Trigonometric Ratios of Allied Angles

Trigonometric Ratios of Allied Angles

We have to remember the values of trigonometric ratios of standard angles like 0°, 30°, 45°, 60°, 90° etc. And, We can establish the relation of the trigonometric ratios of allied angles like 90° + θ, 90° - θ, 180° + θ, 180° - θ, 270° + θ etc. in terms of trigonometric ratios of angle θ. This can be easily done and understood by applying the concept of reflection and rotation of a point on co-ordinate plane.

 

Trigonometric Ratios of Angle θ

Consider an angle θ placed in the standard position. Draw circle with centre at O and radius r.

Let the circle intersect the terminal line at some point P(x, y) as shown in the figure. Draw perpendicular PM from P to the x-axis. Then MP = y, OM = x and OP = r. For the angle θ, the six trigonometric ratios can be defined by the following formula:

sinθ = y/r            cosθ = x/r            tanθ = y/x

cosecθ = r/y        secθ = r/x            cotθ = x/y

 

Trigonometric Ratios of (90° – θ)

Reflect the point P(x, y) in the line y = x. Then the image of P(x, y) under this reflection is P’(y, x). Join OP’. Then ∠XOP’ = 90° - θ.

Now,  

        sin(90° - θ) = x/r = cosθ

        cosec(90° - θ) = r/x = secθ

        cos(90° - θ) = y/r = sinθ

        sec(90° - θ) = r/y = cosecθ

        tan(90° - θ) = x/y = cotθ

        cot(90° - θ) = y/x = tanθ

                               

Trigonometric Ratios of (90° + θ)     

Rotate the point P(x, y) about origin through 90° in anticlockwise (positive) direction. Then the image of P(x, y) under this rotation is P’(-y, x). Join OP’. Then ∠XOP’ = 90° + θ.

Now,  

        sin(90° + θ) = x/r = cosθ

        cosec(90° + θ) = r/x = secθ

        cos(90° + θ) = -y/r = - sinθ

        sec(90° + θ) = r/-y = - cosecθ

        tan(90° + θ) = x/-y = - cotθ

        cot(90° + θ) = -y/x = - tanθ

 

Trigonometric Ratios of (180° – θ)

Reflect the point P(x, y) in the line Y-axis. Then the image of P(x, y) under this reflection is P’(-x, y). Join OP’. Then ∠XOP’ = 180° - θ.

Now,  

        sin(180° - θ) = y/r = sinθ

        cosec(180° - θ) = r/y = cosecθ

        cos(180° - θ) = -x/r = - cosθ

        sec(180° - θ) = r/-x = - secθ

        tan(180° - θ) = y/-x = - tanθ

        cot(180° - θ) = -x/y = - cotθ

 

Trigonometric Ratios of (180° + θ)

Rotate the point P(x, y) about origin through 180° in anticlockwise (positive) direction. Then the image of P(x, y) under this rotation is P’(-x, -y). Join OP’. Then ∠XOP’ = 180° + θ.

Now,  

        sin(180° + θ) = -y/r = - sinθ

        cosec(180° + θ) = r/-y = - cosecθ

        cos(180° + θ) = -x/r = - cosθ

        sec(180° + θ) = r/-x = - secθ

        tan(180° + θ) = -y/-x = tanθ

        cot(180° + θ) = -x/-y = cotθ

 

Trigonometric Ratios of (270° – θ)

Reflect the point P(x, y) in the line y = - x. Then the image of P(x, y) under this reflection is P’(- y, - x). Join OP’. Then ∠XOP’ = 270° - θ.

Now,  

        sin(270° - θ) = -x/r = - cosθ

        cosec(270° - θ) = r/-x = - secθ

        cos(270° - θ) = -y/r = - sinθ

        sec(270° - θ) = r/-y = - cosecθ

        tan(270° - θ) = -x/-y = cotθ

        cot(270° - θ) = -y/-x = tanθ

 

Trigonometric Ratios of (270° + θ)   

Rotate the point P(x, y) about origin through 270° in anticlockwise (positive) direction. Then the image of P(x, y) under this rotation is P’(y, - x). Join OP’. Then ∠XOP’ = 270° + θ.

Now,  

        sin(270° + θ) = -x/r = - cosθ

        cosec(270° + θ) = r/-x = - secθ

        cos(270° + θ) = y/r = sinθ

        sec(270° + θ) = r/y = cosecθ

        tan(270° + θ) = -x/y = - cotθ

        cot(270° + θ) = y/-x = - tanθ

 

Trigonometric Ratios of (360° – θ) or (– θ)

Reflect the point P(x, y) in the line X-axis. Then the image of P(x, y) under this reflection is P’(x, - y). Join OP’. Then ∠XOP’ = 360° - θ. Also, ∠XOP’ in clockwise direction = – θ

Now,   

        sin(180° - θ) = sin(– θ) = -y/r = - sinθ

        cosec(180° - θ) = cosec(– θ) = r/-y = - cosecθ

        cos(180° - θ) = cos(– θ) = x/r = cosθ

        sec(180° - θ) = sec(– θ) = r/x = secθ

        tan(180° - θ) = tan(– θ) = -y/x = - tanθ

        cot(180° - θ) = cot(– θ) = x/-y = - cotθ

 

Signs of the Trigonometric Ratios

We have seen that the trigonometric ratios has different signs in different quadrants. The following table shows the sign of all the trigonometrc ratios of angles in different quadrants.

The sign of the trigonometric ratios of an angle of any magnitude can be remembered from the diagram as shown below.

 

Trigonometric Ratios of any angle

Any allied angle can be expressed in the form (n × 90° ± θ) where n is an integer. We can change the trigonometric ratios of the angle (n × 90° ± θ) into the trigonometric ratio of angle θ.

(i)          If n is even, there will be no change in the trigonometric ratios. i.e.

sin(n × 90° ± θ) ⇒ sinθ

cos(n × 90° ± θ) ⇒ cosθ

etc.

(ii)         If n is odd, then the trigonometric ratios change as follows:

sin(n × 90° ± θ) ⇒ cosθ

cosec(n × 90° ± θ) ⇒ secθ

cos(n × 90° ± θ) ⇒ sinθ

sec(n × 90° ± θ) ⇒ cosecθ

tan(n × 90° ± θ) ⇒ cotθ

cot(n × 90° ± θ) ⇒ tanθ

(iii)        The sign of the trigonometric ratio of the angle (n × 90° ± θ) is determined by taking into consideration that in which quadrant the angle (n × 90° ± θ) lies.

For example: Find the ratio of sin570°.

Solution:

Here, 570° = 6 × 90° + 30°. Therefore 570° lies in 7^th quadrant i.e. in 3^rd quadrant where the ratio of sin will be negative. And, n = 6 is an even number, so there will be no change in trigonometric ratio.

         ∴  sin570° = sin(6 × 90° + 30°) = - sin30° = - ½

 

Trigonometric Ratios of Higher Standard Angles

Trigonometric Ratios of 120°

sin120° = sin(1 × 90° + 30°) = cos30° = √3/2

cosec120° = cosec(1 × 90° + 30°) = sec30° = 2/√3

cos120° = cos(1 × 90° + 30°) = - sin30° = - ½

sec120° = sec(1 × 90° + 30°) = - cosec30° = - 2

tan120° = tan(1 × 90° + 30°) = - cot30° = - √3

cot120° = cot(1 × 90° + 30°) = - tan30° = - 1/√3

 

Trigonometric Ratios of 135°

sin135° = sin(1 × 90° + 45°) = cos45° = 1/√2

cosec135° = cosec(1 × 90° + 45°) = sec45° = √2

cos135° = cos(1 × 90° + 45°) = - sin45° = - 1/√2

sec135° = sec(1 × 90° + 45°) = - cosec45° = - √2

tan135° = tan(1 × 90° + 45°) = - cot45° = - 1

cot135° = cot(1 × 90° + 45°) = - tan45° = - 1

 

Trigonometric Ratios of 150°

sin150° = sin(1 × 90° + 60°) = cos60° = 1/2

cosec150° = cosec(1 × 90° + 60°) = sec60° = 2

cos150° = cos(1 × 90° + 60°) = - sin60° = - √3/2

sec150° = sec(1 × 90° + 60°) = - cosec60° = - 2/√3

tan150° = tan(1 × 90° + 60°) = - cot60° = - 1/√3

cot150° = cot(1 × 90° + 60°) = - tan60° = - √3

 

Trigonometric Ratios of 180°

sin180° = sin(1 × 90° + 90°) = cos90° = 0

cosec180° = cosec(1 × 90° + 90°) = sec90° = ∞

cos180° = cos(1 × 90° + 90°) = - sin90° = - 1

sec180° = sec(1 × 90° + 90°) = - cosec90° = - 1

tan180° = tan(1 × 90° + 90°) = - cot90° = 0

cot180° = cot(1 × 90° + 90°) = - tan90° = ∞

Trigonometric Ratios of Standard Angles:

Worked Out Examples

Example 1: Find the value of tan855°.

Solution:

          Tan855° = tan(9 × 90° + 45°)

                         = - cot45°

                         = - 1

 

Example 2: Prove that: tan25° + sin35° + sec40° = cot65° + cos55° + cosec50°

Solution:

L.H.S. = tan25° + sin35° + sec40°

           = tan(90° - 65°) + sin(90° - 55°) + sec(90° - 50°)

           = cot65° + cos55° + cosec50°

           = R.H.S.  Proved.

 

Example 3: Find the value of A if sin3A = cosA.

Solution:

          sin3A = cosA

or,     sin3A = sin(90° - A)

∴        3A = 90° - A

or,     3A + A = 90°

or,     4A = 90°

or,     A = 90°/4

or,     A = 22.5°

 

Example 4: Prove that: cos (π/8) + cos (3π/8) + cos(5π/8) + cos(7π/8) = 0

Solution:

L.H.S. = cos (π/8) + cos (3π/8) + cos(5π/8) + cos(7π/8)

           = cos(π - 7π/8) + cos(π - 5π/8) + cos(5π/8) + cos(7π/8)

           = - cos(7π/8) - cos(5π/8) + cos(5π/8) + cos(7π/8)

           = 0

  = R.H.S.  Proved.

 

Example 5: If A, B, C are the angles of a triangle, show that: sin(A+B) = sinC

Solution:

          A + B + C = 180° [Sum of angles of a triangle is 180°]

or,     A + B = 180° - C

or,     sin(A + B) = sin(180° - C) [Putting sin to both side]

or,     sin(A + B) = sinC

Proved.

 

You can comment your questions or problems regarding trigonometric ratios of allied or any angles.