Answer:
[tex](a)\ \cos(180 - x)[/tex] --- Never true
[tex](b)\ \cos(90 -x)[/tex] --- Always true
[tex](c)\ \cos(x)[/tex] ---- Sometimes true
[tex](d)\ \cos(2x)[/tex] ---- Sometimes true
Step-by-step explanation:
Given
[tex]\sin(x )[/tex]
Required
Determine if the following expression is always, sometimes of never true
[tex](a)\ \cos(180 - x)[/tex]
Expand using cosine rule
[tex]\cos(180 - x) = \cos(180)\cos(x) + \sin(180)\sin(x)[/tex]
[tex]\cos(180) = -1\ \ \sin(180) =0[/tex]
So, we have:
[tex]\cos(180 - x) = -1*\cos(x) + 0*\sin(x)[/tex]
[tex]\cos(180 - x) = -\cos(x) + 0[/tex]
[tex]\cos(180 - x) = -\cos(x)[/tex]
[tex]-\cos(x) \ne \sin(x)[/tex]
Hence: (a) is never true
[tex](b)\ \cos(90 -x)[/tex]
Expand using cosine rule
[tex]\cos(90 -x) = \cos(90)\cos(x) + \sin(90)\sin(x)[/tex]
[tex]\cos(90) = 0\ \ \sin(90) =1[/tex]
So, we have:
[tex]\cos(90 -x) = 0*\cos(x) + 1*\sin(x)[/tex]
[tex]\cos(90 -x) = 0+ \sin(x)[/tex]
[tex]\cos(90 -x) = \sin(x)[/tex]
Hence: (b) is always true
[tex](c)\ \cos(x)[/tex]
If
[tex]\sin(x) = \cos(x)[/tex]
Then:
[tex]x + x = 90[/tex]
[tex]2x = 90[/tex]
Divide both sides by 2
[tex]x = 45[/tex]
(c) is only true for [tex]x = 45[/tex]
Hence: (c) is sometimes true
[tex](d)\ \cos(2x)[/tex]
If
[tex]\sin(x) = \cos(2x)[/tex]
Then:
[tex]x + 2x = 90[/tex]
[tex]3x = 90[/tex]
Divide both sides by 2
[tex]x = 30[/tex]
(d) is only true for [tex]x = 30[/tex]
Hence: (d) is sometimes true
