We will learn how to solve different types of problems on trigonometric ratios of an angle.
1. Which of the six trigonometric function are positive for x = -10π/3?
Solution:
Given, x = -10π/3
We know that terminal position of x + 2nπ, where n ∈ Z, is the same as that of x.
Here, -10π/3 + 2 × 2π = 2π/3, which lies in the second quadrant.
Note: This process of finding a co-terminal angle or reference number results in an angle or number α, 0 ≤ α < 2π, so that we can determine in which quadrant the given angle or number lies.
Therefore, x = -10π/3 lies in the second quadrant.
Hence, sin x and csc x are positive while the other four trigonometric functions i.e. cos x, tan x, cot x and sec x are negative.
2. Express cos (- 1555°) in terms of the ratio of a positive
angle less than 30°.
Solution:
cos(- 1555°) = cos 1555°, since we know cos (- θ) = cos θ]
= cos (17 × 90° + 25°)
= - sin 25°; since the angle 1555° lies in the second
d quadrant and cos ratio is negative in this quadrant. Again, in the angle 1555° = 17 × 90° + 25°, multiplier
of 90° is 17, which is an odd integer ; for this reason cos ratio has changed
to sin.
Note: The trigonometrical ratio of an angle of any magnitude can always be expressed in terms of ratio of a positive angle less than 30°.
3. If θ = 170° find the sign of (sin θ + cos θ)
Solution:
sin θ = sin 170° = sin (2 × 90° - 10°) = sin 10°
and cos θ = cos 170° =
cos (1 × 90° + 80°)= - sin 80°
Therefore, sin θ + cos θ = sin 10° - sin 80°
Since sin 10° > 0, sin 80° > 0 and sin 80°
> sin 10°, thus sin 10° - sin 80° < 0 (i.e. negative) so, the value of (sin θ +
cos θ) is negative.
4. Find the value of cos
200° sin 160° + sin (- 340°) cos (- 380°).
Solution:
Given, cos 200° sin 160° + sin
(- 340°) cos (- 380°)
= cos (2 × 90° + 20°) sin (1 × 90° + 70°) + (- sin 340°) cos 380°
= - cos 20° cos 70° - sin (3 × 90° + 70°) cos (4 × 90° + 20°)
= - cos 20° cos 700 - (- cos 70°) cos 20°
= - cos 200 cos 70° + cos 70° cos 20°
= 0
● Trigonometric Functions
11 and 12 Grade Math
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