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# maths c3 help pls Watch

1. Hello, if you have got a graph and it inverse graph , then there is this point P on the original graph and the point Q on the inverse graph. The point Q is the reflection of P in y = x. Then we are asked to find out the gradient at Q ...and hence deduce what the gradient at P is (this is the bit that I am finding tricky at the moment to understand). Thanks for all the help in advance!
2. (Original post by laurawoods)
Hello, if you have got a graph and it inverse graph , then there is this point P on the original graph and the point Q on the inverse graph. The point Q is the reflection of P in y = x. Then we are asked to find out the gradient at Q ...and hence deduce what the gradient at P is (this is the bit that I am finding tricky at the moment to understand). Thanks for all the help in advance!
Well, think about the part of the curve around where P is. Now, imagine what it would look like when you reflect that part the line y=x.

You should then see why it's asking you to deduce rather than find.

Does that help?
3. (Original post by laurawoods)
Hello, if you have got a graph and it inverse graph , then there is this point P on the original graph and the point Q on the inverse graph. The point Q is the reflection of P in y = x. Then we are asked to find out the gradient at Q ...and hence deduce what the gradient at P is (this is the bit that I am finding tricky at the moment to understand). Thanks for all the help in advance!
Without knowing further details of what you're expected to know in C3 on your board, observe that since the functions are reflections of one another in y=x, the tangents at each point in a reflected pair of corresponding points must also be reflections of each other in y=x (to get a feel for why this is so without proof, imagine a sketch of a curve with a tangent drawn on it and consider reflecting the whole thing in y=x - the 'new' line in the reflection will have to touch the 'new' curve and hence must be tangential to it at that point). What do you know about the gradients of lines that are reflections of one another in y=x?
4. (Original post by Indeterminate)
Well, think about the part of the curve around where P is. Now, imagine what it would look like when you reflect that part the line y=x.

You should then see why it's asking you to deduce rather than find.

Does that help?
so , will it just be the reciprocal? thanks!
5. (Original post by laurawoods)
so , will it just be the reciprocal? thanks!
Yes
6. (Original post by Indeterminate)
Yes
cool thanks!
7. Can anyone help me with this trig question please:
simplify-
1/(1-cosecx)(1+cosecx)
Can anyone help me with this trig question please:
simplify-
1/(1-cosecx)(1+cosecx)
Really you should start a new thread but I'll give you a bit of help here.

You could do it 2 ways, one is to expand the brackets and use an identity involving cosec^2(x) and cot^2(x), the other is to just expand out the whole thing in terms of sin(x) and simplify like any other equation.

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Updated: April 21, 2013
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