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# C3, Exam question (trig) watch

2. (Original post by joyoustele)
not able to see it...

could you post it in imagepost.org and link from there ?
3. (Original post by joyoustele)
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.
4. (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
5. (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
6. (Original post by joyoustele)
For 9iii you need to use the identity in part (i) and obtain a quadratic in while would just be some constant. From there, you want to show that . This is C1 stuff, except you got tan as the variable.
7. (Original post by RDKGames)
For 9iii you need to use the identity in part (i) and obtain a quadratic in while would just be some constant. From there, you want to show that . This is C1 stuff, except you got tan as the variable.
Ohhh...Thanks
8. (Original post by RDKGames)
For 9iii you need to use the identity in part (i) and obtain a quadratic in while would just be some constant. From there, you want to show that . 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)

9. (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 can be strictly +ve or strictly -ve. In either case, there will be a single solution for each (consider the graph of for , which coveres the +ve and -ve values exactly once.)
10. (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 can be strictly +ve or strictly -ve. In either case, there will be a single solution for each (consider the graph of for , which coveres the +ve and -ve values exactly once.)
11. (Original post by joyoustele)
It's probably referring to CAST diagram. I don't use them so I can't say.
12. (Original post by RDKGames)
It's probably referring to CAST diagram. I don't use them so I can't say.
alright thanks
13. (Original post by joyoustele)
https://dryuc24b85zbr.cloudfront.net...=1493912512633
14. Lol, you've got the poundland picture as your profile

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