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###### math inverse trignometric question..

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1 year ago

I'll file the attachment below..

Reply 1

1 year ago

k must be positive, so x should be positive and in the 1st or 2nd quadrant.. But because there's an inverse of the sin function which has a range -90<=sinx<=90, the only positive values of x that are allowed should be 0<=x<=90.. so the only possible value for x should be x=a.. but ms says its x=a and x=180-a... pretty sure I'm making a stupid mistake here without even realizing it, but i have no idea what that might be.

Original post by Abraham_Otaku

k must be positive, so x should be positive and in the 1st or 2nd quadrant.. But because there's an inverse of the sin function which has a range -90<=sinx<=90, the only positive values of x that are allowed should be 0<=x<=90.. so the only possible value for x should be x=a.. but ms says its x=a and x=180-a... pretty sure I'm making a stupid mistake here without even realizing it, but i have no idea what that might be.

Youre sort of right. There is a single output/solution of arcsin(k) = alpha, so alpha is in quadrant 1 as its positive. However there are multiple solutions to

sin(x) = k

The first two being alpha and 180-alpha as the mark scheme says.

Its the "same" as saying x=sqrt(k) has a single solution/value (positve) but there are two solutions to

x^2 = k

Reply 3

1 year ago

Original post by mqb2766

Youre sort of right. There is a single output/solution of arcsin(k) = alpha, so alpha is in quadrant 1 as its positive. However there are multiple solutions to

sin(x) = k

The first two being alpha and 180-alpha as the mark scheme says.

Its the "same" as saying x=sqrt(k) has a single solution/value (positve) but there are two solutions to

x^2 = k

sin(x) = k

The first two being alpha and 180-alpha as the mark scheme says.

Its the "same" as saying x=sqrt(k) has a single solution/value (positve) but there are two solutions to

x^2 = k

I know there are multiple solutions, but isn't the only solution in range of -90<x<90 x=a? because that's the only solution for which inverse of sinx exists...... idek what I'm doing wrong here

Original post by Abraham_Otaku

I know there are multiple solutions, but isn't the only solution in range of -90<x<90 x=a? because that's the only solution for which inverse of sinx exists...... idek what I'm doing wrong here

There is a single value of arcsin(k). arcsin() is the inverse of sin() only on the limited domain -90..90. However, sin() is not restricted to this domain so there are multiple/infinite number of solutions to

sin(x)=k

one of which is arcsin(k). Its the "same" as there are two solutions to x^2=k, but sqrt(k) is single valued.

(edited 1 year ago)

Reply 5

1 year ago

Original post by mqb2766

Youre sort of right. There is a single output/solution of arcsin(k) = alpha, so alpha is in quadrant 1 as its positive. However there are multiple solutions to

sin(x) = k

The first two being alpha and 180-alpha as the mark scheme says.

Its the "same" as saying x=sqrt(k) has a single solution/value (positve) but there are two solutions to

x^2 = k

sin(x) = k

The first two being alpha and 180-alpha as the mark scheme says.

Its the "same" as saying x=sqrt(k) has a single solution/value (positve) but there are two solutions to

x^2 = k

wait.. I think i get it now.. there are multiple values of x for which sinx=k(as you said).despite one of those values being outside the range of arcsinx, it doesn't really matter, as the values which are outside of range (180-a) would still give the same value for k.. and this value of k gives a value which is within the range of the inverse sine function.. so both values x=a and x=180-a work... is that right?

Reply 6

1 year ago

Original post by mqb2766

There is a single value of arcsin(k). arcsin() is the inverse of sin() only on the limited domain -90..90. However, sin() is not restricted to this domain so there are multiple/infinite number of solutions to

sin(x)=k

one of which is arcsin(k). Its the "same" as there are two solutions to x^2=k, but sqrt(k) is single valued.

sin(x)=k

one of which is arcsin(k). Its the "same" as there are two solutions to x^2=k, but sqrt(k) is single valued.

ahh i see, I understand now. massive thanks.

Original post by Abraham_Otaku

wait.. I think i get it now.. there are multiple values of x for which sinx=k(as you said).despite one of those values being outside the range of arcsinx, it doesn't really matter, as the values which are outside of range (180-a) would still give the same value for k.. and this value of k gives a value which is within the range of the inverse sine function.. so both values x=a and x=180-a work... is that right?

Basically,

* multiple (infinite number of) values of x for which sin(x) = k

* only one of those values is in the range of arcsin(k)

arcsin() is the inverse of sin() restricted to the domain -90..90. For the sin() curve, its the s-shape rising from -1 to 1 (range), so is invertible on that domain.

(edited 1 year ago)

Reply 8

1 year ago

Original post by mqb2766

Basically,

* multiple (infinite number of) values of x for which sin(x) = k

* only one of those values is in the range of arcsin(k)

* multiple (infinite number of) values of x for which sin(x) = k

* only one of those values is in the range of arcsin(k)

thanks again

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