Question

Determine whether each expression below is always, sometimes, or never equivalent to sin x when 0° < x < 90° ? Can someone help me :(

Answer 1

To determine whether each expression is always, sometimes, or never equivalent to sin x when 0° < x < 90°, we need to evaluate the expressions for different values of x.

Step-by-step explanation:

To determine whether each expression is always, sometimes, or never equivalent to sin x when 0° < x < 90°, we need to evaluate the expressions for different values of x. If the expression simplifies to sin x for all values of x in the given range, then it is always equivalent to sin x. If there are some values of x for which the expression is equivalent to sin x and some values for which it is not, then it is sometimes equivalent to sin x. And if it is never equivalent to sin x for any value of x in the given range, then it is never equivalent to sin x.

1. Expression 1: Evaluate the expression for the given values of x and see if it simplifies to sin x. If it does, then the expression is sometimes equivalent to sin x. If not, then it is never equivalent to sin x.
2. Expression 2: Follow the same steps as for Expression 1 to determine if it is always, sometimes, or never equivalent to sin x.
3. Expression 3: Repeat the evaluation process for Expression 3.

Answer 2

`(a)\ \cos(180 - x)` --- Never true

`(b)\ \cos(90 -x)` --- Always true

`(c)\ \cos(x)` ---- Sometimes true

`(d)\ \cos(2x)` ---- Sometimes true

Step-by-step explanation:

Given

`\sin(x )`

Required

Determine if the following expression is always, sometimes of never true

`(a)\ \cos(180 - x)`

Expand using cosine rule

`\cos(180 - x) = \cos(180)\cos(x) + \sin(180)\sin(x)`

`\cos(180) = -1\ \ \sin(180) =0`

So, we have:

`\cos(180 - x) = -1*\cos(x) + 0*\sin(x)`

`\cos(180 - x) = -\cos(x) + 0`

`\cos(180 - x) = -\cos(x)`

`-\cos(x) \\e \sin(x)`

Hence: (a) is never true

`(b)\ \cos(90 -x)`

Expand using cosine rule

`\cos(90 -x) = \cos(90)\cos(x) + \sin(90)\sin(x)`

`\cos(90) = 0\ \ \sin(90) =1`

So, we have:

`\cos(90 -x) = 0*\cos(x) + 1*\sin(x)`

`\cos(90 -x) = 0+ \sin(x)`

`\cos(90 -x) = \sin(x)`

Hence: (b) is always true

`(c)\ \cos(x)`

If

`\sin(x) = \cos(x)`

Then:

`x + x = 90`

`2x = 90`

Divide both sides by 2

`x = 45`

(c) is only true for
`x = 45`

Hence: (c) is sometimes true

`(d)\ \cos(2x)`

If

`\sin(x) = \cos(2x)`

Then:

`x + 2x = 90`

`3x = 90`

Divide both sides by 2

`x = 30`

(d) is only true for
`x = 30`

Hence: (d) is sometimes true