Chapter 2: Trigonometric and Inverse Trigonometric Functions (Set-3)
If sin x = 3/5 and x is in Quadrant I, then cos x equals
A 4/5
B −4/5
C 3/5
D −3/5
Using sin²x + cos²x = 1, we get cos²x = 1 − (3/5)² = 16/25, so cos x = ±4/5. In Quadrant I, cosine is positive, so cos x = 4/5.
If cos x = −5/13 and x is in Quadrant II, then sin x equals
A −12/13
B 5/13
C 12/13
D −5/13
From sin²x + cos²x = 1, sin²x = 1 − (25/169) = 144/169, so sin x = ±12/13. In Quadrant II, sine is positive, so sin x = 12/13.
If tan x = −3/4 and x is in Quadrant IV, then sin x equals
A 3/5
B −4/5
C 4/5
D −3/5
Let opposite = −3 and adjacent = 4 (Quadrant IV makes sine negative, cosine positive). Hypotenuse = 5. Then sin x = opposite/hypotenuse = −3/5.
If sin x = −1/2 and x is in Quadrant IV, then cos x equals
A √3/2
B −√3/2
C 1/2
D −1/2
If sin x = −1/2, the reference angle is 30°. In Quadrant IV, sine is negative and cosine is positive. Therefore cos x = cos 30° = √3/2.
Evaluate sin(150°)
A √3/2
B −1/2
C 1/2
D −√3/2
150° = 180° − 30°. Using sin(180° − θ) = sin θ, sin 150° = sin 30° = 1/2. Sine is positive in Quadrant II, so the value stays positive.
Evaluate cos(150°)
A −√3/2
B √3/2
C −1/2
D 1/2
150° = 180° − 30°. Using cos(180° − θ) = −cos θ, cos 150° = −cos 30° = −√3/2. Cosine is negative in Quadrant II.
Evaluate tan(135°)
A 1
B √3
C −1
D 0
135° = 180° − 45°. Since tan(180° − θ) = −tan θ and tan 45° = 1, we get tan 135° = −1. Tangent is negative in Quadrant II.
Evaluate sin(240°)
A √3/2
B −√3/2
C −1/2
D 1/2
240° = 180° + 60°. Using sin(180° + θ) = −sin θ, sin 240° = −sin 60° = −√3/2. Sine is negative in Quadrant III.
Evaluate cos(300°)
A −1/2
B √3/2
C −√3/2
D 1/2
300° = 360° − 60°. Using cos(360° − θ) = cos θ, cos 300° = cos 60° = 1/2? Wait: cos 60° is 1/2. So cos 300° = 1/2.
The correct value of cos(300°) is
A √3/2
B −1/2
C 1/2
D −√3/2
300° = 360° − 60°. Cosine is positive in Quadrant IV, and cos(360° − θ) = cos θ. Therefore cos 300° = cos 60° = 1/2.
Evaluate sin(210°)
A −1/2
B 1/2
C −√3/2
D √3/2
210° = 180° + 30°. Using sin(180° + θ) = −sin θ, sin 210° = −sin 30° = −1/2. Quadrant III makes sine negative.
If sin x = cos x and x is acute, then x equals
A 30°
B 60°
C 90°
D 45°
sin x = cos x implies tan x = 1 (since cos x ≠ 0 for acute x). The acute angle with tan x = 1 is 45°. This comes from standard triangle ratios.
If sin x = 0 and cos x = −1, then x equals
A 0
B π/2
C π
D 3π/2
sin x = 0 occurs at x = nπ. At x = π, cos π = −1. So the angle matching both conditions is x = π within standard principal angles.
If cos x = 0 and sin x = −1, then x equals
A π/2
B 3π/2
C π
D 0
cos x = 0 at π/2 and 3π/2. Among these, sin(π/2)=1 while sin(3π/2)=−1. Hence x = 3π/2 satisfies both conditions.
Solve sin x = 1/2 for x in [0, 2π]
A π/3, 2π/3
B π/4, 3π/4
C π/6, 5π/6
D π/2, 3π/2
sin x = 1/2 occurs at reference angle 30° which is π/6. Sine is positive in Quadrants I and II, so solutions are π/6 and π − π/6 = 5π/6.
Solve cos x = 1/2 for x in [0, 2π]
A π/3, 5π/3
B π/6, 11π/6
C 2π/3, 4π/3
D π/2, 3π/2
cos x = 1/2 has reference angle 60° which is π/3. Cosine is positive in Quadrants I and IV, giving x = π/3 and x = 2π − π/3 = 5π/3.
Solve tan x = √3 for x in [0, 2π]
A π/6, 7π/6
B 2π/3, 5π/3
C π/4, 5π/4
D π/3, 4π/3
tan x = √3 at reference angle 60° which is π/3. Tangent is positive in Quadrants I and III. Hence solutions are π/3 and π + π/3 = 4π/3.
Solve tan x = −1 for x in [0, 2π]
A π/4, 5π/4
B π/3, 4π/3
C 3π/4, 7π/4
D π/6, 5π/6
tan x = −1 has reference angle 45° (π/4). Tangent is negative in Quadrants II and IV, so x = π − π/4 = 3π/4 and x = 2π − π/4 = 7π/4.
The general solution of sin x = 0 is
A nπ
B π/2 + nπ
C 2nπ + π/2
D (2n+1)π/2
Sine is zero at 0, π, 2π, and so on. These are exactly integer multiples of π. Hence the general solution is x = nπ where n is any integer.
The general solution of cos x = 0 is
A nπ
B 2nπ
C π/2 + nπ
D π + 2nπ
Cosine becomes zero at π/2 and 3π/2, repeating every π. Therefore x = π/2 + nπ captures all such angles where the unit circle x-coordinate is zero.
Solve 2cos²x − 1 = 0 for x in [0, 2π]
A π/6, 5π/6
B π/3, 2π/3
C 0, π, 2π
D π/4, 3π/4, 5π/4, 7π/4
2cos²x − 1 = 0 gives cos²x = 1/2, so cos x = ±1/√2 = ±√2/2. In [0,2π], this occurs at x = π/4, 3π/4, 5π/4, 7π/4.
Solve 1 − 2sin²x = 0 for x in [0, 2π]
A π/6, 5π/6
B π/3, 2π/3
C π/4, 3π/4, 5π/4, 7π/4
D 0, π, 2π
1 − 2sin²x = 0 implies sin²x = 1/2, so sin x = ±√2/2. In [0,2π], sine equals ±√2/2 at π/4, 3π/4, 5π/4, and 7π/4.
If sin 2x = 0, then a correct general solution is
A x = nπ/2
B x = nπ
C x = π/2 + nπ
D x = (2n+1)π/2
sin 2x = 0 means 2x = nπ, because sine is zero at multiples of π. Dividing by 2 gives x = nπ/2. This covers all solutions.
Solve cos 2x = 1 for x in [0, 2π]
A π/2, 3π/2
B π/4, 3π/4
C 0, π, 2π
D π/3, 5π/3
cos 2x = 1 occurs when 2x = 2nπ. So x = nπ. Within [0,2π], n = 0,1,2 give x = 0, π, 2π.
Solve cos 2x = −1 for x in [0, 2π]
A 0, π, 2π
B π/2, 3π/2
C π/4, 3π/4
D π/3, 5π/3
cos 2x = −1 happens when 2x = (2n+1)π. Thus x = (2n+1)π/2. In [0,2π], valid solutions are x = π/2 and 3π/2.
If tan x = sin x, then one valid solution set includes
A x = π/2 + nπ
B x = π/4 + nπ
C x = π/3 + nπ
D x = nπ
tan x = sin x means sin x / cos x = sin x. For cos x ≠ 0, either sin x = 0 or 1/cos x = 1. sin x = 0 gives x = nπ (valid and simplest).
If cos x = sin x and x is in Quadrant II, then x equals
A π/4
B 5π/4
C 3π/4
D 7π/4
cos x = sin x implies tan x = 1, so x = π/4 + nπ. In Quadrant II, the angle with reference angle π/4 is π − π/4 = 3π/4, which satisfies cos x = sin x.
Simplify sin⁻¹(1/2) in principal range
A π/6
B π/3
C π/2
D 0
arcsin(1/2) is the angle in [−π/2, π/2] whose sine is 1/2. Since sin(π/6)=1/2 and π/6 lies in the principal range, the value is π/6.
Simplify cos⁻¹(1/2) in principal range
A π/6
B 2π/3
C π/3
D π/2
arccos(1/2) is the angle in [0,π] with cosine 1/2. cos(π/3)=1/2 and π/3 lies in the principal range, so cos⁻¹(1/2)=π/3.
Principal value of sin⁻¹(−1) is
A π/2
B −π/2
C π
D 0
arcsin returns angles in [−π/2, π/2]. The angle in this interval with sine −1 is −π/2. That is the unique principal value for sin⁻¹(−1).
Principal value of cos⁻¹(−1) is
A 0
B π/2
C π
D 3π/2
arccos returns values in [0,π]. Cosine equals −1 at angle π in this interval. Therefore the principal value cos⁻¹(−1) is π.
Evaluate tan⁻¹(√3) in principal range
A π/3
B π/6
C π/2
D 2π/3
arctan gives values in (−π/2, π/2). Since tan(π/3)=√3 and π/3 lies in the principal interval, tan⁻¹(√3)=π/3.
Evaluate tan⁻¹(−√3) in principal range
A π/3
B −2π/3
C −π/3
D π/6
arctan outputs lie in (−π/2, π/2). Tangent of −π/3 equals −√3, and −π/3 is inside the principal range. So tan⁻¹(−√3)=−π/3.
Simplify cos(sin⁻¹(−3/5))
A −4/5
B 3/5
C −3/5
D 4/5
Let θ = sin⁻¹(−3/5). Then θ lies in [−π/2, π/2], where cosine is nonnegative. Using cos θ = √(1−sin²θ) gives √(1−9/25)=√(16/25)=4/5.
Simplify sin(cos⁻¹(3/5))
A 4/5
B −4/5
C 3/5
D −3/5
Let θ = cos⁻¹(3/5), so θ is in [0,π] and cos θ = 3/5. Then sin θ = √(1−cos²θ)=√(1−9/25)=4/5. Sine is nonnegative on [0,π].
Simplify tan(cos⁻¹(4/5))
A 4/3
B 5/4
C 3/4
D 4/5
Let θ = cos⁻¹(4/5). In [0,π], cos θ = 4/5, so sin θ = √(1−16/25)=3/5 (nonnegative). Thus tan θ = sin θ / cos θ = (3/5)/(4/5)=3/4.
The value of sin⁻¹x + cos⁻¹x equals
A π
B π/2
C 0
D 2π
For x in [−1,1], let θ = sin⁻¹x, so sin θ = x with θ in [−π/2, π/2]. Then cos(π/2 − θ) = sin θ = x, so cos⁻¹x = π/2 − θ, giving the sum π/2.
If sin⁻¹x = π/6, then x equals
A √3/2
B −1/2
C 1/2
D 0
sin⁻¹x = π/6 means x is the sine of π/6. Since sin(π/6)=1/2 and π/6 lies in the principal range of arcsin, x must be 1/2.
If cos⁻¹x = 2π/3, then x equals
A 1/2
B −√3/2
C √3/2
D −1/2
cos⁻¹x = 2π/3 means x = cos(2π/3). Since cos(120°)=−1/2 and 2π/3 is within [0,π], the inverse definition is satisfied, so x = −1/2.
If tan⁻¹x = −π/4, then x equals
A 1
B 0
C −1
D √3
tan⁻¹x = −π/4 means x = tan(−π/4). Since tan is odd and tan(π/4)=1, we get tan(−π/4)=−1. This lies in the principal range of arctan.
Solve sin x = cos x in [0, 2π]
A π/4, 3π/4
B π/4, 5π/4
C 3π/4, 7π/4
D π/2, 3π/2
sin x = cos x implies tan x = 1 (where cos x ≠ 0). So x = π/4 + nπ. In [0,2π], the solutions are π/4 and π/4 + π = 5π/4.
Solve sin x = −cos x in [0, 2π]
A π/4, 5π/4
B π/2, 3π/2
C 3π/4, 7π/4
D π/6, 5π/6
sin x = −cos x gives tan x = −1 when cos x ≠ 0. So x = −π/4 + nπ. In [0,2π], this corresponds to 3π/4 and 7π/4.
If sec x = 2 and x is acute, then cos x equals
A 1/2
B 2
C √3/2
D −1/2
sec x = 1/cos x. If sec x = 2, then cos x = 1/2. Since x is acute, cosine is positive, so cos x = 1/2 is consistent.
If csc x = −2 and x is in Quadrant III, then sin x equals
A 1/2
B −2
C −1/2
D 2
csc x = 1/sin x, so sin x = 1/csc x. With csc x = −2, sin x = −1/2. In Quadrant III, sine is negative, matching the sign.
If cot x = 1 and x is in Quadrant I, then x equals
A π/3
B π/6
C π/2
D π/4
cot x = 1 means tan x = 1 as well, because cot x = 1/tan x. In Quadrant I, the angle with tan x = 1 is π/4 (45°).
The amplitude of y = 3sin x is
A 1/3
B 3
C 2π
D π
In y = A sin x, the amplitude is |A|, the maximum distance from the midline. Here A = 3, so amplitude is 3. This affects vertical stretch, not period.
The period of y = sin(2x) is
A 2π
B π/2
C π
D 4π
For y = sin(kx), period = 2π/|k|. Here k = 2, so period = 2π/2 = π. The graph completes one full cycle in π radians.
The period of y = cos(x/2) is
A 4π
B 2π
C π
D π/2
For y = cos(kx), period is 2π/|k|. Here k = 1/2, so period = 2π ÷ (1/2) = 4π. The cosine wave stretches horizontally.
If tan x = √3 and 0 < x < π/2, then x equals
A π/6
B π/4
C π/3
D π/2
In the first quadrant, tan x = √3 corresponds to the standard angle 60°, which is π/3. The interval 0 < x < π/2 confirms the acute principal solution only.
If sin x = −√3/2 and π < x < 2π, then x can be
A π/3, 2π/3
B π/6, 5π/6
C 0, π
D 4π/3, 5π/3
sin x = −√3/2 has reference angle π/3. In (π,2π), sine is negative in Quadrants III and IV, giving x = π + π/3 = 4π/3 and x = 2π − π/3 = 5π