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Jacob M.
Chemical Engineering Major at Cleveland State University - Part of the Jack, Joseph & Morton Mandel Honors College
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Pre-Calculus
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Question:

Prove the following identities: a) (1 - sin(x)) * (1 + csc(x)) = cos(x)*cot(x) b) sin(x) + sin(x)* [cot(x)]^2 = csc(x) c) sin(x)*cos(x)*tan(x) = 1 - [cos(x)]^2

Jacob M.
Answer:

a) (1 - sin(x)) * (1 + csc(x)) = cos(x)*cot(x) - We will be working with the left side of the equation to prove that it is equivalent to cos(x)*cot(x). The first thing you will want to do is distribute. [1 - sin(x)] * [1 + csc(x)] [1 - sin(x)] * { [1 + [1 / sin(x)] } { 1 + [1/sin(x)] - sin(x) - [sin(x) * (1 / sin(x))] } { 1 + [1/sin(x)] - sin(x) - 1 } [1/sin(x)] - sin(x) - After distribution and simplification, we need to obtain a common denominator to simplify further. For this problem, we will substitute [sin(x)]^2 / [sin(x)] in for sin(x). [1/sin(x)] - sin(x) [1/sin(x)] - [sin(x)]^2 / [sin(x)] {1 - [sin(x)]^2 } / sin(x) - Now we will use the pythagorean identity of 1 - [sin(x)]^2 = [cos(x)]^2 {1 - [sin(x)]^2 } / sin(x) [cos(x)]^2 } / sin(x) - Simplify to get cos(x) * cot(x) [cos(x)]^2 } / sin(x) cos(x) * [cos(x) / sin(x)] cos(x) * cot(x) b) sin(x) + sin(x)* [cot(x)]^2 = csc(x) - We will be working with the left side of the equation to prove that it is equal to csc(x). The first thing we will want to do is take a sin(x) out. sin(x)* { 1 + [cot(x)]^2 } - Next we will use the identity of 1 + [cot(x)]^2 = [csc(x)]^2 sin(x) * [csc(x)]^2 - Then we will put [csc(x)]^2 in terms of sin(x) sin(x) * { 1 / [sin(x)]^2 } - Cancel out sin(x) and show that it is equal to csc(x) sin(x) * { 1 / [sin(x)]^2 } 1 / sin(x) csc(x) c) sin(x)*cos(x)*tan(x) = 1 - [cos(x)]^2 - In this case, we will be working with both sides of the equation. The right side has a pythagorean identity { [sin(x)]^2 + [cos(x)]^2 = 1 }, so we can start there. -Changing the pythagorean identity results in 1 - [cos(x)]^2 = [sin(x)]^2, therefore, the right side of the equation is equivalent to: [sin(x)]^2 -Next, we will simply the left side by putting it all in terms of sin(x) and cos(x). We can change tan(x) to sin(x) / cos(x) sin(x) * cos(x) * [sin(x) / cos(x)] -We can then simplify this expression be canceling out the cos(x) in the numerator and denominator and get: sin(x) * sin(x) -This is then equal to the right side of the equation, proving the identity. [sin(x)]^2 = [sin(x)]^2

Trigonometry
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Question:

You are walking and see a tall tree in the distance. The angle of elevation from where you are standing is 55°. You walk 60 feet closer to the tree and the angle of elevation is 68.2°. How tall is the tree? Round to the nearest foot.

Jacob M.
Answer:

We can set up two equations for this problem. H will represent the height of the tree. And x will represent the distance from the tree after moving in 60 feet. 1) tan(55) = h / (60 + x) - At the original point, the angle of elevation is 55°. The opposite is the height of the tree. The adjacent is the distance from the tree. 2) tan(68.2) = h / x - After moving in 60 feet, the angle of elevation is 68.2°, the adjacent is the height of the tree and you are the x distance from the tree. These two equations can be used to solve for x. Equation 1 can be rearranged into: (60 + x) * tan(55) = h 60tan(55) + xtan(55) = h Equation 2 can be rearranged into: h = xtan(68.2) From here, we will plug equation 2 into equation 1, giving us: 60tan(55) + xtan(55) = xtan(68.2) We will then solve for x: 60tan(55) + xtan(55) = xtan(68.2) 60tan(55) = xtan(68.2) - xtan(55) x = 60tan(55) / [tan(68.2) - tan(55) x = 79.93 ft Now that we have x, we can take equation 2 to solve for the height of the tree. h = xtan(68.2) h = 79.93tan(68.2) h = 199.84 ft Rounded to the nearest foot, the tree is 200 feet tall.

Calculus
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Question:

Find f(x) through f'''(x) for the following function: f''(x) = 5cos(x) f'(π) = 5 f(2π) = 2

Jacob M.
Answer:

Since were are given f''(x), we need to find: f(x), f'(x), and f'''(x) A) Solve for f'(x) - To calculate f'(x), we will take the integral of f''(x) = 5cos(x) ∫ 5 cos(x) dx - Linearity can be applied, so the constant of 5 is moved in front 5 ∫ cos(x) dx - The integral of cos(x) is equal to sin(x). Don't forget the constant of integration. f'(x) = 5 [sin(x)] + C - Since we are looking for f'(x), we will need to plug a number in to find C. The problem gives us f'(π) = 5. Therefore, we will plug in π for x, and solve for C. f'(x) = 5sin(x) + C f'(2) = 5sin(π) + C 5 = 0 + C C = 5 - After C is solved for, we can write the function for f'(x) f'(x) = 5sin(x) + 5 B) Solve for f(x) - We will follow a similar procedure as solving for f'(x) but will use our answer for f'(x) instead of a given equation. -To calculate f(x) we will take the integral of f'(x) = 5sin(x) + 5. ∫ 5sin(x) + 5 dx - The integral of cos(x) is equal to -sin(x) and the integral of 5 is equal to 5x. f(x) = 5cos(x) + 5x + C ***NOTE: Linearity could only be applied if the integral is split into two separate integrals or if a 5 were to be taken out of both terms: ∫ 5[sin(x) + 1] dx = 5 ∫ sin(x) + 1 dx - Since we are looking for the function, we will need to plug in the given values for f(x) to solve for C. From the question: f(2π) = 2 f(x) = 5cos(x) + 5x + C f(2π) = 5cos(2π) + 5(2π) + C 2 = 5 + 10π + C C = 2 - 5 - 10π C = -3 - 10π -After C is solved for, we can write the function for f(x) f(x) = 5cos(x) + 5x -3 - 10π C) Solve for f'''(x) - To calculate f'''(x), we will simply take the derivative of f''(x). Since there is a 5 in front, the 5 remains constant. f''(x) = 5cos(x) - The derivative of cos(x) is equal to -sin(x), therefore: f'''(x) = -5sin(x) D) Rewrite f(x) through f'''(x) f(x) = 5cos(x) + 5x -3 - 10π f'(x) = 5sin(x) + 5 f''(x) = 5cos(x) f'''(x) = -5sin(x)

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