Simple Trig question

Im trying to do one AEA question

Find the exact value of x, x>= 0 for which
arccosx + arccos2x = pi/2

i made arccosx = A and arccos2x = B
and i ended up with

2x^2 + sinAsinB=0

sin^2 + cos^2 = 1
so sin^2= 1- cos^2
so sinA=√(1-cos^2A) right?

and i have no idea what to do now, help please :smile:

Reply 1

spread_logic_not_hate

8

Mobix

Im trying to do one AEA question

Find the exact value of x, x>= 0 for which
arccosx + arccos2x = pi/2

i made arccosx = A and arccos2x = B
and i ended up with

2x^2 + sinAsinB=0

sin^2 + cos^2 = 1
so sin^2= 1- cos^2
so sinA=√(1-cos^2A) right?

and i have no idea what to do now, help please :smile:

Substitute back in for A and B: e.g. A = arccos x, so

\sin A = \sqrt{1-\cos^2A} = \sqrt{1-\cos^2(arccos(x))} = \sqrt{1-x^2}

Reply 2

Mobix

OP

so cos^3(arccos(x)) = x^3??
i dont get how cos^2(arccosx) = x^2

Reply 3

spread_logic_not_hate

8

Mobix

so cos^3(arccos(x)) = x^3??
i dont get how cos^2(arccosx) = x^2

It's a notational thing:

\cos^2(t) \equiv [\cos(t)]^2

and

\cos(arccos(t)) = t

hence

\cos^2(arccos(x)) = [\cos(arccos(x))]^2 = [x]^2 = x^2

Reply 4

Mobix

OP

So cos^2(2) = cos(4) ??

Reply 5

spread_logic_not_hate

8

Mobix

So cos^2(2) = cos(4) ??

Not quite:

\cos^2(2) = [\cos(2)]^2

You've gotta work out the value of the cos function first, then square it. Here you need to work out the value of cos(2) and square that, not square the argument (2) and then evaluate the cos function.

Reply 6

Mobix

OP

Ok i get it thanks

Simple Trig question

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