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Nikitich [7]
3 years ago
5

How to turn a sine function into a cosine function? If possible could someone provide an example with steps? Thanks!

Mathematics
2 answers:
Ganezh [65]3 years ago
5 0
<h2>Answer:</h2>

<u>We can convert sin function into cosine</u><u> by adding Pi/2 or 90° to the x in sin(x)</u><u> to get cosine.</u>

<h2>Step-by-step explanation:</h2>

sine function can be changed to cosine and vice versa by adding 90 degrees and its multiples in domain of function so

For instance

Sin (a+90)= cos a it is +ve as in angle lies in 2nd quad if a is less than 90 and sine is + ve in 2nd quad

Maksim231197 [3]3 years ago
4 0

Answer:

Several ways may be used to <em>turn a sine function into a cosine function</em>, using the fundamental properties of both trigonometric functions. Here I deal with two of them.

One way is using the property cos(x) = sin (90° - x). Other is using the identity sin² (x) + cos² (x) = 1.

You will find a detailed explanation and an example below.

Step-by-step explanation:

  • <u>Form 1:</u>

By the definition of the sine and cosine functions, sin (90° - x) = cos (x), and cos (90° - x) = sin (x).

Hence, starting with the basic function y = sin (x), you can convert it into a cosine function substituting sin(x) with cos (90 - x), obtaining:

y = cos (90 - x)

  • Now see an example:<em>Turn the sine function f(x) = 3 sin (30° - 2x) into a cosine function</em>.

f(x) = 3 sin (30° - 2x) = 3 cos [90° - (30° - 2x) ] = 3 cos [90° - 30° + 2x] =

      = 3 cos [60° + 2x ]

  • <u>Form 2</u>:

Also, you can use the fundamental identity: sin² (x) + cos² (x) = 1

From which: sin² (x) = 1 - cos²(x) ⇒ sin (x) = +/- √ [1 - cos² (x) ]

  • The same example:Turn the sine function f(x) = 3 sin (30° - 2x) into a cosine function.

  • f(x) = 3 sin (30° - 2x) = +/- √[1 - 3 cos² (30° - 2x)]

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Consider the following function. f(x) = 2x3 + 9x2 − 24x (a) Find the critical numbers of f. (Enter your answers as a comma-separ
viktelen [127]

Answer:

(a) The critical number of f(x) are x=-4, 1

(b)

  • Increasing for (-\infty, -4)
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(c)

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Step-by-step explanation:

(a) The critical numbers of a function are given by finding the roots of the first derivative of the function or the values where the first derivative does not exist. Since the function is a polynomial, its domain and the domain of its derivatives is (-\infty, \infty). Thus:

\frac{df(x)}{dx}  = \frac{d(2x^3+9x^2-24x)}{dx} =6 x^2+18x -24\\6 x^2+18x -24=0\\\boxed{x=-4, x=1}

(b)

  • A function f(x) defined on an interval is monotone increasing on (a, b) if for every x_1, x_2 \in (a, b): x_1 implies f(x_1)
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Combining  the domain (-\infty, \infty) with the critical numbers we have the intervals (-\infty, -4), (-4, 1) and (1, \infty). Note that any of the points are included, in the case of the infinity it is by definition and the critical number are never included because the function monotony is not defined in the critical points, i.e. it is not monotone increasing or decreasing. Now, let's check for the monotony in each interval, for this, we check for the sign of the first derivative in each interval. Evaluating in each interval the first derivative (one point is enough), we obtain the monotony of the function to be:

  • Increasing for (-\infty, -4)
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(c) From the values obtained in (a) so the relative extremum are the points (-4, 112) and (1, -13). The y-values are found by evaluating the critical numbers in the original function. Since the first derivative decreases after passing through  x=-4 and increases after passing through the point x=1 we have:

  • relative maximum (-4, 112)
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Answer:

  x = -4, 5/2

Step-by-step explanation:

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A graph is attached. It shows the solutions to be -4 and 5/2.

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Step-by-step explanation:

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