Question

find the values of x between 0 and 2pi where the tangent line to the graph of y=sinxcosx is horizontal.

Answer 1

`\bf \textit{let's recall that }sin^2(\theta)+cos^2(\theta)=1 \\\\\\ sin^2(\theta)=1-cos^2(\theta)\implies sin(\theta )=√(1-cos^2(\theta))\\\\ -------------------------------\\\\`


`\bf y=sin(x)cos(x)\impliedby \textit{now, using the product-rule} \\\\\\ \cfrac{dy}{dx}=cos(x)cos(x)-sin(x)cos(x)\implies 0=cos(x)[cos(x)-sin(x)]\\\\ -------------------------------\\\\ 0=cos(x)\implies \boxed{\measuredangle x= \begin{cases} (\pi )/(2)\\\\ (3\pi )/(2) \end{cases}}\\\\ -------------------------------\\\\`


`\bf 0=cos(x)-sin(x)\implies sin(x)=cos(x)\\\\ thus\qquad √(1-cos^2(x))=cos(x)\implies 1-cos^2(x)=cos^2(x) \\\\\\ 1=2cos^2(x)\implies \cfrac{1}{2}=cos^2(x)\implies \pm\sqrt{\cfrac{1}{2}}=cos(x) \\\\\\ \pm\cfrac{1}{√(2)}=cos(x)\implies \pm\cfrac{√(2)}{2}=cos(x)\implies \boxed{\measuredangle x= \begin{cases} (\pi )/(4)\\\\ (3\pi )/(4)\\\\ (5\pi )/(4)\\\\ (7\pi )/(4) \end{cases}}`