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C3 Trig Question: Can you spot where I went wrong?


I managed to get the whole question right except for part c, but I can't see where I went wrong..

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Reply 1
Avatar for TeeEm
TeeEm
19
how about using the previous parts instead of braking your double angles.
Reply 2
Avatar for davros
davros
16
Original post by creativebuzz

I managed to get the whole question right except for part c, but I can't see where I went wrong..



I can't somment on your attempted 'solution', but as TeeEm says, the obvious / expected thing to do here is to use the result of part (a)!
Original post by TeeEm
how about using the previous parts instead of braking your double angles.


That is what I did..
Reply 4
Avatar for TeeEm
TeeEm
19
Original post by creativebuzz
That is what I did..


I am very busy at the moment but you are in luck since Davros is around
Original post by creativebuzz
That is what I did..


The error is where you said ab=5 so a is 5 and b is 1


However .... There was no need to use the double angle formula at all

You should have used part (a) straight away

And have Rsin(2x + alpha) = 2

Your R and alpha would stay the same as part (a)
(edited 9 years ago)
Original post by TenOfThem
The error is where you said ab=5 so a is 5 and b is 1


However .... There was no need to use the double angle formula at all

You should have used part (a) straight away

And have Rsin(2x + alpha) = 2

Your R and alpha would stay the same as part (a)


Where did I do that?

But thank you, I understand what you mean :smile:

Oh and can you adjust the angle inside Rsin(x+a) for anything? e.g. if the equation was, lets say, 3cos1/2x + 4sin1/2x, would that be Rsin(1/2x+a)?
(edited 9 years ago)
Original post by creativebuzz
Where did I do that?

But thank you, I understand what you mean :smile:

Oh and can you adjust the angle inside Rsin(x+a) for anything? e.g. if the equation was, lets say, 3cos1/2x + 4sin1/2x, would that be Rsin(1/2x+a)?


Yes
Reply 8
Avatar for davros
davros
16
Original post by creativebuzz
Where did I do that?



You basically had 2cosX(5sin(X + 0.64)) = 5

and then somehow 'deduced' that cos X = 5/2 or sin(X + 0.64) =1

- why did you think this was valid?
Original post by davros
You basically had 2cosX(5sin(X + 0.64)) = 5

and then somehow 'deduced' that cos X = 5/2 or sin(X + 0.64) =1

- why did you think this was valid?



Ohh! I don't know! :P

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