Solve \sin(x)+\cos(x)=1

Solve $\mathrm{sin}\left(x\right)+\mathrm{cos}\left(x\right)=1$
You can still ask an expert for help

• 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

Jeffery Autrey
$\mathrm{sin}\left(A+B\right)=\mathrm{sin}A\mathrm{cos}B+\mathrm{sin}B\mathrm{cos}A$
${\mathrm{sin}}^{2}A+{\mathrm{cos}}^{2}A=1$
Compare this equation to
$r\mathrm{sin}\left(x+a\right)=1$
$r\mathrm{sin}x\mathrm{cos}a+r\mathrm{cos}x\mathrm{sin}a=1$
$\mathrm{sin}x+\mathrm{cos}x=1$
$\therefore {\mathrm{cos}}^{2}a+{\mathrm{sin}}^{2}a=\frac{1}{{r}^{2}}+\frac{1}{{r}^{2}}=\frac{2}{{r}^{2}}=1$
${r}^{2}=2⇒r=\sqrt{2}$
And $\mathrm{tan}a=1⇒a=\frac{\pi }{4}$
Therefore,
$2\mathrm{sin}\left(x+\frac{\pi }{4}\right)=1$
$\mathrm{sin}\left(x+\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$
$x+\frac{\pi }{4}=\frac{\pi }{4}+2k\pi ⇒x=2k\pi$
Solutions: $S=\left\{2k\pi ,\frac{\pi }{2}+2k\pi \right\},k\in Z$
Not exactly what you’re looking for?
Laura Worden
${\left(\mathrm{sin}\left(x\right)+\mathrm{cos}\left(x\right)\right)}^{2}={1}^{2}$
${\mathrm{sin}}^{2}\left(x\right)+2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)=1$
Use this identity ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{\theta }=1$
$1+2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)=1$
$2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)=0$
Use the identity $2\mathrm{sin}\theta \mathrm{cos}\theta =\mathrm{sin}2\theta$
$\mathrm{sin}2x=0$
$2x=0,\pi$
$x=2\pi n,\frac{\pi }{2}+2\pi n$