Salvatore Boone

2021-12-27

What is $\mathrm{sin}\left(x\right)+\mathrm{cos}\left(x\right)$ in terms of sine?

Bubich13

Explanation:
Using Pythagorean Identity

$\mathrm{cos}x=±\sqrt{1-{\mathrm{sin}}^{2}x}$
$\mathrm{sin}x+\mathrm{cos}x=\mathrm{sin}x±\sqrt{1-{\mathrm{sin}}^{2}x}$
Using complement / cofunction identity
$\mathrm{cos}x=\mathrm{sin}\left(\frac{\pi }{2}-x\right)$
$\mathrm{sin}x+\mathrm{cos}x=\mathrm{sin}x+\mathrm{sin}\left(\frac{\pi }{2}-2\right)$

lovagwb

Explanation
Suppose that $\mathrm{sin}x+\mathrm{cos}x=R\mathrm{sin}\left(x+\alpha \right)$
Then
$\mathrm{sin}x+\mathrm{cos}x=R\mathrm{sin}x\mathrm{cos}\alpha +R\mathrm{cos}x\mathrm{sin}\alpha$
$=\left(R\mathrm{cos}\alpha \right)\mathrm{sin}x+\left(R\mathrm{sin}\alpha \right)\mathrm{cos}x$
The coefficients of $\mathrm{sin}x$ and of $\mathrm{cos}x$ must be equal so
$R\mathrm{cos}\alpha =1$
$R\mathrm{sin}\alpha =1$

$R=\sqrt{2}$
And now
$\mathrm{cos}\alpha =\frac{1}{\sqrt{2}}$
$\mathrm{sin}\alpha =\frac{1}{\sqrt{2}}$

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

Vasquez

Explanation:
Using pythagorean identity,
${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1$

By taking square root on both the sides,
$\mathrm{cos}x+\mathrm{sin}x=\mathrm{sin}±\sqrt{1-{\mathrm{sin}}^{2}x}$
Using complement or cofunction identity,
$\mathrm{cos}x=\mathrm{sin}\left(\frac{\pi }{2}-x\right)$
$\mathrm{sin}x+\mathrm{cos}x=\mathrm{sin}x+\mathrm{sin}\left(\frac{\pi }{2}-x\right)$
Thus, the expression for $\mathrm{sin}x+\mathrm{cos}x$ in terms of sine is $\mathrm{sin}x+\mathrm{sin}\left(\frac{\pi }{2}–x\right).$

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