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C3, Exam question (trig)

Please help me solve 9iii,
Attachment not found
(edited 6 years ago)
Original post by joyoustele
Please help me solve 9iii,
Attachment not found


not able to see it...

could you post it in imagepost.org and link from there ?
Original post by joyoustele
Please help me solve 9iii,


Just use https://imgur.com/ or something to upload images then link them here. By now it should be obvious that TSR's image upload procedure is terrible.
Reply 3
Original post by the bear
not able to see it...

could you post it in imagepost.org and link from there ?


https://imgur.com/a/gHTjd
Reply 4
Original post by RDKGames
Just use https://imgur.com/ or something to upload images then link them here. By now it should be obvious that TSR's image upload procedure is terrible.


https://imgur.com/a/gHTjd
Original post by joyoustele
Please help me solve 9iii,


For 9iii you need to use the identity in part (i) and obtain a quadratic in tanθ\tan \theta while k2k^2 would just be some constant. From there, you want to show that b24ac>0b^2 - 4ac > 0. This is C1 stuff, except you got tan as the variable.
Reply 6
Original post by RDKGames
For 9iii you need to use the identity in part (i) and obtain a quadratic in tanθ\tan \theta while k2k^2 would just be some constant. From there, you want to show that b24ac>0b^2 - 4ac > 0. This is C1 stuff, except you got tan as the variable.


Ohhh...Thanks
Reply 7
Original post by RDKGames
For 9iii you need to use the identity in part (i) and obtain a quadratic in tanθ\tan \theta while k2k^2 would just be some constant. From there, you want to show that b24ac>0b^2 - 4ac > 0. This is C1 stuff, except you got tan as the variable.


That doesnt work, here is what the Answer says, However I dont understand the bit in red. (I got the equation)

9iii answer.JPG
Original post by joyoustele
That doesnt work, here is what the Answer says, However I dont understand the bit in red. (I got the equation)


It does work it's just a different approach and requires a bit of alternative thinking, though I did misread the question initially so this isn't the best approach.

As for the red bit, the RHS is always positive - this should be obvious - and so the root of that will yield that tanθ\tan \theta can be strictly +ve or strictly -ve. In either case, there will be a single solution for each (consider the graph of tanθ\tan \theta for 0<θ<π0 < \theta < \pi, which coveres the +ve and -ve values exactly once.)
Reply 9
Original post by RDKGames
It does work it's just a different approach and requires a bit of alternative thinking, though I did misread the question initially so this isn't the best approach.

As for the red bit, the RHS is always positive - this should be obvious - and so the root of that will yield that tanθ\tan \theta can be strictly +ve or strictly -ve. In either case, there will be a single solution for each (consider the graph of tanθ\tan \theta for 0<θ<π0 < \theta < \pi, which coveres the +ve and -ve values exactly once.)


What is a quadrant
(edited 6 years ago)
Original post by joyoustele
What is a quadrant


It's probably referring to CAST diagram. I don't use them so I can't say.
Original post by RDKGames
It's probably referring to CAST diagram. I don't use them so I can't say.


alright thanks :smile:

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