Inverse Of Trig Functions
Evaluating Expressions Using Inverses
Simple Trig Equations (1 Solution Max)
More Advanced Trig Equations (Any number of solutions)
Random (Will be a question belonging to any category)
100
Calculate the value of arcsin(0)
The answer is 0. Steps: Rewrite arcsin(0) as sin(x)=0. 0 is the only value that fits, for arcsin has a range of -90° ≤ y ≤90°.
100
Calculate the value of tan(arcsin(x))
The answer is x/√(1-x^2). Steps: Draw a right triangle. We know that the stuff in the parentheses is a side length because of the inverse trig function. Sin is opposite over hypotenuse, so x is the opposite and 1 is the hypotenuse. Use Pythagorean Theorem to find that the adjacent side is √(1-x^2). Tan is opposite over adjacent, so the answer is x/√(1-x^2).
100
Solve sin(x) + 1000 = 1001 for 0° ≤ x < 360°
The answer is 90°. Steps: Subtract 1000 from both sides. Then, take the arcsin of both sides. Finally, we have x = arcsin(1), which equals 90°.
100
Solve 2cos(x) = √2 for 0° ≤ x < 360°
The answers are 45° and 315°. Steps: Divide both sides by 2. Then, recall what reference angles yield a cosine of √2/2. Your answers should be 45° and 315°.
100
What is another was of writing cos^-1(x)?
The answer is arccos. Steps: arccos is a common alternative to cos^-1(x).
200
Calculate the value of arccos(√ 3 / 2)
The answer is π/6. Steps: Rewrite arccos(√ 3 / 2) as cos(x) = √ 3 / 2. √ 3 / 2 is the only value that fits, for arccos has a range of 0° ≤ y ≤ 180°.
200
DAILY DOUBLE! Calculate the value of cos(arcsin(2x))
The answer is √(1-4x^2). Steps: Draw a right triangle. We know that the stuff in the parentheses is a side length because of the inverse trig function. Sin is opposite over hypotenuse so 2x is the opposite and 1 is the hypotenuse. Use Pythagorean Theorem to find that the adjacent side is √(1-4x^2). Cos is adjacent over hypotenuse, so the answer is √(1-4x^2)
200
Solve -cos(x)+1 = 2 for 0° ≤ x < 360°
The answer is 180°. Steps: Subtract 1 from both sides. Then, divide both sides by negative 1. Finally, take the arcos of both sides so that you are left with x = arccos(-1), making x = 180°.
200
Solve sec^2(x) = 4 for 0° ≤ x < 360°
The answers are π/3, 2π/3, 4π/3, and 5π/3. Steps: Take the square root of both sides to get sec(x) = ±2. Sec is 1/cos, so it equals 1/cosx = ±2 which can be rearranged to equal cos(x) = ±1/2. Cos(x) = 1/2 at π/3 and 5π/3. Cos(x) = -1/2 at 2π/3 and 4π/3.
200
Solve csc^4(x)-4csc^2(x)=0 for 0° ≤ x < 360°
The answers are π/6, 5π/6, 7π/6, and 11π/6. Steps: Factor out the csc^2(x) to get (csc^2(x))(csc^2(x-4)) = 0. Using difference of squares, factor out (csc^2(x-4)) into csc(x)-2 and csc(x)+2 to get (csc^2(x))(csc(x)-2)(csc(x)+2)=0 Csc^2(x) can’t equal zero, so it doesn't give any solutions Csc(x)-2 = 0 simplifies to sin(x) = -1/2, which gives π/6 and 5π/6 as solutions. Csc(x)+2 = 0 simplifies to sin(x) = 1/2, which gives 7π/6 and 11π/6 as solutions.
300
Calculate the value of sin(arcsin(3/4))
The answer is 3/4. Steps: Remember that arcsin has a domain of -1 ≤ x ≤ 1 and is the inverse of sin. Therefore, any number that suits this domain requirement will allow you to compose the two functions by canceling them.
300
Calculate the value of csc(arccos(x+1))
The answer is 1/√(1-(x+1)^2). Steps: Draw a right triangle. We know that x+1 in the parentheses is a side length because of the inverse trig function. Cos is adjacent over hypotenuse, so the hypotenuse is 1 and the adjacent side is x + 1. Use pythagorean theorem to find that the opposite side is √(1-(x+1)^2). Csc is hypotenuse over opposite, so the answer is 1/√(1-(x+1)^2).
300
Solve tan^2(x) + 3 = 0 for 0° ≤ x < 360°
There is no solution to this problem. Steps: Subtract 3 from both sides. Then, if we attempt to take the square root from both sides, we realize that we can't take the square root of a negative number, which makes the equation unsolvable.
300
Solve (sin(x))(tan(x - 1))(cos(x - 1)) = 0 for 0° ≤ x < 360°
The answers are 0, π/4, π, and 5π/4. Steps: Set each factor equal to zero. Sin(x) = 0 at 0 and π. Tan(x) = 1 at π/4 and 5π/4. Cosx = 1 at 0.
300
Solve tan^2(x) + 5 = 0 for
There is no solution. Steps: Subtract 5 from both sides. Since we can't take the square root of negative 5, there is no solution to the equation.
400
Calculate the value of cos(arctan(1))
The answer is √ 2 / 2. Steps: First, find arctan(1) by rewriting it as tan(x) = 1. Arctan has a domain of −90° < y < 90°, so the only answer is 45°. Plug 45° back into the original problem, cos(arctan(1)), in place of arctan(1). cos(45)= √ 2 / 2
400
DAILY DOUBLE! θ is an angle of a right triangle and is not equal to 90°. Find the sides of the right triangle given θ=arccot(x/4)
The answer is adjacent=x, opposite=4, and hypotenuse=√x^2+16. Steps: arcot(x/4) means that cotθ=x/4. Cot=adjacent/opposite, so the adjacent side = x, and the opposite side = 4. Now you have the two legs. In order to find the hypotenuse, use the pythagorean theorem.
400
Solve sec(x + 45) = 1 for 0° ≤ x < 360°
The answer is 315. Steps: Take the arcsec of both sides to cancel out sec on the left side. We now have (x + 45) = arcsec(1). The arcsec(1) = 360. Then, substract 45 from both sides, and you will get 315 as your answer.
400
Solve cos(2x) = cos(x) for 0° ≤ x < 360°
The answers are 2π/3 and 4π/3. Steps: Substitute cos(2x) with 2cos^2(x-1) and move everything to one side of the equation to get (2cos^2(x-1))-cos(x) = 0. Factor to get (2cos(x+1))(cos(x-1)) = 0. Set both to zero to get cos(x) = -1/2 and 1. Solve to get x = 2π/3 and 4π/3.
400
Calculate arccos(-√3/2)
The answer is 150°. Steps: Remember that arccos has a range of 0° ≤ y ≤ 180°. Next, we can rearrange the problem to look like this: cos(x)= -√3/2. Now, we recall our reference angles to find the answer.
500
Calculate the value of cos(arctan(1) - arcsin(1))
The answer is √ 2 / 2. Steps: Rewrite arctan(1) as tan(x)=1 and arcsin(1) as sin(x)=1. After you solve both, your answers should be π/4 and π/2, respectively. Replace both answers with their corresponding expressions and then subtract them. Cos(-π/4)=√ 2 / 2
500
Solve sin(A+B) if A=arccos(1/2) and B=arcsin(1/2)
The answer is 1. Steps: Sin(A+B)=sinA x cosB + cosA x sinB. cosA=1/2 and sinB=1/2 since A=arccos1/2 and B=arcsin1/2. Therefore, A=π/3 and B=π/6. Now, plug these values in and solve. Your answer will simplify to sin(π/2), making your answer 1.
500
DAILY DOUBLE! Solve sin^2(x) – sin(x) = 2 for 0° ≤ x < 360°
The answer is 270°. Steps: Subtract 2 from both sides. Next, factor the left side into (sin(x) – 2)(sin(x) + 1). We can ignore the left factor because it yields sin(x)=2, which is impossible since the range of sin is -1 ≤ y ≤ 1. This leaves us with sin(x) + 1 = 0. Subtract 1 from both sides. Finally, recall that sin(270°)=-1
500
Solve sin^2(x) = -cos(2x) for 0° ≤ x < 360°
The answers are π/2 and 3π/2. Steps: Substitute cos(2x) with cos^2(x)-sin^2(x) to get sin^2(x)= -cos^2(x) + sin^2(x). Simplify to get cos^2(x) = 0. Take the square root of both sides to get cos(x) = 0. Cos(x) = 0 at π/2 and 3π/2
500
DAILY DOUBLE! Solve 9 + cot^2(x) = 12 for 0° ≤ x < 360°
The answers are π/6, 5π/6, 7π/6 and 11π/6. Subtract 9 on both sides. Take the square root of both sides to get cot(x) = ±√3. Cot(x) = 1/tan(x), so 1/tan(x) = ±√3 and tan(x) = ±√3/3 Tan(x) = ±√3/3 at π/6, 5π/6, 7π/6 and 11π/6
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