 # Trigonometry Values

## Trigonometric Values Table

Trigonometric values of ratios like sine, cos, tan, cosec, cot, and secant are very useful while solving and dealing with problems related to the measurement of length and angles of a right-angled triangle. 0°, 30°, 45°, 60°, and 90° are the commonly used values of the trigonometric function to solve trigonometric problems.

The concept of trigonometric functions and values is one of the most important parts of Mathematics and also in our day to day life.

### Trigonometric Ratios

Trigonometry values are based on three major trigonometric ratios, Sine, Cosine, and Tangent.

• Sine or sin θ = Side opposite to θ / Hypotenuse = BC / AC

• Cosines or cos θ = Adjacent side to θ / Hypotenuse = AB / AC

• Tangent or tan θ =Side opposite to θ / Adjacent side to θ = BC / AB

Similarly, we will write the trigonometric values for reciprocal properties, Sec, Cosec and Cot ratios.

• Sec θ = 1/Cos θ = Hypotenuse / Adjacent side to angle θ = AC / AB

• Cosec θ = 1/Sin θ = Hypotenuse / Side opposite to angle θ = AC / BC

• Cot θ = 1/tan θ = Adjacent side to angle θ / Side opposite to angleθ = AB / BC

Also,

• Sec θ . Cos θ =1

• Cosec θ . Sin θ =1

• Cot θ . Tan θ =1

## Values of Trigonometric Ratios

 Angle (In Degree) 0° 30° 45° 60° 90° Angle (In Radian) 0 π/6 π/4 π/3 π/2 Sin θ 0 1/2 1/$\sqrt 2$ $\sqrt 3$/2 1 Cos θ 1 $\sqrt 3$/2 1/$\sqrt 2$ 1/2 0 Tan θ 0 1/$\sqrt 3$ 1 $\sqrt 3$ ထ Cot θ ထ $\sqrt 3$ 1 1/$\sqrt 3$ 0 Sec θ 1 2/$\sqrt 3$ $\sqrt 2$ 2 ထ Cosec θ ထ 2 $\sqrt 2$ 2/$\sqrt 3$ 1

### Trigonometry Ratios Formula

• Tan θ = sin θ/cos θ

• Cot θ = cos θ/sin θ

• Sin θ = tan θ/cos θ

• Cos θ = sin θ/tan θ

• Sec θ = tan θ/sin θ

• Cosec θ = cos θ/tan θ

Also,

• sin (90°- θ) = cos θ

• cos (90°- θ) = sin θ

• tan (90°- θ) = cot θ

• cot (90°- θ) = tan θ

• sec (90°- θ) = cosec θ

• cosec (90°- θ) = sec θ

• sin (90°+ θ) = cos θ

• cos (90°+ θ) = -sin θ

• tan (90°+ θ) = -cot θ

• cot (90°+ θ) = -tan θ

• sec (90°+ θ) = -cosec θ

• cosec (90°+ θ) = sec θ

• sin (180°- θ) = sin θ

• cos (180°- θ) = -cos θ

• tan (180°- θ) = -tan θ

• cot (180°- θ) = -cot θ

• sec (180°- θ) = -sec θ

• cosec (180°- θ) = cosec θ

• sin (180°+ θ) = -sin θ

• cos (180°+ θ) = -cos θ

• tan (180°+ θ) = tan θ

• cot (180°+ θ) = cot θ

• sec  (180°+ θ) = -sec θ

• cosec (180°+ θ) = -cosec θ

• sin (360°- θ) = -sin θ

• cos (360°- θ) = cos θ

• tan (360°- θ) = -tan θ

• cot (360°- θ) = -cot θ

• sec (360°- θ) = sec θ

• cosec (360°- θ) = -cosec θ

• sin (360°+ θ) = sin θ

• cos (360°+ θ) = cos θ

• tan (360°+ θ) = -tan θ

• cot (360°+ θ) = -cot θ

• sec (360°+ θ) = sec θ

• cosec (360°+ θ) = -cosec θ

• sin (270°- θ) = -cos θ

• cos (270°- θ) = -sin θ

• sin (270°+ θ) = -cos θ

• cos (270°+ θ) = sin θ

### Questions:

Question 1.  Find the values of another five trigonometric functions if

1. Cos x = -1/2, where x lies in the third quadrant.

Ans.  Since x is in lllrd  Quadrant

sin and cos will be negative

But tan will be positive

Given cos x = -1/v

We know that (image will be uploaded soon)

sin2 x + cos2 x = 1

sin2 x + (-1/2)2 = 1

sin2 x + 1/4 = 1

sin2 x = 1 - 1/4

sin2 x = 4 - 1/4

sin2 x = 3/4

sin x = ±$\sqrt 3$ / 4

sin x = ±$\sqrt 3$ / 2

Since x is in lllrd  Quadrant

sin x is negative lllrd  Quadrant

∴   sin x = $\sqrt 3$ / 2

tan x =  sin x/cos x

=  -√3/2/-1/2 $\frac{-√3/2}{-½}$

=  -√3/2 × 2/-1$\frac{-√3/2}{2/-1}$

tan x  =  [\sqrt{3}\]

Cosec = 1/sin x

= $\frac{1}{-\sqrt 3 / 2}$

Cosec = $\frac{-2}{\sqrt 3}$

Sec x = 1/cosec x

= 1/-1/2 $\frac{1}{-1/2}$

= $\frac{-2}{1}$

Sec x = -2

Cot x = 1/tan x

Cot x = $\frac{1}{\sqrt 3}$

Question 2.  If cos = 4/5 and cos ∅ = 12/13, where and ∅ both lie in the fourth quadrant, find the values of

(i)  cos (θ + ∅ ),          (ii)   sin (θ - ∅),           (iii)    tan (θ + ∅ ).

Solution:       Given:  cos θ= 4/5 and  cos ∅ = 12/13

Since θ lies in the fourth quadrant, we have

cos θ > 0, sin θ < 0 and tan θ < 0.

Again, since ∅ lies in the fourth quadrant, we have

cos ∅  > 0,  sin ∅  < 0  and  tan ∅  < 0.

Now,  sin θ = $\sqrt {1-cos2θ}$ = -$\sqrt{1-\frac{16}{25}}$ = -$\sqrt{\frac{9}{25}}$ = $\frac{3}{5}$

and tan θ = sin θ/cos θ = (-3/5) 5/4 = -3/4

Also, sin ∅ = - $\sqrt{1-cos∅}$ = - $\sqrt{1-\frac{144}{169}}$ = - $\sqrt{\frac{25}{169}}$ = -$\frac{5}{13}$

and tan ∅  = sin ∅/cos ∅ = (-5/13) 13/12 = -5/12

∴  (i)  cos ( + ∅ )  =  cos cos ∅ - sin sin ∅

=  (4/5 12/13) - {(-3/5) (-5/13)}

=  (48/65 - 15/65)  =  33/65

(ii)  sin ( - ∅)  =  sin cos ∅ - sin sin ∅

=  {(-3/5) (12/13)} - {4/5 (-5/13)}

=  (-36/65 + 20/65)  =  -16/65

(iii)   tan ( + ∅ )  =  tan + tan ∅ /1 - tan tan ∅

=  (-3/4) + (-5/12)/1 - {(-3/4) (-5/12)}

=  (-7/6)/1 - 5/16)  =  (-7/6) 16/11

=   -56/33

Question 1 What are the Trigonometric Values?

Ans. The trigonometric values are:

• Sin = Perpendicular/ Hypotenuse

• Cos = Base/ Perpendicular

• Tan = Hypotenuse/ Perpendicular

• Cosec = Hypotenuse/ Perpendicular

• Sec = Hypotenuse/ Base

• Cot = Base/ Hypotenuse

Question 2.  What is the Sine and Cosine Rule for Angles?

Ans. The sine rule is used when we have given either

1.  Two angles and one side
2.  Two sides and non included angle

The cosine rule is used when we have

(a)   Three sides

(b)   Two sides and the included angle