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Trigonometric Functions and the Unit Circle: math test solutions, Grade 11 – download the PDF

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Test solutions Grade 11 : Trigonometric Functions and the Unit Circle — Zyro the alien explorer of Planète Maths

Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

1 Radians and degrees / 3 pts

  1. \(72\times\dfrac{\pi}{180}=\dfrac{2\pi}{5}\). (1 pt)
  2. \(\dfrac{5\pi}{12}\times\dfrac{180}{\pi}=75^\circ\). (1 pt)
  3. \(\dfrac{7\pi}{9}\times\dfrac{180}{\pi}=140^\circ\). (1 pt)

2 Exact values / 4 pts

  1. Quadrant II, reference angle \(\dfrac{\pi}{3}\), sine positive: \(\dfrac{\sqrt{3}}{2}\). (1 pt)
  2. Quadrant III, reference angle \(\dfrac{\pi}{6}\), cosine negative: \(-\dfrac{\sqrt{3}}{2}\). (1 pt)
  3. Quadrant III, reference angle \(\dfrac{\pi}{4}\), tangent positive: \(1\). (1 pt)
  4. The angle is in quadrant IV with reference angle \(\dfrac{\pi}{6}\), sine negative: \(-\dfrac{1}{2}\). (1 pt)

3 Using the identity / 3 pts

  1. \(\cos^2\theta=1-\dfrac{64}{289}=\dfrac{225}{289}\), so \(\cos\theta=\pm\dfrac{15}{17}\) (1 pt). Cosine is negative in quadrant III, so \(\cos\theta=-\dfrac{15}{17}\) (1 pt).
  2. \(\tan\theta=\dfrac{-8/17}{-15/17}=\dfrac{8}{15}\) (1 pt).

4 Reading a sinusoid / 3 pts

  1. \(4x-\pi=4\left(x-\dfrac{\pi}{4}\right)\), so \(y=2\sin\left(4\left(x-\dfrac{\pi}{4}\right)\right)+3\) (1 pt, with the factoring shown).
  2. Amplitude 2, period \(\dfrac{2\pi}{4}=\dfrac{\pi}{2}\), phase shift \(\dfrac{\pi}{4}\) to the right, midline \(y=3\) (1 pt for all four values).
  3. Maximum \(3+2=5\), minimum \(3-2=1\) (1 pt).

5 Equations / 3 pts

  1. \(\sin x=\dfrac{\sqrt{3}}{2}\), reference angle \(\dfrac{\pi}{3}\), quadrants I and II: \(x=\dfrac{\pi}{3}\) or \(x=\dfrac{2\pi}{3}\). (1 pt)
  2. Factor: \((2\cos x-1)(\cos x+1)=0\) (1 pt). Then \(\cos x=\dfrac{1}{2}\) gives \(x=\dfrac{\pi}{3},\dfrac{5\pi}{3}\), and \(\cos x=-1\) gives \(x=\pi\) (1 pt for the three solutions).

6 A floating buoy / 4 pts

  1. Period \(\dfrac{2\pi}{\pi/5}=10\) seconds; maximum \(2+1.5=3.5\) m; minimum \(2-1.5=0.5\) m. (1 pt)
  2. \(h(2.5)=2+1.5\sin\dfrac{\pi}{2}=3.5\) m. (1 pt)
  3. \(1.5\sin\dfrac{\pi t}{5}=0\), so \(\dfrac{\pi t}{5}=0,\ \pi,\ 2\pi\) and \(t=0,\ 5,\ 10\) seconds. (1 pt)
  4. \(3.5\div0.3048\approx11.48\) ft. (1 pt)
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