# Range of trig function

• Jan 2nd 2011, 01:59 PM
atreyyu
Range of trig function
Find the range of $f(x)=\sin ^2 x (\sin (2x) +1)$.

Ideas?
• Jan 2nd 2011, 02:03 PM
dwsmith
What do you know about sine?
• Jan 2nd 2011, 02:14 PM
atreyyu
Well, enough to know that even though the range of sine is <-1;1>, the range of f will not be $1^2\cdot(1+1)$...
What else would be any help?
• Jan 2nd 2011, 02:18 PM
dwsmith
Quote:

Originally Posted by atreyyu
Find the range of $f(x)=\sin ^2 x (\sin 2x +1)$.

Ideas?

Also, what do you mean here?

$\sin{(2x+1)} \ \mbox{or} \ \sin{(2x)}+1$
• Jan 2nd 2011, 02:33 PM
dwsmith
$\sin{2x}=2\sin{x}\cos{x}$

$2\sin^2{x}\sin{x}\cos{x}+\sin^2{x}$

$\sin{x}\cos{x}\neq\pm 1$

$0\leq\sin^2{x}\leq 1$

If $\sin^2{x}=1$, then $2\sin^2{x}\sin{x}\cos{x}=0$.

What is the max value $2\sin^2{x}\sin{x}\cos{x}$ this can be?

Once you determine the x value that obtains the max, x_0, of $2\sin^2{x}\sin{x}\cos{x}$, then $f(x_0)=2\sin^2{x_0}\sin{x_0}\cos{x_0}+\sin^2{x_0}= y_0$

The question then remains is $y_0>1\mbox{?}$
• Jan 2nd 2011, 02:37 PM
mr fantastic
Quote:

Originally Posted by atreyyu
Find the range of $f(x)=\sin ^2 x (\sin (2x) +1)$.

Ideas?

Are you expected to use calculus to find turning points?
• Jan 2nd 2011, 06:29 PM
SammyS
Quote:

Originally Posted by atreyyu
Find the range of $f(x)=\sin ^2 x (\sin (2x) +1)$.

Ideas?

It's clear that f(x) has a minimum value of zero; at x=0 for instance.

$\displaystyle \sin(2x) = 2(\sin x)(\cos x)$ and $\displaystyle 1= \sin^2 x+\cos^2 x$

Then, $\displaystyle \sin(2x)+1 = \sin^2+2(\sin x)(\cos x)+\cos^2 x$

$\displaystyle \text{ } = (\sin x+ \cos x)^2$
At this point we have that:

$\displaystyle f(x)=(\sin ^2 x) (\sin (2x) +1)=(\sin ^2 x)(\sin x+ \cos x)^2\,.$

A not so well known trig identity gives:

$\sin x+ \cos x=\sqrt{2}\ \sin\left(x+{{\pi}\over{4}}\right)$

We now have: $\displaystyle f(x)=(\sin ^2 x)\left(\sqrt{2}\ \sin\left(x+{{\pi}\over{4}}\right)\right)^2=2(\sin ^2 x) \sin^2\left(x+{{\pi}\over{4}}\right) \,.$

Now use a product to sum identity on: $\displaystyle (\sin x) \sin\left(x+{{\pi}\over{4}}\right)$
$\displaystyle = {{1}\over{2}}\left[\cos\left({{\pi}\over{4}}\right) - \cos\left((2x+{{\pi}\over{4}}\right) \right]$

$\displaystyle = {{1}\over{2}}\left[\left({{1}\over{\sqrt{2}}}\right) - \cos\left((2x+{{\pi}\over{4}}\right) \right]$

Finally, $\displaystyle f(x)=2\cdot {{1}\over{4}}\left[\left({{1}\over{\sqrt{2}}}\right) - \cos\left((2x+{{\pi}\over{4}}\right) \right]^2$

$\displaystyle ={{1}\over{2}}\left[\left({{1}\over{\sqrt{2}}}\right) - \cos\left((2x+{{\pi}\over{4}}\right) \right]^2$
I suggest finding the minimum and maximum values of the expression inside the brackets. Square those then divide by 2.