) (0.5 pt)

(c) \(\tan\left(-\dfrac{\pi}{3}\right)=-\sqrt3\), so the answer is \(-\dfrac{\pi}{3}\). (0.5 pt)

(d) Opposite 5, hypotenuse 13, adjacent 12: \(\dfrac{12}{13}\). (1 pt)

(e) \(\sin\dfrac{7\pi}{6}=-\dfrac12\), so the answer is \(-\dfrac{\pi}{6}\). (0.5 pt)

6 Hours of daylight / 3 pts

(a) Maximum \(12+3.5=15.5\) h, minimum \(12-3.5=8.5\) h, period 365 days. (1 pt)

(b) \(D(172)=12+3.5\sin\left(\dfrac{2\pi\cdot92}{365}\right)\approx12+3.5(0.99992)\approx15.50\) h. (1 pt)

(c) Need \(\sin\theta\ge\dfrac{2}{3.5}\approx0.5714\) where \(\theta=\dfrac{2\pi(t-80)}{365}\). With \(\arcsin0.5714\approx0.6082\), this holds for \(0.6082\le\theta\le\pi-0.6082\), an interval of length \(\pi-1.2164\approx1.9252\). The number of days is \(\dfrac{1.9252}{2\pi}\cdot365\approx112\) days. (1 pt)