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babymother [125]
4 years ago
14

Write the ratios for sin A and cos A. The triangle is not drawn to scale

Mathematics
1 answer:
Bumek [7]4 years ago
5 0

Answer:

sin(A)=\frac{48}{50}, cos(A)=\frac{14}{50}

Step-by-step explanation:

Part a) Find the sin(A)

we know that

In the right triangle ABC

sin(A)=\frac{BC}{AB} ----> the sine of angle (A)  is the opposite side angle (A) divided by the hypotenuse

substitute the values

sin(A)=\frac{48}{50}

Part b) Find the cos(A)

we know that

In the right triangle ABC

cos(A)=\frac{AC}{AB} ----> the cosine of angle (A)  is the adjacent side angle (A) divided by the hypotenuse

substitute the values

cos(A)=\frac{14}{50}

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a)  \displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1

b)  \displaystyle \frac{dy}{dx} \bigg| \limits_{x = \frac{\pi}{2}} = -1

General Formulas and Concepts:

<u>Pre-Calculus</u>

  • Unit Circle

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]  

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹  

Derivative Rule [Product Rule]:                                                                             \displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Derivative Rule [Quotient Rule]:                                                                           \displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Derivative Rule [Chain Rule]:                                                                                 \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Trigonometric Differentiation

Logarithmic Differentiation

Step-by-step explanation:

a)

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle y = ln \bigg( \frac{1 - x}{\sqrt{1 + x^2}} \bigg)

<u>Step 2: Differentiate</u>

  1. Logarithmic Differentiation [Chain Rule]:                                                     \displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 - x}{\sqrt{1 + x^2}}} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]
  2. Simplify:                                                                                                         \displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{d}{dx}[\frac{1 - x}{\sqrt{1 + x^2}}]
  3. Quotient Rule:                                                                                               \displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{(1 - x)'\sqrt{1 + x^2} - (1 - x)(\sqrt{1 + x^2})'}{(\sqrt{1 + x^2})^2}
  4. Basic Power Rule [Chain Rule]:                                                                     \displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \frac{-\sqrt{1 + x^2} - (1 - x)(\frac{x}{\sqrt{x^2 + 1}})}{(\sqrt{1 + x^2})^2}
  5. Simplify:                                                                                                         \displaystyle \frac{dy}{dx} = \frac{-\sqrt{x^2 + 1}}{x - 1} \cdot \bigg( \frac{x(x - 1)}{(x^2 + 1)^\bigg{\frac{3}{2}}} - \frac{1}{\sqrt{x^2 + 1}} \bigg)
  6. Simplify:                                                                                                         \displaystyle \frac{dy}{dx} = \frac{x + 1}{(x - 1)(x^2 + 1)}

<u>Step 3: Find</u>

  1. Substitute in <em>x</em> = 0 [Derivative]:                                                                     \displaystyle \frac{dy}{dx} \bigg| \limit_{x = 0} = \frac{0 + 1}{(0 - 1)(0^2 + 1)}
  2. Evaluate:                                                                                                         \displaystyle \frac{dy}{dx} \bigg| \limits_{x = 0} = -1

b)

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle y = ln \bigg( \frac{1 + sinx}{1 - cosx} \bigg)

<u>Step 2: Differentiate</u>

  1. Logarithmic Differentiation [Chain Rule]:                                                     \displaystyle \frac{dy}{dx} = \frac{1}{\frac{1 + sinx}{1 - cosx}} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]
  2. Simplify:                                                                                                         \displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{d}{dx}[\frac{1 + sinx}{1 - cosx}]
  3. Quotient Rule:                                                                                               \displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{(1 + sinx)'(1 - cosx) - (1 + sinx)(1 - cosx)'}{(1 - cosx)^2}
  4. Trigonometric Differentiation:                                                                       \displaystyle \frac{dy}{dx} = \frac{-[cos(x) - 1]}{sin(x) + 1} \cdot \frac{cos(x)(1 - cosx) - sin(x)(1 + sinx)}{(1 - cosx)^2}
  5. Simplify:                                                                                                         \displaystyle \frac{dy}{dx} = \frac{-[cos(x) - sin(x) - 1]}{[sin(x) + 1][cos(x) - 1]}

<u>Step 3: Find</u>

  1. Substitute in <em>x</em> = π/2 [Derivative]:                                                                 \displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = \frac{-[cos(\frac{\pi}{2}) - sin(\frac{\pi}{2}) - 1]}{[sin(\frac{\pi}{2}) + 1][cos(\frac{\pi}{2}) - 1]}
  2. Evaluate [Unit Circle]:                                                                                   \displaystyle \frac{dy}{dx} \bigg| \limit_{x = \frac{\pi}{2}} = -1

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e

4 0
3 years ago
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