cos2x = 1/2 means 2x = ±π/3 + 2nπ. Dividing by 2 gives x = ±π/6 + nπ. In [0,2π], this gives π/6, 5π/6, 7π/6, 11π/6.
In [0, 2π], solve 2sin²x − sin x − 1 = 0
A π/6, 5π/6
B π/2, 7π/6, 11π/6
C 0, π, 2π
D π/3, 5π/3
Let s = sin x. Then 2s² − s − 1 = 0 ⇒ (2s+1)(s−1)=0, so s=1 or s=−1/2. In [0,2π], sinx=1 at π/2 and sinx=−1/2 at 7π/6, 11π/6.
In [0, 2π], solve 2cos²x + cos x − 1 = 0
A π/6, 5π/6
B 0, π, 2π
C π/4, 3π/4
D π/3, 5π/3, π
Let c = cos x. Then 2c² + c − 1 = 0 ⇒ (2c−1)(c+1)=0, so c=1/2 or c=−1. In [0,2π], cosx=1/2 at π/3, 5π/3 and cosx=−1 at π.
In [0, 2π], solve cos 2x = cos x
A π/2, 3π/2
B π/6, 11π/6
C 0, 2π, 2π/3, 4π/3
D π/3, 5π/3
cos2x = 2cos²x − 1. So 2cos²x − 1 = cosx ⇒ 2c² − c − 1 = 0 ⇒ (2c+1)(c−1)=0. Thus cosx=1 gives 0,2π and cosx=−1/2 gives 2π/3, 4π/3.
In [0, 2π], solve sin x + cos x = 0
A 3π/4, 7π/4
B π/4, 5π/4
C π/6, 5π/6
D π/3, 4π/3
sin x + cos x = 0 implies sin x = −cos x. Dividing by cos x (where cosx ≠ 0) gives tan x = −1. In [0,2π], tanx=−1 at 3π/4 and 7π/4.
Principal value of sin⁻¹(sin 5π/6)
A 5π/6
B π/6
C −π/6
D −5π/6
sin(5π/6)=1/2. arcsin returns a value in [−π/2, π/2]. The angle in that range with sine 1/2 is π/6. So the principal value is π/6.
Principal value of cos⁻¹(cos 4π/3)
A 4π/3
B π/3
C 2π/3
D 5π/3
cos(4π/3)=−1/2. arccos returns values in [0,π]. The angle in [0,π] with cosine −1/2 is 2π/3. Hence the principal value is 2π/3.
Principal value of tan⁻¹(tan 3π/4)
A 3π/4
B π/4
C −3π/4
D −π/4
tan(3π/4)=−1. arctan returns values in (−π/2, π/2). The angle in that range with tangent −1 is −π/4. So the principal value is −π/4.
Evaluate sin⁻¹(√3/2) in principal range
A π/3
B 2π/3
C π/6
D 5π/6
arcsin gives an angle in [−π/2, π/2]. Since sin(π/3)=√3/2 and π/3 lies in the principal interval, sin⁻¹(√3/2)=π/3.
Evaluate cos⁻¹(−√2/2) in principal range
A π/4
B 5π/4
C 3π/4
D 7π/4
arccos returns values in [0,π]. cos(3π/4)=−√2/2 and 3π/4 lies within [0,π]. Therefore cos⁻¹(−√2/2)=3π/4.
Evaluate tan⁻¹(1/√3) in principal range
A π/3
B π/6
C π/4
D π/2
arctan returns values in (−π/2, π/2). Since tan(π/6)=1/√3 and π/6 is in the principal interval, tan⁻¹(1/√3)=π/6.
For x in [−1,1], cos⁻¹(−x) equals
A π − cos⁻¹x
B cos⁻¹x − π
C π + cos⁻¹x
D cos⁻¹(1−x)
Let θ = cos⁻¹x, so θ is in [0,π] and cosθ=x. Then cos(π−θ)=−cosθ=−x. Since π−θ is also in [0,π], cos⁻¹(−x)=π−cos⁻¹x.
For x in [−1,1], sin⁻¹(−x) equals
A π − sin⁻¹x
B π + sin⁻¹x
C −sin⁻¹x
D sin⁻¹x − π
arcsin returns an angle in [−π/2, π/2], where sine is odd. If sinθ=x, then sin(−θ)=−x and −θ is still in the principal range. So sin⁻¹(−x)=−sin⁻¹x.
For x > 0, tan⁻¹x + tan⁻¹(1/x) equals
A π
B 0
C π/4
D π/2
Let θ = tan⁻¹x with x>0, so θ is in (0, π/2). Then tan(π/2−θ)=1/tanθ=1/x. Both angles are in principal ranges, so tan⁻¹(1/x)=π/2−θ, giving the sum π/2.
Evaluate tan⁻¹2 + tan⁻¹3 in principal form
A π/4
B π/2
C 3π/4
D π
Use arctan addition: tan(α+β)=(2+3)/(1−6)=5/−5=−1, so α+β differs from arctan(−1)=−π/4 by π because both angles are positive and 2·3>1. Hence α+β=−π/4+π=3π/4.
The value of tan⁻¹1 + tan⁻¹2 + tan⁻¹3 equals
A π
B π/2
C 3π/4
D π/4
A known identity is arctan1 + arctan2 + arctan3 = π. It follows by combining arctan1 + arctan2 first (adjusting by π because product > 1) and then adding arctan3.
For real x, simplify sec(tan⁻¹x)
A 1/√(1+x²)
B √(1+x²)
C x/√(1+x²)
D √(1−x²)
Let θ = tan⁻¹x, so tanθ = x = opposite/adjacent. Take adjacent=1, opposite=x, then hypotenuse is √(1+x²). secθ = hypotenuse/adjacent = √(1+x²).
For x ≠ 0, simplify csc(tan⁻¹x)
A x/√(1+x²)
B 1/√(1+x²)
C √(1+x²)/x
D √(1+x²)
If θ = tan⁻¹x, take a right triangle with opposite=x, adjacent=1, hypotenuse=√(1+x²). Then sinθ = x/√(1+x²). So cscθ = 1/sinθ = √(1+x²)/x.
For x in [−1,1] and x ≠ 0, simplify tan(cos⁻¹x)
A √(1−x²)/x
B x/√(1−x²)
C √(1+x²)
D 1/√(1−x²)
Let θ = cos⁻¹x, so cosθ = x and θ is in [0,π]. Then sinθ = √(1−x²) (nonnegative). So tanθ = sinθ/cosθ = √(1−x²)/x, valid for x ≠ 0.
For x in [−1,1], simplify sin(2cos⁻¹x)
A 1−2x²
B 2x/(1+x²)
C √(1−x²)
D 2x√(1−x²)
Let θ = cos⁻¹x, so cosθ=x and sinθ=√(1−x²). Then sin(2θ)=2sinθ cosθ = 2(√(1−x²))(x) = 2x√(1−x²). This ensures the correct sign using principal θ.
For x in [−1,1], simplify cos(2sin⁻¹x)
A 2x√(1−x²)
B 1+2x²
C 1−2x²
D 2x²−1
Let θ = sin⁻¹x, so sinθ=x. Using cos(2θ)=1−2sin²θ gives cos(2sin⁻¹x)=1−2x². This is a clean way to remove inverse-trig from the expression.
Simplify sec²x − tan²x where defined
A 0
B 1
C sec x
D tan x
From the identity sec²x = 1 + tan²x, rearrange to get sec²x − tan²x = 1. This relationship often helps reduce complicated expressions to constants.
Express sin x cos x using a double-angle form
A (1/2)sin 2x
B sin 2x
C (1/2)cos 2x
D cos 2x
The double-angle identity says sin2x = 2sinx cosx. Dividing both sides by 2 gives sinx cosx = (1/2)sin2x. This is helpful in simplifying products.
Identify the simplest equivalent of cos²x − sin²x
A sin 2x
B 1 − sin²x
C cos 2x
D 1 − cos²x
One standard double-angle identity is cos2x = cos²x − sin²x. It is derived from cosine addition rules and is used to convert squared terms into a single cosine.
Solve tan²x − 3 = 0 in [0, 2π]
A π/6, 5π/6
B 0, π, 2π
C π/4, 3π/4
D π/3, 2π/3, 4π/3, 5π/3
tan²x − 3 = 0 gives tanx = ±√3. tanx=√3 at π/3 and 4π/3; tanx=−√3 at 2π/3 and 5π/3. These are all within [0,2π].
In [0, 2π], solve 4cos²x − 3 = 0
A π/4, 3π/4
B π/6, 5π/6, 7π/6, 11π/6
C 0, π, 2π
D π/3, 2π/3
4cos²x − 3 = 0 gives cos²x = 3/4, so cosx = ±√3/2. In [0,2π], cosx=√3/2 at π/6, 11π/6 and cosx=−√3/2 at 5π/6, 7π/6.
In [0, 2π], solve cos x = sin 2x
A π/3, 5π/3
B 0, π, 2π
C π/6, 5π/6, π/2, 3π/2
D π/4, 7π/4
cosx = sin2x = 2sinx cosx. Bring terms: cosx(1−2sinx)=0. So either cosx=0 giving π/2, 3π/2, or sinx=1/2 giving π/6, 5π/6 in [0,2π].
If x is in (π, 2π) and cos x = −1/2, then sin(x/2) equals
A √3/2
B −√3/2
C 1/2
D −1/2
Use sin²(x/2) = (1−cosx)/2. With cosx=−1/2, sin²(x/2)=(1+1/2)/2=3/4, so |sin(x/2)|=√3/2. Since x/2 is in (π/2, π), sine is positive there.
For x = 5π/3, the value of cos(x/2) is
A √3/2
B −1/2
C 1/2
D −√3/2
If x=5π/3, then x/2 = 5π/6 which lies in Quadrant II. cos(5π/6) = −cos(π/6) = −√3/2. The sign is negative because cosine is negative in Quadrant II.
Simplify (sin x + cos x)² into a standard trig form
A 1 − sin 2x
B 1 + cos 2x
C 1 + sin 2x
D 1 − cos 2x
Expand: (sinx+cosx)² = sin²x + cos²x + 2sinx cosx. Since sin²x+cos²x=1 and 2sinx cosx = sin2x, the result is 1 + sin2x.
Simplify (sin x − cos x)² into a standard trig form
A 1 + sin 2x
B 1 − sin 2x
C 1 − cos 2x
D 1 + cos 2x
Expand: (sinx−cosx)² = sin²x + cos²x − 2sinx cosx. The first two add to 1, and 2sinx cosx = sin2x, so the expression becomes 1 − sin2x.
Derivative of cos⁻¹x for |x| < 1 is
A 1/√(1−x²)
B 1/(1+x²)
C −1/√(1−x²)
D −1/(1+x²)
The standard result is d/dx(arccos x) = −1/√(1−x²). It comes from implicit differentiation of cos y = x and solving for dy/dx carefully.
Solve cos x < 0 for x in [0, 2π]
A (π/2, 3π/2)
B (0, π/2)
C (3π/2, 2π)
D (0, 2π)
Cosine is the x-coordinate on the unit circle. It is negative in Quadrants II and III, which corresponds to angles strictly between π/2 and 3π/2. Endpoints are excluded since cos is zero there.
Solve tan x > 0 for x in [0, 2π]
A QII and QIV
B QI and QIII
C QI only
D QIII only
tanx = sinx/cosx is positive when sine and cosine have the same sign. That happens in Quadrant I (both positive) and Quadrant III (both negative). Exclude angles where tan is undefined.
If sin x = 2cos x and 0 ≤ x < 2π, then x equals
A π/3, 4π/3
B π/6, 7π/6
C π/4, 5π/4
D tan⁻¹2, π+tan⁻¹2
sinx = 2cosx implies tanx = 2 (where cosx ≠ 0). Tangent equals 2 in Quadrant I and III. So solutions are x = arctan2 and x = π + arctan2 within [0,2π).
A kite is 50 m high and the string is 100 m long. The angle of elevation is
A 45°
B 60°
C 30°
D 90°
The string is the hypotenuse and height is the opposite side. So sinθ = 50/100 = 1/2. The angle with sine 1/2 is 30° (taking the acute angle for elevation).
A tower is 20 m high and the angle of elevation from a point is 60°. The horizontal distance is
A 20/√3
B 10√3
C 20√3
D 40/√3
tanθ = height/distance. With θ = 60°, tan60° = √3 = 20/d. So d = 20/√3. This is a direct use of tangent in heights and distances.
For y = 2cos x − 1, the range of y is
A [−1, 3]
B [−2, 2]
C [−1, 1]
D [−3, 1]
cosx ranges from −1 to 1. Multiply by 2 gives −2 to 2, then subtract 1 gives −3 to 1. So the minimum is −3 and maximum is 1.
For y = sin(x − π/2), the simplest equivalent is
A cos x
B −sin x
C −cos x
D sin x
Use the identity sin(a−b)=sin a cos b − cos a sin b. With b=π/2, cos(π/2)=0 and sin(π/2)=1. So sin(x−π/2)=0·sinx − 1·cosx = −cosx.
For y = cos(x + π/2), the simplest equivalent is
A sin x
B −sin x
C −cos x
D cos x
Use cos(a+b)=cos a cos b − sin a sin b. With b=π/2, cos(π/2)=0 and sin(π/2)=1. So cos(x+π/2)=0·cosx − 1·sinx = −sinx.
In [0, 2π], solve 1 + tan²x = 2
A π/4, 3π/4, 5π/4, 7π/4
B π/6, 5π/6
C 0, π, 2π
D π/3, 4π/3
1+tan²x=2 gives tan²x=1, so tanx=±1. In [0,2π], tanx=1 at π/4 and 5π/4, and tanx=−1 at 3π/4 and 7π/4.
In [0, 2π], solve sin x = √3/2
A π/6, 5π/6
B 4π/3, 5π/3
C π/3, 2π/3
D π/2, 3π/2
sinx = √3/2 has reference angle π/3 (60°). Sine is positive in Quadrants I and II, so solutions in [0,2π] are x = π/3 and x = π − π/3 = 2π/3