hard questionn in maths A2

lahabz2012

how could you go about solving that and finding the x root value please help? it wi
inverse functions
thank you guys and appreciate your time

(edited 10 years ago)

Reply 1

brianeverit

9

Original post by lahabz2012

COS^-1 (x) = sin^-1 (x)

how could you go about solving that and finding the x root value please help? it will be 0.7 but how ??
inverse functions
thank you guys and appreciate your time

This is surely just equivalent to saying sin x=cos x

Reply 2

EpsilonSigma

2

Original post by brianeverit

This is surely just equivalent to saying sin x=cos x

That would only work for (sin(x))^-1 + (cos(x))^-1 (or as a fraction,

\frac{1}{cos(x)} + \frac{1}{sin(x)}

), not when you have arcsin and arccos.

Use this:

arccos(x) = \frac{Pi}{2} - arcsin(x)

Then it's just an equation with one trigonometric function in that you can solve.

That is, of course, assuming that the notation used by lahabz is what I'm thinking it is, and not what I said it wasn't :P

(edited 10 years ago)

Reply 3

Stanno

9

So:

arccos(x) = arcsin(x)
sin(arccos(x)) = x
sin^2(arccos(x)) = x^2
1 - cos^2(arccos(x)) = x^2
...
Think you should be able to go from there :wink:

.

(edited 10 years ago)

Reply 4

brianeverit

9

Original post by EpsilonSigma

That would only work for (sin(x))^-1 + (cos(x))^-1 (or as a fraction,

\frac{1}{cos(x)} + \frac{1}{sin(x)}

), not when you have arcsin and arccos.

Use this:

arccos(x) = \frac{Pi}{2} - arcsin(x)

Then it's just an equation with one trigonometric function in that you can solve.

That is, of course, assuming that the notation used by lahabz is what I'm thinking it is, and not what I said it wasn't :P

I assumed that by sin ^-1(x) he meant arcsin x

Reply 5

lahabz2012

OP

Guys am not quite sure wht the arcsin means but I do AQA c3 and I never met it . And it can't be a fraction ?

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Reply 6

lahabz2012

OP

ImageUploadedByStudent Room1386187916.087811.jpg

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Reply 7

Stanno

9

Original post by lahabz2012

Guys am not quite sure wht the arcsin means but I do AQA c3 and I never met it . And it can't be a fraction ?

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arcsin, arccos and arctan are the 'proper' notation for what will be denoted on your calculator as sin^-1, cos^-1 and tan^-1.

Reply 8

username456717

19

I would multiply both sides by sin x. Then follow from there, getting tanx and a number..

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Reply 9

Titus20

8

You can say that sin(x)=cos(x) and solve from there

Reply 10

Stanno

9

Original post by nebelbon

I would multiply both sides by sin x. Then follow from there, getting tanx and a number..

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Original post by Titus20

You can say that sin(x)=cos(x) and solve from there

I think the OP is referring to arcsin and arccos rather than 1/sin and 1/cos

Reply 11

TenOfThem

18

Original post by EpsilonSigma

.

Use this:

\arccos (x) = \frac{\pi}{2} - \arcsin (x)

Totally agree that this is the way to go

Reply 12

zoxe

7

Original post by TenOfThem

Totally agree that this is the way to go

I think an easier way would be to solve

x = sin(arccos(x))

and note that

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

Reply 13

TenOfThem

18

Original post by zoxe

I think an easier way would be to solve

x = sin(arccos(x))

and note that

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

Why would that be easier

Reply 14

Titus20

8

Original post by Stanno

I think the OP is referring to arcsin and arccos rather than 1/sin and 1/cos

I know. Solving sin(x)=cos(x) still finds the solutions.

If you consider it graphically, you can see that sin(x) and cos(x) both have the same inverse value when they are equal to each other.

Reply 15

Phichi

11

To clarify this all:

sin(x)=cos(x)

IS NOT the same as

arccos(x)=arcsin(x)

The only way to utilize the first equation would be to solve it, find the ANGLE at which its true, then sub that in either arccos(x) or arcsin(x).

The value of x in the first equation is an angle, in the second its not.

If you don't wish to take this approach, I believe somebody further up quoted a nice method:

arccos(x)=arcsin(x)

cos(arccos(x))=cos(arcsin(x))

x=cos(arcsin(x))

x^2=cos^2(arcsin(x))

Replace the cos^2 with 1 - sin^2, and note that sin^2(arcsin(x)) = x^2.

Reply 16

GeorgeL3

5

Original post by lahabz2012

COS^-1 (x) = sin^-1 (x)

how could you go about solving that and finding the x root value please help? it will be 0.7 but how ??
inverse functions
thank you guys and appreciate your time

To solve cos^-1x = sin^-1x, first consider this:

Let y = sin^-1x
siny = x
sin²y = x²1 - cos²y = x²cos²y = 1 - x²
cosy = √(1 - x²)
cosy = √(1 - sin²y)

Now back to cos^-1x = sin^-1x,
cos(sin^-1x) = x
Using our statement that we found earlier:
√[1 - sin²(sin^-1x)] = x
Since sin²(sin^-1x) = sin(sin^-1x) * sin(sin^-1x) = x²So √(1 - x²) = x
1 - x² = x²2x²= 1
x = √(1/2)

Reply 17

lahabz2012

OP

Original post by GeorgeL3

To solve cos^-1x = sin^-1x, first consider this:

Let y = sin^-1x
siny = x
sin²y = x²1 - cos²y = x²cos²y = 1 - x²
cosy = √(1 - x²)
cosy = √(1 - sin²y)

Now back to cos^-1x = sin^-1x,
cos(sin^-1x) = x
Using our statement that we found earlier:
√[1 - sin²(sin^-1x)] = x
Since sin²(sin^-1x) = sin(sin^-1x) * sin(sin^-1x) = x²So √(1 - x²) = x
1 - x² = x²2x²= 1
x = √(1/2)

Got it thank u very much . Are u an alevel student !

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Reply 18

GeorgeL3

5

Original post by lahabz2012

Got it thank u very much . Are u an alevel student !

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That's okay, glad it helped.
I did AQA maths and further maths last year but am now working for a year before university so not any more!
Also, I wouldn't worry too much if you didn't get it before since I'm pretty sure we only looked at that technique in FP2, not C3.

Reply 19

Liamnut

13

Disclaimer: This is probably wrong because I don't get 0.7.

arcsinx=arccosx

So,

sin(arccosx)=x

Let arccosx=y

Therefore,

Siny=x/1

Therefore,

Cosy=sqrt(1-x^2)

Sin^2(y)+Cos^2(y)=1 => Siny=sqrt(1-cos^2(y))

Cos^2(y)=1-x^2

Therefore,

x=sqrt(1-1+x^2)

So,

x^2=x^2

Wrong.

The point when sin(x)=cos(x) is a reflection, in the line y=x, of the point where arcsin(x)=arccos(x). So, the x coordinates and y coordinates will be swapped round maybe

(edited 10 years ago)

hard questionn in maths A2

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