I was trying to find the least value of |sin⁡x|+|cos⁡x| and applied the above inequality...

lugreget9

lugreget9

Answered

2022-01-01

I was trying to find the least value of |sinx|+|cosx| and applied the above inequality as :
|sinx+cosx||sinx|+|cosx|
2|sinx|+|cosx|
But the range of the given function is [1,2].

Answer & Explanation

jean2098

jean2098

Expert

2022-01-02Added 38 answers

It doesn't hold iff x and y have the same sign. As for your problem remember that if a1 then a2a so
|sinx|+|cosx||sinx|2+|cosx|2=1
and for upper bound (remember (a+b)22(a2+b2)):
|sinx|+|cos|x2(|sinx|2+|cosx|2)=2
yotaniwc

yotaniwc

Expert

2022-01-03Added 34 answers

|sinx+cosx||sinx|+|cosx| is correct.
But, for example, if x=0 then both sides are 1; thus we cannot conclude 2|sinx|+|cosx|
Vasquez

Vasquez

Expert

2022-01-08Added 457 answers

Let {f(x)=sinx+cosxg(x)=|sin(x)|+|cos(x)|
Notice that g(x+π2)=|cos(x)|+|sin(x)|=g(x) so
g is periodic of period π2 and we can study it on [0,π2].
But on this interval both sin and cos are positive, so we can get rid of absolute values and g(x)=f(x) on [0,π2].
You now can use the addition formulas to transform f to f(x)=2sin(x+π4) and see that f(x) has values between 0 and 2 while g(x) has values between 1 and 2.

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