trig (1 Viewer)

[fullonoob]

find the max and min values of the expression (2-sin@)/cos@ for 0<_@<_pi/4
and state the values of @ for which they occur
i obtain pi / 6 but am not sure how to get @ = 0 as well, and you have to double diff to show max min right, full working please

 

[Trebla]

find the max and min values of the expression (2-sin@)/cos@ for 0<_@<_pi/4
and state the values of @ for which they occur
i obtain pi / 6 but am not sure how to get @ = 0 as well, and you have to double diff to show max min right, full working please

Let y = (2 - sin @) / cos @ = 2sec @ - tan @
dy/d@ = 2sec @tan @ - sec²@ = sec @(2tan @ - sec @)
For dy/d@ = 0 (sec @ is non-zero)
2tan @ = sec @
=> @ = π/6
(Check this gives a local minimum)
Since there are no other turning points in the domain this is also the global minimum

Since there is no local maximum, you find the global maximum on the boundaries of the interval (provided there is continuity and differentiability in that interval). So check @ = 0 and @ = π/4 and you'll find @ = 0 gives the global maximum

 

[fullonoob]

oh i see but is there a method to find the local max from the restricted domain. I only can obtain @ = 0 as the max from visual inspection and subbing in values. Is there an algebraic method

 

[Trebla]

oh i see but is there a method to find the local max from the restricted domain. I only can obtain @ = 0 as the max from visual inspection and subbing in values. Is there an algebraic method

If there is a maximum turning point then there is obviously a local maximum. If there are none, the global maximum will always be at one of the boundary points given there are no discontinuities in the domain.