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

Answer:

$(a)\ \cos(180 - x)$ --- 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:

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

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

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

$-\cos(x) \ne \sin(x)$

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