University:
Washington State University
Course:
MATH 172 | Calculus II
Academic year:

2024
Views:

101

Pages:

4
Author:

Bailey Ayala

Question 1 Please, rewrite the expression as an algebraic expression in x. sin(tan⁻¹x) Answer 1 Let θ = tan⁻¹(x), then tanθ = x Draw a right triangle with θ as one of it's acute angles: AC = √AB² + DC² AC = √x² + 1² = √1 + x² sin (tan⁻¹(x)) = x/√1+x² Answer 2 Step 1: Recall the identity tan⁻¹x = arctan(x). Step 2: Use the definition of the tangent function: tan(θ) = sin(θ)/cos(θ). In this case, θ = tan⁻¹x. So we can rewrite tan⁻¹x as arctan(x) = sin(arctan(x))/cos(arctan(x)) Step 3: Apply the identity sin(arctan(x)) = x/√1+x². This can be derived using the right triangle definition of the trigonometric functions. Step 4: Apply the identity cos(arctan(x)) = 1/√1+x². Again, this can be derived using the right triangle definition of the trigonometric functions. Step 5: Substitute the values of sin(arctan(x)) and cos(arctan(x)) back into the expression sin(arctan(x))/cos(arctan(x)) Combining all the steps, we get: sin(tan⁻¹x) = sin(arctan(x)) = sin(arctan(x))/cos(arctan(x)) = (x/√1+x²)/(1/√1+x²) = x/√1+x² * √1+x²/1 = x√1+x²/√1+x² = x Therefore, the expression sin(tan⁻¹x) can be rewritten as the algebraic expression x. Answer 3 To rewrite the expression sin(tan⁻¹x) as an algebraic expression in x, we can use trigonometric identities. First, we'll start by using the identity tan⁻¹x = arctan(x): sin(tan⁻¹x) = sin(arctan(x)) Next, we can use the identity sin(arctan(x)) = x/√1+x² sin(tan⁻¹x) = sin(arctan(x)) = x/√1+x² Therefore, the algebraic expression in x equivalent to sin(tan⁻¹x) is x/√1+x² Answer 4 Explanation: tan⁻¹x = arctanx Using this identity, we have: sin(tan⁻¹x) = sin(arctanx) Now, let's use another trigonometric identity to express sin(arctanx) in terms of x. The identity is: sin(arctanx) = x/√1+x² Therefore, the algebraic expression in x equivalent to sin(tan⁻¹x) is: x/√1 + x² Question 2 Find the Opposite Side (O) of the triangle if the Adjacent Side (A) is 18 and the Hypotenuse (H) is 47. [Image of a right triangle with angle x, adjacent side labeled 18, and hypotenuse labeled 47] Find the Opposite Side (O) of the triangle if the Adjacent Side (A) is 18 and the Hypotenuse (H) is 47. Answer The given triangle we have is [Image of a right triangle with angle x, adjacent side labeled 18, and hypotenuse labeled 47, opposite side labeled O] We have to find the opposite side O. Solution: To find the opposite side, we will use Pythagoras theorem. Hypotenuse² = Base² + Perpendicular² → 47² = 18² + Opposite² → Opposite² = 47² - 18² = 2209 - 324 = 1885 Opposite = √1885 = 43.416 ~ 43.42 Therefore, the opposite side is 43.42. Question 3 Find the value of each trigonometric ratio to the nearest ten-thousandth tan 50° Answer Given: tan 50° tan 50° = sin 50/cos 50 tan 50° = 0.76604/0.64278 tan 50° =1.1928 Question 4 Find the radian measure of the central angle of a circle of radius r that intercepts an arc of length s. r = 14 feet, s = 8 feet. Answer θ = s/r = 8/14 = 4/7 radians Question 5 A Major League baseball diamond has four bases forming a square whose sides measure 90 feet each. The pitcher's mound is 60.5 feet from home plate on a line joining home plate and second base. What is the distance from the pitcher's mound to first base to the nearest tenth of a foot? Answer 1 a² = b²+c²-2 x b x c x cos A x² = 60.5² + 90² - 2 × 60.5 x 90 x cos 45 x = 63.7 Answer 2 A² = B²+C²-2.B.C.COSA x² = 60.5² + 90² - 2 * 60.5 * 90 * cos 45 X = 63.7