SAT Mathematics – Trigonometry Test for practice Free

This problem can be solved using the Law of Cosines: d² = 10² + 20² - 2(10)(20)cos(120°), d = √700 = 10√7.

The period of cos(bt) is 2π/|b|. Here, b = 2π, so the period is 2π/(2π) = 1 second.

Since tan(x) = 3/4 and x is in the third quadrant, both sin(x) and cos(x) are negative. sin(x) = -3/5, cos(x) = -4/5, so sin(x) + cos(x) = -7/5.

The angle 240° is in the third quadrant (where tangent is positive). The reference angle is 60°, so tan 240° = tan 60° = √3.

The area of a triangle is given by (1/2)ab sin C. Here, Area = (1/2)(7)(8)sin(60°) = 28(√3/2) = 14√3.

This expression is the double-angle formula for cosine: cos(2θ) = cos² θ - sin² θ. Thus, cos(π/4) = √2/2.

Using the double-angle identity for sine: 2sin(x)cos(x) = cos(x), solving gives cos(x) = 0 or sin(x) = 1/2. Thus, the solutions are x = π/6, π/2, 5π/6, 3π/2.

For polar coordinates (r, θ), x = rcos(θ) and y = rsin(θ). Here, x = 4cos(5π/6) = -2√3, y = 4sin(5π/6) = 2.

sin(45°) = √2/2, so sin²(45°) = (√2/2)² = 1/2.

The transformation involves a vertical stretch by 3 (3sin(x)), horizontal compression by 2 (3sin(2x)), and a vertical shift of -1, resulting in y = 3sin(2x) - 1.

The tangent function equals 1 when the reference angle is π/4. In the given interval, tan x = 1 at x = π/4 and x = 5π/4.

arcsin(x) returns an angle in the range [-π/2, π/2]. The angle whose sine is -√3/2 in this range is -π/3.

The angle 240° is in the third quadrant (where cosine is negative). The reference angle is 60°, so cos 240° = -cos 60° = -0.5.

The maximum value of sin(4t) is 1. Therefore, the maximum height is 3(1) + 8 = 11 inches.

Using the sum formula for sines: sin(x+π/2) = cos(x) = b.

The dot product u • v = (2)(-1) + (3)(4) = -2 + 12 = 10.

The cosine of 30° is √3/2.

The x-component of v is given by vₓ = |v|cos(θ) = 10cos(120°) = 10(-1/2) = -5.

At x = 60°, the reference triangle gives tan 60° = √3.

The tangent function is negative in the second and fourth quadrants. tan x = -1 at x = 3π/4 and x = 7π/4.

Using the product-to-sum identity: 2sin(x)cos(3x) = sin(4x) - sin(2x).

Using the half-angle formula for sine: sin(θ/2) = √[(1 - cos(θ))/2] = √[(2/3)/2] = √(1/3) = √3/3.

After rotating 240°, the rider has moved past the top of the circle. The angle from the vertical is 240° - 180° = 60°. The vertical distance from the center of the circle is 50cos(60°) = 25 feet. Since the center of the circle is 50 feet above the ground, the rider's height is 50 + 25 = 75 feet.

Squaring both equations, we get x² = 16cos²(t) and y² = 16sin²(t). Adding these equations gives x² + y² = 16.

The amplitude is 60 (the radius). The period is 2 minutes = 120 seconds, so b = π/60. The rider starts at the bottom, so the equation is -60cos(πt/60) + 60.

If tan x = 1, then x = 45° (or π/4 radians). Therefore, sin 2x = sin 90° = 1.

The tangent of 0° is 0.

The sine function equals 0.5 at x = π/6 and x = 5π/6 within the given interval.

Tangent is negative in the second and fourth quadrants. The angles in the interval [0, 2π] where tan(x) = -1 are 3π/4 and 7π/4.

This equation is a quadratic in sin(x). Solving the quadratic gives the solutions x = 7π/6, 11π/6, π/2.

Using the Law of Sines: sin(A)/a = sin(B)/b, sin(30)/5 = sin(B)/10, sin(B) = 1, B = 90 degrees.

sin(45°) = cos(45°) = √2/2, so their sum is (√2/2) + (√2/2) = √2.

Let L be the length of the ladder. We can use the cosine function: cos(60°) = Adjacent/Hypotenuse = 7/L, 1/2 = 7/L, L = 14 feet.

Since tan(x) = -1 and x is in the fourth quadrant, sin(x) = -√2/2.

sin(90°) = 1, cos(0°) = 1, so 1 - 1 = 0.

This expression is the double-angle formula for cosine: cos(2θ) = cos² θ - sin² θ. Thus, cos(π/4) = √2/2.

Using the Law of Sines and Law of Cosines, the solution to angle Q is approximately 62°.

tan(x) is defined as sin(x)/cos(x).

Using the angle sum formula for sine: sin(75°) = (√6 + √2)/4.

Since sin(x) = 1/3, the adjacent side is negative. tan(x) = -√2/4.

Using the sum formula for sines: sin(x + π) = sin(x)cos(π) + cos(x)sin(π) = -sin(x).

Using the Law of Cosines, the length of side a is √61.

The horizontal distance is maximized when sin(2θ) is maximized. The maximum value of the sine function is 1, which occurs when 2θ = 90°, or θ = 45°.

This is a direct application of the double-angle identity for sine: sin(2x) = 2sin(x)cos(x).

sin(60°) = √3/2, tan(30°) = √3/3, so (√3/2) * (√3/3) = 3/6 = 1/2. The previous answer was incorrect.

The period of a function of the form cos(bx) is given by 2π/|b|. In this case, b = 2, so the period is 2π/2 = π.

In the third quadrant, both sine and cosine are negative. Using the Pythagorean theorem, we find sin θ = -12/13.

The sine function is negative in the third and fourth quadrants. For sin x = -0.707, x ≈ 225° and x ≈ 315°.

Using the angle difference formula for cosine: cos(15°) = cos(45° - 30°) = (√6 + √2)/4.

Using the Pythagorean theorem and the second quadrant properties, tan θ = -3/4.

Since x is in the third quadrant, x/2 will be in the second quadrant where tan is negative. tan(x/2) = -√((1-cos(x))/(1+cos(x))) = -1/2.