Trigonometric Functions
Questions with Answers

A set of trigonometry questions related to trigonometric functions is presented below. Complete explanations, solutions, and final answers are provided.

Question 1

Find the exact value of \(\sin(x/2)\) if \(\sin x = \dfrac{1}{4}\) and \(\dfrac{\pi}{2} < x < \pi\).

Solution to Question 1

• To find \(\sin(x/2)\), we use the half-angle formula \[ \sin\left(\dfrac{x}{2}\right) = \pm \sqrt{\dfrac{1-\cos x}{2}} \]
• Since \(\dfrac{\pi}{2} < x < \pi\), then \(\dfrac{\pi}{4} < x/2 < \dfrac{\pi}{2}\). Hence \(x/2\) is in quadrant I and \(\sin(x/2)\) is positive. Therefore \[ \sin\left(\dfrac{x}{2}\right) = \sqrt{\dfrac{1-\cos x}{2}} \]
• Given \(\sin x = \dfrac{1}{4}\), we use the identity \(\sin^2 x + \cos^2 x = 1\). Since \(x\) is in quadrant II, \(\cos x\) is negative: \[ \cos x = -\sqrt{1-\sin^2 x} = -\sqrt{1-\dfrac{1}{16}} = -\dfrac{\sqrt{15}}{4} \]
• Substitute \(\cos x\) into the half-angle formula: \[ \sin\left(\dfrac{x}{2}\right) = \sqrt{\dfrac{1+\dfrac{\sqrt{15}}{4}}{2}} \]
• This simplifies to \[ \sin\left(\dfrac{x}{2}\right) = \dfrac{1}{4}\sqrt{8-2\sqrt{15}} \]

Question 2

\(x\) is in quadrant III. Approximate \(\sin(2x)\) if \(\cos x = -0.2\). Round your answer to two decimal places.

Solution to Question 2

• Use the double-angle identity \[ \sin(2x) = 2\sin x\cos x \]
• Since \(x\) is in quadrant III, \(\sin x\) is negative. Using \(\sin^2 x + \cos^2 x = 1\): \[ \sin x = -\sqrt{1-(-0.2)^2} \]
• Substitute into the double-angle formula: \[ \sin(2x) = 2\big[-\sqrt{1-(-0.2)^2}\big](-0.2) \approx 0.39 \]

Question 3

\(\tan x = 4\) and \(x\) is in quadrant III. Find the exact value of \(\cos x\).

Solution to Question 3

• Use the identity \[ 1+\tan^2 x = \sec^2 x \]
• Since \(x\) is in quadrant III, \(\sec x\) is negative: \[ \sec x = -\sqrt{1+4^2} = -\sqrt{17} \]
• Therefore \[ \cos x = \dfrac{1}{\sec x} = -\dfrac{1}{\sqrt{17}} \]

Question 4

\(\cos(2x)=0.6\) and \(2x\) is in quadrant I. Find the exact value of \(\csc x\).

Solution to Question 4

• Use \[ \cos(2x)=2\cos^2 x-1 \] which gives \[ \cos^2 x=0.8 \]
• Using \(\sin^2 x+\cos^2 x=1\): \[ \sin x = \sqrt{1-0.8} \]
• Hence \[ \csc x = \dfrac{1}{\sqrt{1-0.8}} \]

Question 5

Find the exact value of \(\cos 15^\circ\).

Solution to Question 5

• Using the half-angle formula \[ \cos\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{1+\cos x}{2}} \]
• \[ \cos 15^\circ = \sqrt{\dfrac{1+\cos 30^\circ}{2}} = \dfrac{1}{2}\sqrt{2+\sqrt{3}} \]

Question 6

Find the exact value of \(\tan(-22.5^\circ)\).

Solution to Question 6

• \[ \tan(-22.5^\circ)=-\tan(22.5^\circ) \]
• Using \[ \tan\left(\dfrac{x}{2}\right)=\dfrac{\sin x}{1+\cos x} \] with \(x=45^\circ\): \[ \tan(22.5^\circ)=\sqrt{2}-1 \] \[ \tan(-22.5^\circ)= 1 - \sqrt{2} \]

Question 7

\(x\) and \(y\) are angles in quadrants I and III respectively. If \(\cos x=a\) and \(\sin y=b\), find \(\cos(x+y)\) in terms of \(a\) and \(b\).

Solution to Question 7

• Using \[ \cos(x+y)=\cos x\cos y-\sin x\sin y \]
• Since \(x\) is in quadrant I: \[ \sin x=\sqrt{1-a^2} \]
• Since \(y\) is in quadrant III: \[ \cos y=-\sqrt{1-b^2} \]
• \[ \cos(x+y)=-a\sqrt{1-b^2}-b\sqrt{1-a^2} \]

Question 8

\(x\) is in quadrant III and \(\sin x= - \dfrac{1}{3}\). Find \(\sin(3x)\) and \(\cos(3x)\).

Solution to Question 8

• Using \[ \sin(3x)=3\sin x-4\sin^3 x \]
• \[ \sin(3x)=3\left(-\dfrac{1}{3}\right)-4\left(-\dfrac{1}{3}\right)^3 = - \dfrac{23}{27} \]

• Using \[ \cos(3x)= 4\cos ^{3}(x)-3\cos (x) \]
• Evaluate \( \cos(x) \) \[ \cos(x)= - \sqrt{1 - (- \dfrac{1}{3})^2} = - \dfrac{2\sqrt 2}{3} \]
• \[ \cos(3x)= 4 \left(-\dfrac{2\sqrt 2}{3}\right)^3-3\left(-\dfrac{2\sqrt 2}{3}\right) = - \dfrac{10 \sqrt 2}{27} \]

Question 9

Reduce the power of the expression \[ 4\sin^3 x+4\cos^3 x \]

Solution to Question 9

• Using power-reducing identities: \[ \sin^3 x=\dfrac{3}{4}\sin x-\dfrac{1}{4}\sin(3x),\quad \cos^3 x=\dfrac{3}{4}\cos x+\dfrac{1}{4}\cos(3x) \]
• \[ 4\sin^3 x+4\cos^3 x=3\sin x-\sin(3x)+3\cos x+\cos(3x) \]

Question 10

Factor the expression \[ \sin x+\sin(2x) \]

Solution to Question 10

• Using \(\sin(2x)=2\sin x\cos x\): \[ \sin x+\sin(2x)=\sin x(1+2\cos x) \]

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