Salvatore Boone

Answered

2021-12-27

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

Answer & Explanation

Bubich13

Expert

2021-12-28Added 36 answers

Explanation:

Using Pythagorean Identity

${\mathrm{sin}}^{2}x+{\mathrm{cos}}^{2}x=1,\text{}\text{so}\text{}{\mathrm{cos}}^{2}x=1-{\mathrm{sin}}^{2}x$

$\mathrm{cos}x=\pm \sqrt{1-{\mathrm{sin}}^{2}x}$

$\mathrm{sin}x+\mathrm{cos}x=\mathrm{sin}x\pm \sqrt{1-{\mathrm{sin}}^{2}x}$

Using complement / cofunction identity

$\mathrm{cos}x=\mathrm{sin}(\frac{\pi}{2}-x)$

$\mathrm{sin}x+\mathrm{cos}x=\mathrm{sin}x+\mathrm{sin}(\frac{\pi}{2}-2)$

Using Pythagorean Identity

Using complement / cofunction identity

lovagwb

Expert

2021-12-29Added 50 answers

Explanation

Suppose that$\mathrm{sin}x+\mathrm{cos}x=R\mathrm{sin}(x+\alpha )$

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

$R\mathrm{cos}\alpha =1$

$R\mathrm{sin}\alpha =1$

Squaring and adding, we get

${R}^{2}{\mathrm{cos}}^{2}\alpha +{R}^{2}{\mathrm{sin}}^{2}\alpha =2\text{}\text{so}\text{}{R}^{2}({\mathrm{cos}}^{2}\alpha +{\mathrm{sin}}^{2}\alpha )=2$

$R=\sqrt{2}$

And now

$\mathrm{cos}\alpha =\frac{1}{\sqrt{2}}$

$\mathrm{sin}\alpha =\frac{1}{\sqrt{2}}$

$\text{So}\text{}\alpha ={\mathrm{cos}}^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$

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

Suppose that

Then

The coefficients of

Squaring and adding, we get

And now

Vasquez

Expert

2022-01-08Added 457 answers

Explanation:

Using pythagorean identity,

By taking square root on both the sides,

Using complement or cofunction identity,

Thus, the expression for

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