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

I was trying to find the least value of $$\displaystyle{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}$$ and applied the above inequality as :
$$\displaystyle{\left|{\sin{{x}}}+{\cos{{x}}}\right|}\leq{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}$$
$$\displaystyle\Rightarrow\sqrt{{2}}\leq{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}$$
But the range of the given function is $$\displaystyle{\left[{1},\sqrt{{2}}\right]}$$.

## Want to know more about Trigonometry?

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

jean2098
It doesn't hold iff x and y have the same sign. As for your problem remember that if $$\displaystyle{a}\leq{1}$$ then $$\displaystyle{a}^{{2}}\leq{a}$$ so
$$\displaystyle{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}\geq{\left|{\sin{{x}}}\right|}^{{2}}+{\left|{\cos{{x}}}\right|}^{{2}}={1}$$
and for upper bound (remember $$\displaystyle{\left({a}+{b}\right)}^{{2}}\leq{2}{\left({a}^{{2}}+{b}^{{2}}\right)}$$):
$$\displaystyle{\left|{\sin{{x}}}\right|}+{\left|{\cos}\right|}{x}\leq\sqrt{{{2}{\left({\left|{\sin{{x}}}\right|}^{{2}}+{\left|{\cos{{x}}}\right|}^{{2}}\right)}}}=\sqrt{{2}}$$
###### Not exactly what you’re looking for?
yotaniwc
$$\displaystyle{\left|{\sin{{x}}}+{\cos{{x}}}\right|}\leq{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}$$ is correct.
But, for example, if x=0 then both sides are 1; thus we cannot conclude $$\displaystyle\sqrt{{2}}\leq{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}$$
Vasquez

Let $$\displaystyle{\left\lbrace\begin{matrix} f{{\left({x}\right)}}= \sin{{x}}+ \cos{{x}}\\ g{{\left({x}\right)}}={\left| \sin{{\left({x}\right)}}\right|}+{\left| \cos{{\left({x}\right)}}\right|}\end{matrix}\right.}$$
Notice that $$g(x+\frac{\pi}{2})=|\cos(x)|+|\sin(x)|=g(x)$$ so
g is periodic of period $$\frac{\pi}{2}$$ and we can study it on $$[0,\frac{\pi}{2}]$$.
But on this interval both sin and cos are positive, so we can get rid of absolute values and g(x)=f(x) on $$[0,\frac{\pi}{2}].$$
You now can use the addition formulas to transform f to $$f(x)=\sqrt2 \sin(x+\frac{\pi}{4})$$ and see that $$f(x)$$ has values between 0 and $$\sqrt2$$ while g(x) has values between 1 and $$\sqrt2$$.