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
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 7th quadrant i.e. in 3rd
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.
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problems regarding trigonometric ratios of allied or any angles.
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