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Oct 02 P3 maths
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Sure someone will... 
for Q5 (iii):
to find the greatest value of a fraction, we need to find the least value of the denominator.
so the denominator we have here is (4sinθ -3cosθ +6) which, as we found out in (i) can also be written as [5sin(θ-36.9)] so now the denominator is [ 5sin(θ-36.9) + 6 ] since the least value for sin(θ-36.9) is -1, the least value for 5sin(θ-36.9) is -5. so the least value for the whole denominator then is (-5 + 6 = 1). take the reciprocal (well, it doesn't matter in this case but in other questions it might) and the greatest value of 1/(4sinθ -3cosθ +6) is 1.
i hope that's the right answer, i didn't check...
to find the greatest value of a fraction, we need to find the least value of the denominator.
so the denominator we have here is (4sinθ -3cosθ +6) which, as we found out in (i) can also be written as [5sin(θ-36.9)] so now the denominator is [ 5sin(θ-36.9) + 6 ] since the least value for sin(θ-36.9) is -1, the least value for 5sin(θ-36.9) is -5. so the least value for the whole denominator then is (-5 + 6 = 1). take the reciprocal (well, it doesn't matter in this case but in other questions it might
) and the greatest value of 1/(4sinθ -3cosθ +6) is 1.i hope that's the right answer, i didn't check...