Find domain and range of $\log_2\left(\sin(2x)+\cos(2x)\right)$

Question

I'm having trouble finding domain and range of this function.

Can somebody give me a HINT please? Thanks.

Answer 1

\begin{align} \cos(2x)+\sin(2x) &=\cos(2x)\,r\cos \phi + \sin(2x)\,r\sin \phi \\ &= r\cos(2x-\phi) \end{align}

\begin{cases} r\cos \phi =1\\ r\sin \phi =1 \end{cases}

Finding $r$: \begin{align} r^2 (\sin^2\phi +\cos^2 \phi) &= 1^2+1^2\\ r &=\sqrt{2} \end{align}

Finding $\phi$: \begin{align} \frac{r\sin \phi}{r\cos \phi} &= \frac{1}{1}\\ \tan \phi &= 1\\ \phi &= 45^\circ \end{align}

Return to the main. Argument of logarithm must be positive.

\begin{align} \cos(2x)+\sin(2x)&>0\\ \sqrt{2}\cos(2x-45^\circ) &>0\\ \cos(2x-45^\circ) &>0 \end{align}

\begin{align} -90^\circ + 360^\circ k < 2x-45^\circ < 90^\circ + 360^\circ k\\ -45^\circ + 360^\circ k < 2x < 135^\circ + 360^\circ k\\ -22.5^\circ + 180^\circ k < x < 67.5^\circ + 180^\circ k \end{align}

Answer 2

$$ \log_2(\sin(2x)+\cos(2x)), \\ \bbox[lightgreen] { (\sin(2x)+\cos(2x))>0, \\ -1\le\sin(2x)\le{1}, \\ -1\le\cos(2x)\le{1} }. $$ Also don't forget to use this formula: $$ \bbox[pink] { a\sin(x)+b\cos(x)= \sqrt{a^2+b^2}\sin\left(x+\arctan\left(\frac{b}{a}\right)\right) }. $$

Answer 3

Hint: one way to find the range can be to try expressing $sin(2x)+cos(2x)$ as a single (transformed) sine function