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Double identities involving inverse
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KingRich
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#1
Okay, this section is absolutely kicking my ass. It’s hard trying to recall the single identities let alone double identities and now I have recall inverse functions.
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Can someone explain why arcos x=A
I know if it were arcos 1/2 for example then it
Cos A=1/2, then A=60
Is it simply =A because we have no value of x, so as to represent unknown A?
This is probably a bit silly to ask but I like to verify my understanding.
Thanks in advance
Can someone explain why arcos x=A
I know if it were arcos 1/2 for example then it
Cos A=1/2, then A=60
Is it simply =A because we have no value of x, so as to represent unknown A?
This is probably a bit silly to ask but I like to verify my understanding.
Thanks in advance
Last edited by KingRich; 1 month ago
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mqb2766
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#2
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#2
Probably help to post the original question/example, but I guess they're just using a variable A to represent the angle.
Are you expected to draw a unit circle / triangle? If so cos(A) will be the base (1/2) as the hypotenuse is 1
So
tan(A) = sqrt(1-A^2)/A = sqrt(3)
for instance.
Edit - as above, theyre just using it to say the expression arccos(x) represents some angle A.
Are you expected to draw a unit circle / triangle? If so cos(A) will be the base (1/2) as the hypotenuse is 1
So
tan(A) = sqrt(1-A^2)/A = sqrt(3)
for instance.
Edit - as above, theyre just using it to say the expression arccos(x) represents some angle A.
Last edited by mqb2766; 1 month ago
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KingRich
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#3
(Original post by mqb2766)
Probably help to post the original question/example, but I guess they're just using a variable A to represent the angle.
Are you expected to draw a unit circle / triangle? If so cos(A) will be the base (1/2) as the hypotenuse is 1
So
tan(A) = sqrt(1-A^2)/A = sqrt(3)
for instance.
Edit - as above, theyre just using it to say the expression arccos(x) represents some angle A.
Probably help to post the original question/example, but I guess they're just using a variable A to represent the angle.
Are you expected to draw a unit circle / triangle? If so cos(A) will be the base (1/2) as the hypotenuse is 1
So
tan(A) = sqrt(1-A^2)/A = sqrt(3)
for instance.
Edit - as above, theyre just using it to say the expression arccos(x) represents some angle A.
I’m confused how they’ve expanded the function here.
is this basically the same
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T!gger34
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#4
KingRich
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#5
(Original post by dextrous63)
Yes
Yes
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mqb2766
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#6
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#6
(Original post by KingRich)
Damn book likes to make things confusing
Damn book likes to make things confusing
cos(2A) = cos^2(A)-1 = x^2-1
I guess they've done it like this as x is frequently used as a variable name for an angle (rather than a dimensionless side ratio in this case) so you could be thinking about cos(2x) = cos^2(x)-1... which is obviously wrong for this example.
Also, you should know that
cos^2(A) = (cos(A))^2
The left hand side is just a common way of writing the right hand side. You have to be careful with
cos^(-1)(x) = 1/cos(x)
and is not the inverse arccos(x).
Last edited by mqb2766; 1 month ago
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KingRich
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#7
(Original post by mqb2766)
Its a bit verbose and you/they could simplify it as
cos(2A) = cos^2(A)-1 = x^2-1
I guess they've done it like this as x is frequently used as a variable name for an angle (rather than a dimensionless side ratio in this case) so you could be thinking about cos(2x) = cos^2(x)-1... which is obviously wrong for this example.
Also, you should know that
cos^2(A) = (cos(A))^2
The left hand side is just a common way of writing the right hand side. You have to be careful with
cos^(-1)(x) = 1/cos(x)
and is not the inverse arccos(x).
Its a bit verbose and you/they could simplify it as
cos(2A) = cos^2(A)-1 = x^2-1
I guess they've done it like this as x is frequently used as a variable name for an angle (rather than a dimensionless side ratio in this case) so you could be thinking about cos(2x) = cos^2(x)-1... which is obviously wrong for this example.
Also, you should know that
cos^2(A) = (cos(A))^2
The left hand side is just a common way of writing the right hand side. You have to be careful with
cos^(-1)(x) = 1/cos(x)
and is not the inverse arccos(x).
I’m not sure how they’ve introduced the trig Pythagoras identity sin²x+cos²x=1, due to the 2 seemingly disappearing from the original function.
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mqb2766
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#8
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#8
(Original post by KingRich)
Can you explain part b for me. I see the identity for sin 2A=2(SinaCosa)
I’m not sure how they’ve introduced the trig Pythagoras identity sin²x+cos²x=1, due to the 2 seemingly disappearing from the original function.
Can you explain part b for me. I see the identity for sin 2A=2(SinaCosa)
I’m not sure how they’ve introduced the trig Pythagoras identity sin²x+cos²x=1, due to the 2 seemingly disappearing from the original function.
sin(2A) = 2 sin(A) cos(A) = 2 sin(A) x
using pythagoras
sin(A) = sqrt(1 - cos^2(A)) = sqrt(1-x^2)
so
sin(2A) = 2 x sqrt(1-x^2)
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KingRich
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#9
(Original post by mqb2766)
Its just a combination of the double angle sin and pythagoras, as you say, so
sin(2A) = 2 sin(A) cos(A) = 2 sin(A) x
using pythagoras
sin(A) = sqrt(1 - cos^2(A)) = sqrt(1-x^2)
so
sin(2A) = 2 x sqrt(1-x^2)
Its just a combination of the double angle sin and pythagoras, as you say, so
sin(2A) = 2 sin(A) cos(A) = 2 sin(A) x
using pythagoras
sin(A) = sqrt(1 - cos^2(A)) = sqrt(1-x^2)
so
sin(2A) = 2 x sqrt(1-x^2)
So, we derive 2SinAx from the double identity.
I take it we take 2 out of the equation by dividing by 2 first, so we’re left with sin A.
Taking sin A, we have to introduce cos by using Pythagoras identity. Hence,
So, sin²a=1-cos²a
Sin²a=1-x² due to cos a=x
Hence, sin a=√1-x²
I believe I made sense of that now lol
Last edited by KingRich; 1 month ago
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T!gger34
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mqb2766
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#11
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#11
(Original post by KingRich)
Mmm, let’s see if I follow correctly.
So, we derive 2SinAx from the double identity.
I take it we take 2 out of the equation by dividing by 2 first, so we’re left with sin A.
Taking sin A, we have to introduce cos by using Pythagoras identity. Hence,
So, sin²a=1-cos²a
Sin²a=1-x² due to cos a=x
Hence, sin a=√1-x²
I believe I made sense of that now lol
Mmm, let’s see if I follow correctly.
So, we derive 2SinAx from the double identity.
I take it we take 2 out of the equation by dividing by 2 first, so we’re left with sin A.
Taking sin A, we have to introduce cos by using Pythagoras identity. Hence,
So, sin²a=1-cos²a
Sin²a=1-x² due to cos a=x
Hence, sin a=√1-x²
I believe I made sense of that now lol
2 x sin(A)
expression. The pythagoras identity which gives
sin(A) = sqrt(1-x^2)
does not follow from the double angle identity, so dont think of it as dividing the double angle identity by 2. Just imagine that its a seperate piece of work to get sin(A) in terms of x.
Its a bit like if you wanted tan(A). Youd use the standard trig identity
tan(A) = sin(A)/cos(A) = sin(A)/x
Then using the pythagoras identity to get sin(A) = sqrt(1-x^2), youd then sub into the first identity to get
tan(A) = sqrt(1-x^2)/x
Last edited by mqb2766; 1 month ago
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