I was trying to find the least value of |sin⁡x|+|cos⁡x| and applied the above inequality as : |sin⁡x+cos⁡x|≤|sin⁡x|+|cos⁡x| ⇒2≤|sin⁡x|+|cos⁡x| But the range of the given function is [1,2].

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jean2098 · Accepted Answer

It doesn&#039;t hold iff x and y have the same sign. As for your problem remember that if a≤1 then a2≤a so|sin⁡x|+|cos⁡x|≥|sin⁡x|2+|cos⁡x|2=1and for upper bound (remember (a+b)2≤2(a2+b2)):|sin⁡x|+|cos|x≤2(|sin⁡x|2+|cos⁡x|2)=2

yotaniwc · Accepted Answer

Vasquez · Accepted Answer

Let {f(x)=sin⁡x+cos⁡xg(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.

I was trying to find the least value of |\sin

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