Question 1088220: the max value of sin(x+π/6)+cos(x+π÷6)in the intervals (0,π÷2)is attained at Found 2 solutions by htmentor, natolino_2017:Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! If all we are looking for is the maximum value, then a shift in x of pi/6 will not change the amplitude, only the phase.
Therefore, we only need to find the maximum value of the function sin(x) + cos(x)
The max. value is obtained where df/dx = 0 = cos(x) - sin(x)
Thus tan(x) = 1 -> x = pi/4
f(pi/4) = 1/sqrt(2) + 1/sqrt(2) = 2/sqrt(2) = sqrt(2)
Ans: sqrt(2)
You can put this solution on YOUR website! let f(x) = sin(x+π/6)+cos(x+π÷6)
so f'(x) = cos(x+π/6)-sin(x+π÷6) = 0
tan(x+π/6) = 1 so x = π÷12
Which is stationary point.
also frontier's poitns are candidate.
f(0) = 1/2 +sqrt(3)/2 = (1+sqrt(3))/2.
f(π÷2) = sin(2π/3)+cos(2π/3) = sqrt(3)/2 -1/2 = (sqrt(3)-1)/2 (min).
f(π÷12) = sin(π/4)+cos(π/4) = sqrt(2)/2+sqrt(2)/2 = sqrt(2) (max).
Max is sqrt(2), Min is (sqrt(3)-1)/2, all this not using
sinusoid's properties.
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