Usa a right triangle to write each expression as an algebraic expression. Assume the referenced angle is theta, and x is positive and in the domain of the given inverse trig. function. Label the triangle.
1. sec^sq(tan^-1x)
2. sin(arccot x/(sqrt)1-x^2)
1. If you mean sec^2[tan^-1x], use the fact that
sec (tan^-1x)= sqrt(x^2+1)/sqrt(1).
Imagine a right triangle with sides x, 1 and sqrt(1+x^2). Side x is opposite the angle with tangent equal to x.
Square the secant and you get x^2 + 1
2. For the angle in question, as part of a right triangle:
adjacent side (not hypotentuse) = x
opposite side = sqrt(1-x^2)
hypotenuse = 1
sin = sqrt(1-x^2)
The last line is your answer.
To write each expression as an algebraic expression using a right triangle, we can start by drawing a right triangle and labeling its sides based on the given trigonometric functions.
1. For the expression sec^2(tan^-1(x)):
- Start by drawing a right triangle.
- Label one of the acute angles as theta (θ).
- Let the adjacent side be x, the opposite side be 1, and the hypotenuse be r (the radius).
- Since tan(θ) is opposite over adjacent, we have tan(θ) = x/1 = x.
- Therefore, the opposite side is x, and the adjacent side is 1.
- Using Pythagoras' theorem, we can find the hypotenuse r: r^2 = 1^2 + x^2.
- Simplifying, we have r^2 = 1 + x^2. Taking the square root of both sides, we get r = sqrt(1 + x^2).
- Next, sec(θ) is defined as the hypotenuse over the adjacent side: sec(θ) = r/1 = r.
- Finally, we raise sec(θ) to the power of 2, giving us sec^2(θ) = (sqrt(1 + x^2))^2 = 1 + x^2.
Therefore, the algebraic expression is 1 + x^2.
2. For the expression sin(arccot(x) / sqrt(1 - x^2)):
- Draw a right triangle.
- Label one of the acute angles as theta (θ).
- Let the adjacent side be x (cotangent of θ), the opposite side be 1 (arccotangent of x), and the hypotenuse be r (the radius).
- Since cot(θ) = x (adjacent over opposite), we have x = adjacent/opposite = x/1 = x.
- Therefore, the adjacent side is x, and the opposite side is 1.
- Using Pythagoras' theorem, we can find the hypotenuse r: r^2 = x^2 + 1^2 = x^2 + 1.
- Taking the square root of both sides, we get r = sqrt(x^2 + 1).
- Next, sin(θ) is defined as the opposite side over the hypotenuse: sin(θ) = 1/r = 1/sqrt(x^2 + 1).
- Finally, we divide sin(θ) by sqrt(1 - x^2), resulting in sin(arccot(x) / sqrt(1 - x^2)) = (1/sqrt(x^2 + 1)) / sqrt(1 - x^2).
Therefore, the algebraic expression is (1/sqrt(x^2 + 1)) / sqrt(1 - x^2).