- #1
KnowledgeIsPower
- 90
- 1
I can do pretty much all the proofs except these two, though I've tried to make a start. Note that in the questions the angle is given as theta, but for the sake of my keyboard I've changed it to angle 'x'.
Q15: Prove:
Tan2xsecx = 2sinxsec2x.
So I need to show the two sides are equal.
First i worked on the left hand side.
Using the double angle conversion formulae it becomes:
=2sinxsec2x
=(2tanx)/(1-tan^2 x)secx
=(2tanx)/(1-tan^2 x)1/cosx
=(2tanx)/(1-tan^2 x)cosx
=(2tanx)/(cosx-tan^2cosx)
I find it difficult to find exactly the right form to convert back to double angles via manipulation so i converted the right hand side to single angles (to see what form i needed it in).
=2sinxsec2x
=2sinx multiplied by 1/(cos2x)
=2sinx multipled by 1/(cos^2x - sin^2x)
=2sinx/(cos^2x - sin^2x)
So far i have:
=(2tanx)/(cosx-tan^2cosx) = 2sinx/(cos^2x - sin^2x)
Is that correct so far and how can i go on to show they are exactly the same?
And the other one:
Q17: Prove that:
1/(cosx-sinx) - 1/(cosx+sinx) = 2sinxsec2x
The LHS consisted entirely of single angles so i left that for the time being and worked on the RHS.
=2sinxsec2x
=2sinx multiplied by 1/(cos2x)
=2sinx multiplied by 1/(cos^2x - sin^2x)
=2sinx/(cos^2x - sin^2x)
=2sinx/(1-sin^2x-sin^2x)
=2sinx/(1-2sin^x)
And i have no idea where to go from there.
Any help on either of the questions would be much appreciated.
Q15: Prove:
Tan2xsecx = 2sinxsec2x.
So I need to show the two sides are equal.
First i worked on the left hand side.
Using the double angle conversion formulae it becomes:
=2sinxsec2x
=(2tanx)/(1-tan^2 x)secx
=(2tanx)/(1-tan^2 x)1/cosx
=(2tanx)/(1-tan^2 x)cosx
=(2tanx)/(cosx-tan^2cosx)
I find it difficult to find exactly the right form to convert back to double angles via manipulation so i converted the right hand side to single angles (to see what form i needed it in).
=2sinxsec2x
=2sinx multiplied by 1/(cos2x)
=2sinx multipled by 1/(cos^2x - sin^2x)
=2sinx/(cos^2x - sin^2x)
So far i have:
=(2tanx)/(cosx-tan^2cosx) = 2sinx/(cos^2x - sin^2x)
Is that correct so far and how can i go on to show they are exactly the same?
And the other one:
Q17: Prove that:
1/(cosx-sinx) - 1/(cosx+sinx) = 2sinxsec2x
The LHS consisted entirely of single angles so i left that for the time being and worked on the RHS.
=2sinxsec2x
=2sinx multiplied by 1/(cos2x)
=2sinx multiplied by 1/(cos^2x - sin^2x)
=2sinx/(cos^2x - sin^2x)
=2sinx/(1-sin^2x-sin^2x)
=2sinx/(1-2sin^x)
And i have no idea where to go from there.
Any help on either of the questions would be much appreciated.