Fundamentals of trigonometry

Unit 4 Fundamentals of trigonometry

Key unit competence: Apply the trigonometric circle and identities in determining trigonometric ratios and solving related problems in daily life.

Key ideas

• Angles are measured in degrees or radians: $180^{\circ }=\pi$ rad. Convert: $x^{\circ }=x\times \frac{\pi }{180}$ rad.
• On the unit circle, a point at angle $\theta$ is $(\cos \theta ,\sin \theta )$; $\tan \theta =\frac{\sin \theta }{\cos \theta }$; $\sec \theta =\frac{1}{\cos \theta }$, $\csc \theta =\frac{1}{\sin \theta }$, $\cot \theta =\frac{1}{\tan \theta }$.
• Signs by quadrant (CAST): I all positive; II sine positive; III tangent positive; IV cosine positive.
• Special angles: $\sin 30^{\circ }=\frac{1}{2}$, $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$, $\tan 30^{\circ }=\frac{\sqrt{3}}{3}$; $\sin 45^{\circ }=\cos 45^{\circ }=\frac{\sqrt{2}}{2}$; $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$, $\cos 60^{\circ }=\frac{1}{2}$, $\tan 60^{\circ }=\sqrt{3}$; $\sin 90^{\circ }=1$, $\cos 90^{\circ }=0$.
• Identities: ${\sin }^2\theta +{\cos }^2\theta =1$; $1+{\tan }^2\theta ={\sec }^2\theta$; $1+{\cot }^2\theta ={\csc }^2\theta$.
• Reduction: $\sin (180^{\circ }-\theta )=\sin \theta$, $\cos (180^{\circ }-\theta )=-\cos \theta$, $\sin (-\theta )=-\sin \theta$, $\cos (-\theta )=\cos \theta$, $\sin (90^{\circ }-\theta )=\cos \theta$.
• Sine law: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Cosine law: $a^2=b^2+c^2-2bc\cos A$. Area $=\frac{1}{2}bc\sin A$.

Formulas to know

Common misconceptions

• "${\sin }^2\theta$ means $\sin ({\theta }^2)$." No: it means $(\sin \theta {)}^2$.
• "$\cos 120^{\circ }=\cos 60^{\circ }$." No: $\cos 120^{\circ }=-\cos 60^{\circ }=-\frac{1}{2}$ (second quadrant).
• "Use the sine law when you know three sides." No: with three sides (or two sides and the included angle) use the cosine law.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Convert $225^{\circ }$ to radians and give $\sin$, $\cos$, $\tan$ of $225^{\circ }$.

   Show answer

   $\frac{5\pi }{4}$; $\sin =-\frac{\sqrt{2}}{2}$, $\cos =-\frac{\sqrt{2}}{2}$, $\tan =1$.

   Common mistake: Forgetting the signs in quadrant III.

2. Find the area of a triangle with sides 7, 8 and included angle $150^{\circ }$.

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   $\frac{1}{2}\times 7\times 8\times \sin 150^{\circ }=28\times \frac{1}{2}=14$.

   Common mistake: Using $\cos 150^{\circ }$.

3. Given $\tan \theta =\frac{3}{4}$ with $\theta$ in quadrant III, find $\sin \theta$ and $\cos \theta$.

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   Hypotenuse 5; both negative in QIII: $\sin \theta =-\frac{3}{5}$, $\cos \theta =-\frac{4}{5}$.

   Common mistake: Giving positive values.

4. A tower is observed from two points $A$ and $B$ 50 m apart on level ground in line with it. The angles of elevation are $30^{\circ }$ (from $A$) and $45^{\circ }$ (from $B$, nearer). Find the height.

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   $h=x\tan 45^{\circ }=x$ and $h=(x+50)\tan 30^{\circ }$. So $x=\frac{x+50}{\sqrt{3}}$, $x(\sqrt{3}-1)=50$, $h=\frac{50}{\sqrt{3}-1}\approx 68.3$ m.

   Common mistake: Using the same horizontal distance for both angles.

5. Prove that $\tan \theta +\cot \theta =\sec \theta \csc \theta$.

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   $\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{1}{\sin \theta \cos \theta }$.

   Common mistake: Adding numerators without a common denominator.

6. Sketch $y=\cos x$ for $0\le x\le 2\pi$, stating period, maximum, minimum and zeros.

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   Period $2\pi$; max 1 at $0$ and $2\pi$; min $-1$ at $\pi$; zeros at $\frac{\pi }{2}$ and $\frac{3\pi }{2}$.

   Common mistake: Starting the cosine graph at 0.

7. In triangle $ABC$, $A=40^{\circ }$, $B=75^{\circ }$ and $a=10$ cm. Find $b$.

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   $b=\frac{10\sin 75^{\circ }}{\sin 40^{\circ }}\approx \frac{9.659}{0.643}\approx 15.0$ cm.

   Common mistake: Using the cosine law with insufficient data.

8. Find the largest angle of the triangle with sides $a=5$, $b=6$, $c=9$.

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   The largest angle is opposite $c$: $\cos C=\frac{25+36-81}{60}=-\frac{1}{3}$, $C\approx 109.5^{\circ }$.

   Common mistake: Computing an angle opposite a shorter side.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find the area of triangle $ABC$ with $b=6$, $c=10$ and $A=150^{\circ }$.	a. $1$
   2. Simplify $(1+{\tan }^{2}x){\cos }^{2}x$.	b. $1+\cos x$
   3. Simplify $\frac{{\sin }^{2}x}{1-\cos x}$.	c. $7$
   4. In triangle $ABC$, $b=5$, $c=8$ and $A=60^{\circ }$. Find $a$.	d. $15$

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   1 → d ($15$) · 2 → a ($1$) · 3 → b ($1+\cos x$) · 4 → c ($7$)

   • Step 1. Area $=\frac{1}{2}bc\sin A$.
     Step 2. $\frac{1}{2}\times 6\times 10\times \sin 150^{\circ }=30\times \frac{1}{2}=15$.
   • Step 1. $1+{\tan }^{2}x={\sec }^{2}x=\frac{1}{{\cos }^{2}x}$.
     Step 2. $\frac{1}{{\cos }^{2}x}\times {\cos }^{2}x=1$.
   • Step 1. ${\sin }^{2}x=1-{\cos }^{2}x=(1-\cos x)(1+\cos x)$.
     Step 2. Cancel $1-\cos x$: $1+\cos x$.
   • Step 1. Cosine law: $a^2=25+64-2(5)(8)\cos 60^{\circ }$.
     Step 2. $a^2=89-40=49$, so $a=7$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $4\pi$ cm  ·  $71.4$ m  ·  $35.0$ m  ·  $\frac{1+\cos x}{\sin x}$
   1. From a point 50 m from the foot of a tower, the angle of elevation of the top is $35^{\circ }$. The height of the tower is about .
   2. An arc of a circle of radius 6 cm subtends $\frac{2\pi }{3}$ rad at the centre. Its length is .
   3. The expression $\frac{\sin x}{1-\cos x}$ is equal to .
   Show answer

   1: $35.0$ m · 2: $4\pi$ cm · 3: $\frac{1+\cos x}{\sin x}$

   • Step 1. $\tan 35^{\circ }=\frac{h}{50}$.
     Step 2. $h=50\tan 35^{\circ }\approx 50\times 0.700=35.0$ m.
   • Step 1. $l=r\theta$ (with $\theta$ in radians).
     Step 2. $l=6\times \frac{2\pi }{3}=4\pi \approx 12.6$ cm.
   • Step 1. Multiply top and bottom by $1+\cos x$: $\frac{\sin x(1+\cos x)}{1-{\cos }^{2}x}$.
     Step 2. $1-{\cos }^{2}x={\sin }^{2}x$; cancel $\sin x$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find the exact value of $\sin \frac{3\pi }{4}\cos \frac{\pi }{3}+\cos \frac{3\pi }{4}\sin \frac{\pi }{3}$.	a. ${\sin }^{2}x-{\cos }^{2}x$
   2. Simplify ${\sin }^{4}x-{\cos }^{4}x$.	b. $-\sqrt{3}$
   3. Simplify $\sin (180^{\circ }-x)+\cos (90^{\circ }-x)$.	c. $2\sin x$
   4. Find the exact value of $\tan \frac{5\pi }{3}$.	d. $\frac{\sqrt{2}-\sqrt{6}}{4}$

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   1 → d ($\frac{\sqrt{2}-\sqrt{6}}{4}$) · 2 → a (${\sin }^{2}x-{\cos }^{2}x$) · 3 → c ($2\sin x$) · 4 → b ($-\sqrt{3}$)

   • Step 1. $\sin \frac{3\pi }{4}=\frac{\sqrt{2}}{2}$, $\cos \frac{3\pi }{4}=-\frac{\sqrt{2}}{2}$.
     Step 2. $\frac{\sqrt{2}}{2}\cdot \frac{1}{2}-\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{2}-\sqrt{6}}{4}$.
   • Step 1. Difference of squares: $({\sin }^{2}x-{\cos }^{2}x)({\sin }^{2}x+{\cos }^{2}x)$.
     Step 2. ${\sin }^{2}x+{\cos }^{2}x=1$.
   • Step 1. $\sin (180^{\circ }-x)=\sin x$.
     Step 2. $\cos (90^{\circ }-x)=\sin x$.
     Step 3. Sum: $2\sin x$.
   • Step 1. $\frac{5\pi }{3}=300^{\circ }$: quadrant IV, reference angle $60^{\circ }$.
     Step 2. Tangent is negative in QIV: $-\tan 60^{\circ }=-\sqrt{3}$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $8$  ·  $60^{\circ }$  ·  $4$  ·  $2$
   1. The maximum value of $3\sin x+1$ is .
   2. The minimum value of $5-3\cos x$ is .
   3. A triangle has sides 5, 7 and 8. The angle opposite the side 7 is .
   Show answer

   1: $4$ · 2: $2$ · 3: $60^{\circ }$

   • Step 1. The largest value of $\sin x$ is 1.
     Step 2. $3(1)+1=4$.
   • Step 1. $-1\le \cos x\le 1$, so $-3\le -3\cos x\le 3$.
     Step 2. $5-3\cos x$ ranges from 2 to 8: the minimum is 2 (at $\cos x=1$).
   • Step 1. $\cos \theta =\frac{5^2+8^2-7^2}{2\times 5\times 8}=\frac{40}{80}=\frac{1}{2}$.
     Step 2. $\theta =60^{\circ }$.

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