# 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.

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

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.

**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.

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

Yes

**dextrous63**)Yes

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mqb2766

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#6

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#6

(Original post by

Damn book likes to make things confusing

**KingRich**)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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#7

(Original post by

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).

**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).

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

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.

**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.

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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#9

(Original post by

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)

**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)

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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#11

(Original post by

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

**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

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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