 # Trigonometric Ratios Of Standard Angles

Trigonometric ratios are Sine, Cosine, Tangent, Cotangent, Secant and Cosecant. The standard angles for these trigonometric ratios are 0°, 30°, 45°, 60° and 90°. These angles can also be represented in the form of radians such as 0, π/6, π/4, π/3, and π/2. These angles are most commonly and frequently used in trigonometry. Learning the values of these trigonometry angles is very necessary to solve various problems.

Trigonometric Ratios Formulas:

The six trigonometric ratios are basically expressed in terms of the right-angled triangle. ∆ABC is a right-angled triangle, right-angled at (shown in figure 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 Table (Standard Angles)

 Angle = ∠C 0° 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

The above table shows the important angles for all the six trigonometric ratios. Let us learn here how to derive these values.

## Derivation of Trigonometric Ratios for Standard Angles

Value of Trigonometric Ratios for Angle equal to 45 degrees 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$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

$cot~45°$ = $\frac{1}{tan~45°}$ = $1$

Value of Trigonometric Ratios for Angle equal to 30 and 60 degrees In figure 3, $∆PQR$ 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

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

$cos~60°$ = $\frac{1}{2}$

$tan~60°$ = $\sqrt{3}$       $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

$cot~60°$ = $\frac{1}{\sqrt{3}}$

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

$sec~60°$ = $2$

Value of Trigonometric Ratios for Angle equal to 0 and 90 degrees

In $∆ABC$ 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:

### Solved Examples

Question 1: What is the value of tan 30+sin 60?

Solution: tan 30 = 1/√3 and sin 60 = √3/2

Adding both the values we get;

1/√3  + √3/2

Rationalising the denominator gives:

(2+√3.√3)/2√3

2+3/2√3

5/2√3

Question 2: What is the value of sin45 – cos 45?

Solution: Sin 45 = 1/√2 and cos 45 = 1/√2

Therefore, on putting the values we get:

1/√2 – 1/√2 = 0

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