Trigonometric Ratios of an Angle

We will learn how to find the values of trigonometric ratios of an angle. The questions are related to find the values of trigonometric functions of a real number x (i.e., sin x, cos x, tan x, etc.) at any values of x.

1. Find the values of cos (\(\frac{-11\Pi}{3}\))

Solution:

cos (\(\frac{-11\Pi}{3}\)) = cos (\(\frac{11\Pi}{3}\)), since cos (- θ) = cos θ

= cos (\(\frac{11 × 180°}{3}\))

= cos (\(\frac{1980°}{3}\))

= cos 660°

= cos (7 × 90° + 30°)

= sin 30°, [Since the angle 660° lies in the 4th quadrant and cos ratio is positive in this quadrant. Again, in  the angle 660° = 7 × 90° + 30°, multiplier of 90° is 7, which is an odd integer ; for this reason cos ratio has changed to sin.]

= 1/2

 

2. Find the values of cot (- 855°)

Solution:

cot (- 855°) = - cot 855° [since, cot (-θ) = - cot θ]

= - cot (9  × 90° + 45°)

= - (- tan 45°) [Since the angle 855° = 9  ×  90° + 45° lies in the second quadrant and only sin and csc ratios are positive in the second quadrant, thus cot ratio has become negative.  Again, in 855° = 9 x 90° + 45°, the number 9 i.e., an odd integer appears as a multiplier of 90°; for this reason cot ratio has changed to tan.]

= tan 45°

= 1.

 

3. Find the values of csc (-1650°)

Solution:

csc (-1650°) =  - csc 1650°, [since,  csc (-θ) = - csc θ]

= - csc (18 × 90° + 30°)

= - (- csc 30°), [Since, the angle 1650° lies in the 3th quadrant and csc ratio is negative in this quadrant. Again, in 1650° = 18 × 90° + 30°, multiplier of 90° is 18, which is an even integer; for this reason csc ratio remains unaltered.]

= csc 30°

= 2

 

4. If sin 49° = 3/4, find the value of sin 581°.                              

Solution:

sin 581° = sin (7 × 90° - 49°)

= - cos 49°, [Since the angle 581° = 7 × 90° - 49° lies in the 3rd quadrant and only tan and cot ratios are positive in the 3rd quadrant, thus sin ratio has become negative.  Again, in 581° = 7 × 90° - 49°, the number 7 i.e., an odd integer appears as a multiplier of 90°; for this reason sin ratio has changed to cos.]

= - √(1- sin\(^{2}\) 49°)

= - \(\sqrt{1 - (\frac{3}{4})^{2}}\)

= = - \(\sqrt{1 - \frac{9}{16}}\)

= - \(\sqrt{\frac{16 - 9}{16}}\), [since,  sin 49° = ¾]

= \(\frac{√7}{4}\)





11 and 12 Grade Math

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