Chapter 2: Trigonometric and Inverse Trigonometric Functions (Set-2)
For any angle x, the reciprocal of sec x is
A sin x
B tan x
C csc x
D cos x
By definition, sec x = 1/cos x wherever cos x ≠ 0. Therefore its reciprocal is cos x. This simple identity is often used to convert expressions into sine–cosine form.
For any angle x, the reciprocal of csc x is
A cos x
B sin x
C cot x
D sec x
Cosecant is defined as csc x = 1/sin x when sin x ≠ 0. Taking the reciprocal gives sin x. This helps in simplifying and rewriting trigonometric expressions.
If tan x is defined, then cot x equals
A tan x
B sin x
C 1/tan x
D cos x
Cotangent is the reciprocal of tangent for all x where tan x exists. So cot x = 1/tan x. This is useful when converting equations into a single trigonometric function.
If cos x ≠ 0, then tan x · cos x equals
A cos x
B sin x
C 1
D tan x
Using tan x = sin x / cos x, multiplying by cos x cancels the denominator and gives sin x. This identity is handy when simplifying expressions in solving trig equations.
If sin x ≠ 0, then cot x · sin x equals
A cos x
B sin x
C 1
D tan x
Since cot x = cos x / sin x, multiplying by sin x cancels sin x and leaves cos x. Such cancellations quickly simplify many trigonometric proofs and expressions.
Value of cos 90° is
A 1
B −1
C Undefined
D 0
On the unit circle, 90° corresponds to the point (0,1). Cosine is the x-coordinate, which is 0. This is also why sec 90° is undefined.
Value of sin 180° is
A 1
B 0
C −1
D Undefined
At 180° on the unit circle, the point is (−1,0). Sine is the y-coordinate, which is 0. This gives a key zero of the sine graph.
Value of tan 0° is
A 1
B Undefined
C 0
D −1
tan 0° = sin 0° / cos 0°. Since sin 0° = 0 and cos 0° = 1, the ratio is 0. This is a basic starting point for tangent.
Value of tan 60° is
A 1
B √3
C 1/√3
D 0
In a 30°–60°–90° triangle, sides are in ratio 1 : √3 : 2. For 60°, opposite is √3 and adjacent is 1, so tan 60° = √3.
Value of cos 30° is
A 1/2
B 0
C 1
D √3/2
Using the 30°–60°–90° triangle ratios, for 30° the adjacent side is √3 and hypotenuse is 2. So cos 30° = √3/2, a standard exact value.
Value of sin 60° is
A 1/2
B 1
C √3/2
D 0
For a 60° angle in a 30°–60°–90° triangle, opposite is √3 and hypotenuse is 2. Hence sin 60° = √3/2, used frequently in simplifications.
In radians, 180° equals
A 2π
B π
C π/2
D 3π/2
A complete straight angle is 180°, which corresponds to π radians by definition of radian measure. This conversion is essential when working with trig functions in calculus.
In radians, 90° equals
A π/2
B π
C 2π
D π/3
Since 180° equals π radians, half of that angle (90°) equals π/2 radians. This is widely used in principal values and standard trigonometric evaluations.
A full revolution equals
A π radians
B π/2 radians
C 2π radians
D 3π radians
One complete rotation around the unit circle is 360°, which equals 2π radians. This value defines periodicity and is used to write general solutions for trig equations.
If θ is acute, then sin θ is
A Negative
B Positive
C Zero
D Undefined
An acute angle lies between 0° and 90°. On the unit circle, the y-coordinate is positive in the first quadrant, so sine is positive for all acute angles.
If θ is in Quadrant II, then cos θ is
A Positive
B Zero
C Undefined
D Negative
Quadrant II has x-coordinate negative and y-coordinate positive on the unit circle. Since cosine equals the x-coordinate, cos θ is negative for angles in Quadrant II.
If θ is in Quadrant III, then tan θ is
A Negative
B Zero
C Positive
D Undefined
In Quadrant III, both sine and cosine are negative. Tangent equals sin/cos, so the ratio of two negatives is positive. This sign rule helps choose correct solutions.
If θ is in Quadrant IV, then sin θ is
A Positive
B Negative
C Zero
D Undefined
In Quadrant IV, the y-coordinate on the unit circle is negative while x is positive. Since sine equals y-coordinate, sin θ is negative for Quadrant IV angles.
The expression sin x / cos x is undefined when
A cos x = 0
B sin x = 0
C x = 0
D x = π
A fraction becomes undefined when its denominator is zero. Since sin x / cos x has denominator cos x, the expression is undefined exactly at angles where cos x = 0.
Solutions of cos x = 0 are
A nπ
B 2nπ
C π/2 + nπ
D π/3 + nπ
Cosine becomes zero at 90° and 270°, which are π/2 and 3π/2 radians. These repeat every π, giving the general solution x = π/2 + nπ.
Solutions of sin x = 1 are
A 3π/2 + 2nπ
B π/2 + 2nπ
C nπ
D π + 2nπ
Sine reaches 1 at the top of the unit circle, which is angle π/2. Since sine has period 2π, all solutions are x = π/2 + 2nπ.
Solutions of cos x = 1 are
A (2n+1)π
B π/2 + nπ
C π/3 + 2nπ
D 2nπ
Cosine equals 1 at angle 0 (point (1,0)) and repeats every 2π. Therefore the general solution is x = 2nπ, where n is any integer.
Solutions of tan x = 0 are
A π/2 + nπ
B 2nπ + π/2
C nπ
D (2n+1)π/2
Tangent is zero when sine is zero and cosine is nonzero. That occurs at x = 0, π, 2π, etc. Hence x = nπ gives all solutions for tan x = 0.
The equation sin x = −1 has solutions
A π/2 + 2nπ
B 3π/2 + 2nπ
C nπ
D π + 2nπ
Sine equals −1 at the bottom of the unit circle, angle 3π/2. With period 2π, it repeats every full revolution, so x = 3π/2 + 2nπ.
The equation cos x = −1 has solutions
A π + 2nπ
B 2nπ
C π/2 + nπ
D π/3 + 2nπ
Cosine equals −1 at angle π (point (−1,0)). Since cosine repeats every 2π, all solutions are x = π + 2nπ. This is often used in graph-based reasoning.
If sin x = 0 in [0, 2π], then x can be
A π/2, 3π/2
B π/3, 2π/3
C 0, π, 2π
D π/6, 5π/6
In one full cycle from 0 to 2π, sine crosses the x-axis at 0, π, and 2π. These are the only points where the y-coordinate is exactly zero.
If cos x = 0 in [0, 2π], then x can be
A 0, π, 2π
B π/2, 3π/2
C π/3, 2π/3
D π/6, 5π/6
Cosine is the x-coordinate on the unit circle. It becomes zero at points (0,1) and (0,−1), corresponding to π/2 and 3π/2 within [0,2π].
Solve 2sin x cos x = 0 in [0, 2π]
A π/6, 5π/6
B π/3, 2π/3
C π/4, 3π/4
D 0, π/2, π, 3π/2, 2π
2sin x cos x = 0 implies sin x = 0 or cos x = 0. In [0,2π], sin x = 0 at 0, π, 2π and cos x = 0 at π/2, 3π/2.
Solve sin²x = 1 in [0, 2π]
A 0, π, 2π
B π/6, 5π/6
C π/2, 3π/2
D π/3, 2π/3
sin²x = 1 means |sin x| = 1, so sin x = 1 or sin x = −1. In [0,2π], these occur at π/2 and 3π/2 only.
Solve cos²x = 1 in [0, 2π]
A π/2, 3π/2
B 0, π, 2π
C π/6, 5π/6
D π/3, 2π/3
cos²x = 1 means |cos x| = 1, so cos x = 1 or cos x = −1. In [0,2π], that happens at 0, π, and 2π.
Half-angle identity for sin²(x/2) is
A (1−cos x)/2
B (1+cos x)/2
C (1−sin x)/2
D (1+sin x)/2
From cos x = 1 − 2sin²(x/2), rearranging gives sin²(x/2) = (1 − cos x)/2. It helps reduce powers and solve equations involving half angles.
Half-angle identity for cos²(x/2) is
A (1−cos x)/2
B (1+sin x)/2
C (1−sin x)/2
D (1+cos x)/2
Using cos x = 2cos²(x/2) − 1, rearrange to get cos²(x/2) = (1 + cos x)/2. This is commonly used in integration and simplification.
Product-to-sum identity for 2sin A cos B is
A cos(A+B)+cos(A−B)
B sin(A+B)−sin(A−B)
C sin(A+B)+sin(A−B)
D cos(A−B)−cos(A+B)
The identity 2sin A cos B = sin(A+B) + sin(A−B) converts products into sums. It is useful for simplifying expressions and solving equations involving mixed angles.
Product-to-sum identity for 2cos A cos B is
A cos(A+B)−cos(A−B)
B cos(A+B)+cos(A−B)
C sin(A+B)+sin(A−B)
D sin(A+B)−sin(A−B)
The identity 2cos A cos B = cos(A+B) + cos(A−B) rewrites a product into a sum. This helps when comparing expressions or simplifying trig products.
Product-to-sum identity for 2sin A sin B is
A cos(A−B)+cos(A+B)
B sin(A+B)+sin(A−B)
C sin(A+B)−sin(A−B)
D cos(A−B)−cos(A+B)
The identity 2sin A sin B = cos(A−B) − cos(A+B) is derived from cosine addition rules. It is useful in transforming products into sums for simplification.
Convert sin x + sin y into product form
A 2cos((x+y)/2)sin((x−y)/2)
B 2sin((x+y)/2)cos((x−y)/2)
C 2sin((x−y)/2)sin((x+y)/2)
D 2cos((x−y)/2)cos((x+y)/2)
Sum-to-product gives sin x + sin y = 2sin((x+y)/2)cos((x−y)/2). It is helpful when factoring expressions and solving equations by converting sums into products.
Convert cos x + cos y into product form
A 2sin((x+y)/2)sin((x−y)/2)
B 2sin((x+y)/2)cos((x−y)/2)
C 2cos((x+y)/2)cos((x−y)/2)
D 2cos((x+y)/2)sin((x−y)/2)
Sum-to-product states cos x + cos y = 2cos((x+y)/2)cos((x−y)/2). This form is useful to factor and identify zeros of the expression.
Principal value of cos⁻¹(0) is
A 0
B π
C 3π/2
D π/2
arccos(0) is the angle in the principal range [0,π] whose cosine is 0. Cosine becomes 0 at π/2 within that interval, so the principal value is π/2.
Principal value of sin⁻¹(0) is
A π/2
B 0
C π
D −π
arcsin(0) is the angle in the principal range [−π/2, π/2] with sine 0. The angle 0 satisfies this and lies in the range, so it is the principal value.
Principal value of tan⁻¹(1) is
A π/2
B π/3
C π/4
D 3π/4
arctan(1) is the angle in (−π/2, π/2) whose tangent equals 1. tan(π/4)=1, and π/4 lies in the principal interval, so it is the answer.
Principal value of tan⁻¹(−1) is
A −π/4
B π/4
C −π/2
D 3π/4
arctan(−1) must lie in the principal interval (−π/2, π/2). Since tan(−π/4)=−1 and −π/4 is in the interval, it is the principal value.
For x in [−1,1], simplify tan(sin⁻¹x)
A √(1−x²)/x
B x/√(1−x²)
C √(1+x²)
D x√(1−x²)
Let θ = sin⁻¹x, so sin θ = x and θ is in [−π/2, π/2]. Then cos θ = √(1−x²). Hence tan θ = sin θ / cos θ = x/√(1−x²).
For x in [−1,1], simplify cos(tan⁻¹x)
A x/√(1+x²)
B √(1+x²)
C √(1−x²)
D 1/√(1+x²)
If θ = tan⁻¹x, then tan θ = x with θ in (−π/2, π/2). Take a right triangle with opposite x and adjacent 1, hypotenuse √(1+x²). So cos θ = 1/√(1+x²).
For x in [−1,1], simplify sin(tan⁻¹x)
A 1/√(1+x²)
B √(1+x²)
C x/√(1+x²)
D √(1−x²)
With θ = tan⁻¹x, tan θ = x = opposite/adjacent. Use opposite = x, adjacent = 1, so hypotenuse is √(1+x²). Then sin θ = opposite/hypotenuse = x/√(1+x²).
A correct inverse property is
A sin(sin⁻¹x)=x for |x|≤1
B sin⁻¹(sin x)=x for all x
C cos⁻¹(cos x)=x for all x
D tan⁻¹(tan x)=x for all x
sin⁻¹x is defined only for x in [−1,1], and it returns an angle in the principal range. Applying sin to that principal angle gives back the original x.
When can sin⁻¹(sin x) equal x
A x in [0, π]
B x in [−π/2, π/2]
C x in (−π, π]
D x in [0, 2π)
arcsin returns values only in [−π/2, π/2]. So sin⁻¹(sin x) equals x only when x already lies in that interval. Outside it, the principal value differs.
Derivative of sin⁻¹x is
A 1/√(1+x²)
B √(1−x²)
C √(1+x²)
D 1/√(1−x²)
The standard calculus result is d/dx(arcsin x) = 1/√(1−x²), valid for |x|<1. It comes from implicit differentiation of sin y = x and solving for dy/dx.
Derivative of tan⁻¹x is
A 1/(1−x²)
B √(1+x²)
C 1/(1+x²)
D 1/√(1+x²)
A known derivative formula is d/dx(arctan x) = 1/(1+x²) for all real x. It follows from differentiating tan y = x and using sec²y = 1+tan²y.
Solve the inequality sin x ≥ 0 for x in [0, 2π]
A [π, 2π]
B [0, π]
C [0, π/2]
D [π/2, 3π/2]
On the unit circle, sin x is the y-coordinate. It is nonnegative in Quadrants I and II and equals zero at 0 and π. Thus sin x ≥ 0 on [0, π] within [0,2π].
For a right triangle with hypotenuse 10 and angle θ, if sin θ = 3/5 then opposite side is
A 6
B 8
C 4
D 10
sin θ = opposite/hypotenuse. With sin θ = 3/5 and hypotenuse 10, opposite = (3/5)×10 = 6. This is a basic application of trig ratios in triangles