# trigonometry a-level maths question... HELP!

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#1
If someone can go through this question and tell me the answer that would be great! (or how to get to the answer)

f(x) = 2 cos^-1(x) -1

1. Find the range and domain
2. find the exact value of f(-0.5)
3. Find f^-1(x)
0
2 months ago
#2
(Original post by grapefruits)
If someone can go through this question and tell me the answer that would be great! (or how to get to the answer)

f(x) = 2 cos^-1(x) -1

1. Find the range and domain
2. find the exact value of f(-0.5)
3. Find f^-1(x)
If you are already familiar with function/range/domain questions, please post your initial thoughts on this one. If you are not yet familiar with this type of question, I would suggest trying some more basic examples before tackling this one.
0
2 months ago
#3
ok so the domain is -1 to 1 because thats the domain of arccos (cos^-1) ye. the range is -1 to 2pi-1 , because the range of arccos is 0 to pi (so double that and subtract one).

f(-0.5) = 2 * arccos(0.5) - 1 = 2 * pi/3 - 1 = 2pi/3 - 1 ~= 1.094

get f^-1(x) as follows:
y=f(x)=2arccos(x)-1
y+1=2arccos(x)
(y+1)/2=arccos(x)
cos((y+1)/2)=x
f^-1(x)=cos((x+1)/2)
0
2 months ago
#4
(Original post by jakesimian)
ok so the domain is -1 to 1 because thats the domain of arccos (cos^-1) ye. the range is -1 to 2pi-1 , because the range of arccos is 0 to pi (so double that and subtract one).

f(-0.5) = 2 * arccos(0.5) - 1 = 2 * pi/3 - 1 = 2pi/3 - 1 ~= 1.094

get f^-1(x) as follows:
y=f(x)=2arccos(x)-1
y+1=2arccos(x)
(y+1)/2=arccos(x)
cos((y+1)/2)=x
f^-1(x)=cos((x+1)/2)
All looks good. Just one point: you should leave the answer to Q2 in exact form.
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2 months ago
#5
(Original post by old_engineer)
All looks good. Just one point: you should leave the answer to Q2 in exact form.
ye course mate i like to just put decimal approximations so i can check with it looks approximately right on the graphical calculator. of course in an exam give the exact form a fat underline so the examiner knows u know what ur on about. ur right tho my bad i should have clarified
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#6
Thank you so much @old_engineer and @username4261208. It all makes sense and I was able to then solve similar problems.
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