Trigonometric Ratios of 0°, 30°, 45°, 60° and 90°, Class 10

Deriv­ing the val­ues#

The ratios of 45°#

In $\triangle ABC$ with $\angle B = 90^\circ$, sup­pose $\angle A = 45^\circ$. The angle sum of a tri­an­gle is $180^\circ$, so $\angle C = 45^\circ$ too. Sides oppo­site equal angles are equal, so $AB = BC$. Call each of them $a$. By Pythago­ras' the­o­rem,

$$AC^2 = a^2 + a^2 = 2a^2, \qquad AC = \sqrt{2}\,a.$$

There­fore

$$\displaystyle \sin 45^\circ = \frac{BC}{AC} = \frac{a}{\sqrt{2}\,a} = \frac{1}{\sqrt{2}}, \qquad \cos 45^\circ = \frac{AB}{AC} = \frac{1}{\sqrt{2}}, \qquad \tan 45^\circ = \frac{BC}{AB} = 1.$$

Tak­ing rec­i­p­ro­cals, $\operatorname{cosec} 45^\circ = \sqrt{2}$, $\sec 45^\circ = \sqrt{2}$ and $\cot 45^\circ = 1$.

The ratios of 30° and 60°#

Take an equi­lat­eral tri­an­gle $ABC$ with each side $2a$. Every angle is $60^\circ$. Draw the per­pen­dic­u­lar $AD$ from $A$ to $BC$.

Tri­an­gles $ABD$ and $ACD$ are con­gru­ent by RHS: $AB = AC$, $AD$ is com­mon, and $\angle ADB = \angle ADC = 90^\circ$. Hence $BD = DC = a$ and $\angle BAD = \angle CAD = 30^\circ$. In $\triangle ABD$, right-angled at $D$,

$$AD = \sqrt{AB^2 - BD^2} = \sqrt{4a^2 - a^2} = \sqrt{3}\,a.$$

For $\angle BAD = 30^\circ$ the oppo­site side is $BD = a$ and the adja­cent side is $AD = \sqrt{3}\,a$:

$$\displaystyle \sin 30^\circ = \frac{a}{2a} = \frac{1}{2}, \qquad \cos 30^\circ = \frac{\sqrt{3}\,a}{2a} = \frac{\sqrt{3}}{2}, \qquad \tan 30^\circ = \frac{a}{\sqrt{3}\,a} = \frac{1}{\sqrt{3}}.$$

For $\angle ABD = 60^\circ$ the roles swap: the oppo­site side is $AD$ and the adja­cent side is $BD$:

$$\displaystyle \sin 60^\circ = \frac{\sqrt{3}}{2}, \qquad \cos 60^\circ = \frac{1}{2}, \qquad \tan 60^\circ = \sqrt{3}.$$

The ratios of 0° and 90° as lim­it­ing cases#

Imag­ine a right tri­an­gle $ABC$ in which $\angle A$ keeps shrink­ing while the hypotenuse $AC$ stays the same length. The oppo­site side $BC$ gets shorter and shorter, and the adja­cent side $AB$ gets closer and closer to $AC$. As $\angle A$ approaches $0^\circ$, $BC$ approaches $0$ and $AB$ approaches $AC$. This moti­vates the def­i­n­i­tions

$$\sin 0^\circ = 0, \qquad \cos 0^\circ = 1, \qquad \tan 0^\circ = 0.$$

Since $\sin 0^\circ = 0$, the ratio $\displaystyle \operatorname{cosec} 0^\circ = \frac{1}{0}$ is not defined, and so is $\cot 0^\circ$. Also $\sec 0^\circ = 1$.

Now let $\angle A$ grow towards $90^\circ$. Then $\angle C$ shrinks towards $0^\circ$, the side $BC$ approaches $AC$ and $AB$ approaches $0$. So

$$\sin 90^\circ = 1, \qquad \cos 90^\circ = 0, \qquad \operatorname{cosec} 90^\circ = 1, \qquad \cot 90^\circ = 0,$$

while $\tan 90^\circ$ and $\sec 90^\circ$ are not defined because they would need divi­sion by $\cos 90^\circ = 0$.

The com­plete table#

Pat­terns that help you remem­ber#

• The sine row can be writ­ten as $\displaystyle \frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$. Note that $\displaystyle \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
• The cosine row is the sine row read back­wards.
• Each tan entry is sine divided by cosine in the same col­umn.
• The last three rows are the rec­i­p­ro­cals of the first three.
• As $A$ increases from $0^\circ$ to $90^\circ$, $\sin A$ increases from 0 to 1 and $\cos A$ decreases from 1 to 0.

Method: find­ing an unknown angle#

1. Rearrange the equa­tion so that one ratio stands alone, for exam­ple $\displaystyle \sin (A - B) = \frac{1}{2}$.
2. Look along the cor­rect row of the table for that value and note the angle.
3. Write the argu­ment equal to that angle: $A - B = 30^\circ$.
4. If there are two equa­tions, solve the result­ing pair of lin­ear equa­tions.
5. Check the answer against any con­di­tion given (such as $A \gt B$ or $A + B \le 90^\circ$) and sub­sti­tute back.

Worked exam­ples#

Exam­ple 1: a direct eval­u­a­tion#

Eval­u­ate $\cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ$.

Solu­tion.

$$\displaystyle \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}.$$

Exam­ple 2: squares of ratios#

Eval­u­ate $4 \sin^2 60^\circ + 3 \tan^2 30^\circ - 8 \sin 45^\circ \cos 45^\circ$.

Solu­tion.

$$\displaystyle \begin{aligned} &= 4 \left(\frac{\sqrt{3}}{2}\right)^2 + 3 \left(\frac{1}{\sqrt{3}}\right)^2 - 8 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \\ &= 4 \cdot \frac{3}{4} + 3 \cdot \frac{1}{3} - 8 \cdot \frac{1}{2} \\ &= 3 + 1 - 4 = 0. \end{aligned}$$

Exam­ple 3: show­ing two expres­sions are equal#

Show that $\displaystyle \frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ} = \tan 30^\circ$.

Solu­tion. The numer­a­tor is $\displaystyle \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$. The denom­i­na­tor is $\displaystyle 1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 2$. So

$$\displaystyle \text{LHS} = \frac{2/\sqrt{3}}{2} = \frac{1}{\sqrt{3}} = \tan 30^\circ = \text{RHS}.$$

Exam­ple 4: ratio­nal­is­ing the answer#

Eval­u­ate $\displaystyle \frac{\cos 45^\circ}{\operatorname{cosec} 60^\circ + \cot 45^\circ}$ and write the answer with a ratio­nal denom­i­na­tor.

Solu­tion. $\displaystyle \operatorname{cosec} 60^\circ + \cot 45^\circ = \frac{2}{\sqrt{3}} + 1 = \frac{2 + \sqrt{3}}{\sqrt{3}}$. Hence

$$\displaystyle \begin{aligned} \frac{\cos 45^\circ}{\operatorname{cosec} 60^\circ + \cot 45^\circ} &= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2 + \sqrt{3}} \\ &= \frac{\sqrt{3}\,(2 - \sqrt{3})}{\sqrt{2}\,(2 + \sqrt{3})(2 - \sqrt{3})} \\ &= \frac{2\sqrt{3} - 3}{\sqrt{2} \times (4 - 3)} = \frac{2\sqrt{3} - 3}{\sqrt{2}} \\ &= \frac{(2\sqrt{3} - 3)\sqrt{2}}{2} = \frac{2\sqrt{6} - 3\sqrt{2}}{2}. \end{aligned}$$

Exam­ple 5: two angles from two equa­tions#

If $\displaystyle \sin (A + B) = \frac{\sqrt{3}}{2}$ and $\displaystyle \cos (A - B) = \frac{\sqrt{3}}{2}$, where $0^\circ \lt A + B \le 90^\circ$ and $A \gt B$, find $A$ and $B$.

Solu­tion. From the table, $\displaystyle \sin 60^\circ = \frac{\sqrt{3}}{2}$, so $A + B = 60^\circ$. Also $\displaystyle \cos 30^\circ = \frac{\sqrt{3}}{2}$, so $A - B = 30^\circ$.

$$\begin{aligned} (A + B) + (A - B) &= 60^\circ + 30^\circ \;\Rightarrow\; 2A = 90^\circ \;\Rightarrow\; A = 45^\circ, \\ B &= 60^\circ - 45^\circ = 15^\circ. \end{aligned}$$

Check: $\displaystyle \sin (45^\circ + 15^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\displaystyle \cos (45^\circ - 15^\circ) = \cos 30^\circ = \frac{\sqrt{3}}{2}$. Both equa­tions hold.

Exam­ple 6: a mul­ti­ple angle#

Find the acute angle $x$ if $2 \sin 3x = \sqrt{3}$, given $0^\circ \lt 3x \lt 90^\circ$.

Solu­tion. $\displaystyle \sin 3x = \frac{\sqrt{3}}{2} = \sin 60^\circ$, so $3x = 60^\circ$ and $x = 20^\circ$.

A sim­i­lar one: if $\sqrt{3} \tan 2\theta - 3 = 0$ with $2\theta$ acute, then $\displaystyle \tan 2\theta = \frac{3}{\sqrt{3}} = \sqrt{3}$, so $2\theta = 60^\circ$ and $\theta = 30^\circ$.

Com­mon mis­takes#

• Swap­ping $\sin 30^\circ$ and $\sin 60^\circ$. Remem­ber that sine grows with the angle, so the smaller value, $\displaystyle \frac{1}{2}$, belongs to $30^\circ$.
• Writ­ing $\tan 90^\circ = 0$ or $\operatorname{cosec} 0^\circ = 0$. These are not defined.
• Squar­ing only the num­ber and not the surd: $\displaystyle \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$, not $\displaystyle \frac{\sqrt{3}}{4}$.
• Assum­ing $\sin (A - B) = \sin A - \sin B$. Trigono­met­ric ratios do not split over addi­tion or sub­trac­tion.
• Writ­ing $2 \sin 3x = \sqrt{3}$ as $\displaystyle \sin x = \frac{\sqrt{3}}{6}$. Find $3x$ first, then divide by 3.

Try these#

1. Eval­u­ate $\sin^2 30^\circ + \cos^2 45^\circ + \tan^2 60^\circ$. Answer: $\displaystyle \frac{15}{4}$.
2. Eval­u­ate $2 \cos^2 60^\circ + 3 \sec^2 30^\circ - 2 \tan^2 45^\circ$. Answer: $\displaystyle \frac{5}{2}$.
3. If $\tan (A + B) = 1$ and $\displaystyle \tan (A - B) = \frac{1}{\sqrt{3}}$, with $0^\circ \lt A + B \le 90^\circ$ and $A \gt B$, find $A$ and $B$. Answer: $A = 37.5^\circ$, $B = 7.5^\circ$.
4. Find $x$ if $\displaystyle \cos 2x = \frac{1}{2}$ and $2x$ is acute. Answer: $x = 30^\circ$.
5. Eval­u­ate $\displaystyle \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ}$. Answer: $\displaystyle \frac{1}{2}$.

Key terms#

Stan­dard angles

The angles 0°, 30°, 45°, 60° and 90°, whose ratios have exact val­ues.

Isosce­les right tri­an­gle

A right tri­an­gle with two equal legs; its acute angles are both 45°.

Equi­lat­eral tri­an­gle

A tri­an­gle with all sides equal and all angles 60°; its alti­tude splits it into two 30°–60°–90° tri­an­gles.

Not defined

A value that would require divi­sion by zero, such as tan 90° or cosec 0°.

Lim­it­ing case

The value a ratio approaches as the angle approaches 0° or 90°.

Ratio­nal­is­ing

Mul­ti­ply­ing numer­a­tor and denom­i­na­tor by a suit­able surd to remove the root from the denom­i­na­tor.

Com­mon ques­tions#

Do I have to derive the table in the exam?#

Usu­ally you only need the val­ues, but a deriva­tion for 45° or for 30° and 60° is some­times asked as a short-answer ques­tion. Know­ing the tri­an­gles also helps you recover a value you have for­got­ten.

Should I write 1/√2 or √2/2?#

Both are cor­rect and equal. If a ques­tion asks for a ratio­nalised answer, use $\displaystyle \frac{\sqrt{2}}{2}$.

Is sin 2A equal to 2 sin A?#

No. For exam­ple, $\displaystyle \sin 60^\circ = \frac{\sqrt{3}}{2}$ but $2 \sin 30^\circ = 1$.

Why is tan 90° not defined but sin 90° is?#

$\sin 90^\circ = 1$ is a proper num­ber. $\tan 90^\circ$ would be $\displaystyle \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0}$, and divi­sion by zero has no value.

Which angle has equal sine and cosine?#

Among acute angles, only $45^\circ$, since both equal $\displaystyle \frac{1}{\sqrt{2}}$ there.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Sharma, R. D. Math­e­mat­ics for Class 10. Dhan­pat Rai Pub­li­ca­tions.
3. Loney, S. L. Plane Trigonom­e­try. Cam­bridge Uni­ver­sity Press.