Trigonometric Ratios Of Standard Angles

Trigonometric Ratios:

There are six trigonometric ratios that relates sides of right angle triangle to its angle. They are defined by parameters namely height, base and perpendicular.

Trigonometric Ratios

\(∆ABC\) is a right angled triangle, right angled at \(B\) (shown in fig. 1). The six trigonometric ratios for \(∠C\) are defined as:

\(sin~∠C\) = \(\frac{AB}{AC}\)

\(cosec~∠C\) = \(\frac{1}{sin~∠C}\)

\(cos~∠C\) = \(\frac{BC}{AC}\)

\(sec~∠C\) = \(\frac{1}{cos~∠C}\)

\(tan~∠C\) = \(\frac{sin~∠C}{cos~∠C}\)\(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(cot~∠C\) = \(\frac{1}{tan~∠C}\)

The standard angles for which trigonometric ratios can be easily determined are \(0°,30°,45°,60°\) and \(90°\). The values are determined using properties of triangles. The two acute angles of a right-angled triangle are complementary.

Trigonometric Ratios of Standard Angles:

  • For 45°:

Trigonometric Ratios

In \(⊿ABC\), if \(∠C\) = \(45°\), then \(∠A\) = \(45°\). Since the angles are equal, \(⊿ABC\) becomes a right angled isosceles triangle. So, \(AB\) = \(BC\). Assume \(AB\) = \(BC\) = \(a\) units.

Using Pythagoras theorem ,

\(AC^2\) = \(AB^2~+~BC^2\)

\(AC^2\) = \(a^2~+~a^2\)

\(AC\) = \(a\sqrt{2} ~units\)

\(∠C\) = \(45°\)

\(∴~ sin~∠C\) = \(sin~45°\) = \(\frac{AB}{AC}\) = \(\frac{a}{a\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\) \(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(cosec~45°\) = \(\frac{1}{sin~45°}\) = \(\sqrt{2}\)

\(cos~∠C\) = \(cos~45°\) = \(\frac{BC}{AC}\) = \(\frac{a}{a\sqrt{2}}\) = \(\frac{1}{\sqrt{2}}\)   \(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(sec~45°\) = \(\frac{1}{cos~45°}\) = \(\sqrt{2}\)

\(tan~45°\) = \(\frac{sin~45°}{cos~45°}\) = \(\frac{\frac{a}{\sqrt{2}}}{\frac{a}{\sqrt{2}}}\) = \(1\)\(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(cot~45°\) = \(\frac{1}{tan~45°}\) = \(1\)

  • For \(30°\) and \(60°\):

Trigonometric Ratios

In fig. 2, \(∆PQR\) is equilateral. The perpendicular from any vertex on the opposite side is coincident with the angle bisector of that particular vertex. Also, the perpendicular bisects the opposite side. If a perpendicular \(PS\) is dropped on \(QR\), then \(∠QPS\) = \(∠SPR\) = \(30°\) and \(QS\) = \(SR\). Assume \(PQ\) = \(QR\) = \(RP\) = \(2a\) units.

\(⇒~QS\) = \(SR\) = \(a\) units

In \(∆PSQ\), by Pythagoras theorem,

\(PQ^2\) = \(QS^2~+~PS^2\)

\(PS^2\) = \((2a)^2~+~a^2\)

\(PS\) = \(\sqrt{3a^2}\) = \(\sqrt{3} a\)

\(∠SPQ\) = \(30°\)

\(sin~∠SPQ\) = \(sin~30°\) = \(\frac{SQ}{PQ}\) = \(\frac{a}{2a}\) = \(\frac{1}{2}\)   \(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(cosec~30°\) = \(\frac{1}{sin~30°}\) = \(2\)

\(cos~∠SPQ\) = \(cos~30°\) = \(\frac{PS}{PQ}\) = \(\frac{\sqrt{3} a}{2a}\) = \(\frac{\sqrt{3}}{2}\)    \(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(sec~30°\) = \(\frac{1}{cos~30°}\) = \(\frac{2}{\sqrt{3}}\)

\(tan~30°\) = \(\frac{sin~30°}{cos~30°}\) = \(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\) = \(\frac{1}{\sqrt{3}}\)    \(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(cot~30°\) = \(\frac{BC}{AB}\) = \(\sqrt{3}\)

Similarly, ratios of 60° are determined by finding the ratios of \(∠SQP\) as

\(sin~60°\) = \(\frac{\sqrt{3}}{2}\)      \(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(cos~60°\) = \(\frac{1}{2}\)

\(tan~60°\) = \(\sqrt{3}\)       \(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(sec~60°\) = \(\frac{2}{\sqrt{3}}\)

\(cosec~60°\) = \(\frac{2}{\sqrt{3}}\)      \(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\)

\(cot~60°\) = \(2\)

For \(0°\) and \(90°\)

Visualization of angles

In fig. 4, \(∆ABC\) is a right angled triangle. If the length of side \(BC\) is continuously decreased, then value of \(∠A\) will keep on decreasing. Similarly, value of \(∠C\) is increasing as length of \(BC\) is decreasing. When BC = 0, ∠A = 0 , ∠C = 90° and AB = AC.

Taking ratios for \(∠A\) = \(0°\)

\(sin~∠A\) = \(sin~0°\) = \(\frac{BC}{AC}\) = \(\frac{0}{AC}\) = \(0\)

\(cosec~0°\) = \(\frac{1}{sin~0°}\) = \(\frac{1}{0}\) = Not Defined

\(cos~∠A\) = \(cos~0°\) = \(\frac{AB}{AC}\) = \(\frac{AC}{AC}\) = \(1\)

\(sec~0°\) = \(\frac{1}{cos~0°}\) = \(\frac{1}{1}\) = \(1\)

\(tan~0°\) = \(\frac{sin~0°}{cos~0°}\) = \(\frac{0}{1}\) = \(0\)

\(cot~0°\) = \(\frac{1}{tan~0°}\) = \(\frac{1}{0}\) = Not Defined

Taking ratios for \(∠C\) = \(90°\)

\(sin~∠C\) = \(sin~90°\) = \(\frac{AB}{AC}\) = \(\frac{AC}{AC}\) = \(1\)

\(cosec 90°\) = \(\frac{1}{sin~90°}\) = \(\frac{1}{1}\) = \(1\)

\(cos~∠C\) = \(cos~90°\) = \(\frac{BC}{AC}\) = \(\frac{0}{AC}\) = \(0\)

\(sec~90°\) = \(\frac{1}{cos~90°}\) = \(\frac{1}{0}\) = Not Defined

\(tan~90°\) = \(\frac{sin~90°}{cos~90°}\) = \(\frac{1}{0}\) = Not Defined

\(cot~90°\) = \(\frac{1}{tan~90°}\) = \(\frac{0}{1}\) = \(0\)

Following is the trigonometric ratios table which contains all the trigonometric ratios of standard angles:

Table 1: Trigonometric Ratios Table

∠C

30°

45°

60°

90°

\(sin~C\)

0

\(\frac{1}{2}\)

\(\frac{1}{\sqrt{2}}\)

\(\frac{\sqrt{3}}{2}\)

1

\(cos~C\)

1

\(\frac{\sqrt{3}}{2}\)

\(\frac{1}{\sqrt{2}}\)

\(\frac{1}{2}\)

0

\(tan~C\)

0

\(\frac{1}{\sqrt{3}}\)

1

\(\sqrt{3}\)

Not Defined

\(cosec~C\)

Not Defined

2

\(\sqrt{2}\)

\(\frac{2}{\sqrt{3}}\)

1

\(sec~C\)

1

\(\frac{2}{\sqrt{3}}\)

\(\sqrt{2}\)

2

Not Defined

\(cot~C\)

Not Defined

\(\sqrt{3}\)

1

\(\frac{1}{\sqrt{3}}\)

0

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Practise This Question

Use the Principle of Mathematical Induction, check if, for any positive integer n, 6n1 is divisible by 5. True or False?