Question

Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction pi over 4 endfraction)) minus 2 y = negative 3 sine (2 (x startfraction pi over 4 endfraction)) minus 2 y = 3 cosine (2 (x startfraction pi over 4 endfraction)) minus 2 y = negative 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2

Answer 1

The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

How to convert sine of an angle to some angle of cosine?

We can use the fact that:

$\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)$

to convert the sine to cosine.

Which trigonometric functions are positive in which quadrant?

• In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
• In second quadrant(π/2 < θ < π), only sin and cosec are positive.
• In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
• In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

$y= 3\cos(2(x + \pi/2)) - 2$

The options are:

1. $y= 3\sin(2(x + \pi/4)) - 2$
2. $y= -3\sin(2(x + \pi/4)) - 2$
3. $y= 3\cos(2(x + \pi/4)) - 2$
4. $y= -3\cos(2(x + \pi/2)) - 2$

Checking all the options one by one:

• Option 1: $y= 3\sin(2(x + \pi/4)) - 2$

$y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2$

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

• Option 2: $y= -3\sin(2(x + \pi/4)) - 2$

This option if would be true, then from option 1 and this option, we'd get:
$-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0$

which isn't true for all values of x.

Thus, this option is not same as the given function.

• Option 3: $y= 3\cos(2(x + \pi/4)) - 2$

The given function is $y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2$

This option's function simplifies as:

$y= 3\cos(2(x + \pi/4)) - 2 = 3\cos(2x + \pi/2) -2 = -3\sin(2x) - 2$

Thus, this option isn't true since $\sin(2x) \neq \cos(2x)$ always (they are equal for some values of x but not for all).

• Option 4: $y= -3\cos(2(x + \pi/2)) - 2$

The given function simplifies to:$y= 3\cos(2(x + \pi/2)) - 2 = 3\cos(2x + \pi) -2 = -3\cos(2x) -2$

The given option simplifies to:

$y= -3\cos(2(x + \pi/2)) - 2 = -3\cos(2x + \pi ) -2\\y = 3\cos(2x) -2$

Thus, this function is not same as the given function.

Thus, the function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

Learn more about sine to cosine conversion here:

brainly.com/question/1421592

Answer 2

Answer:

a

Step-by-step explanation: