# What is \sin(x)+\cos(x) in terms of sine?

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

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Bubich13
Explanation:
Using Pythagorean Identity
$$\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1},\ \text{ so }\ {{\cos}^{{2}}{x}}={1}-{{\sin}^{{2}}{x}}$$
$$\displaystyle{\cos{{x}}}=\pm\sqrt{{{1}-{{\sin}^{{2}}{x}}}}$$
$$\displaystyle{\sin{{x}}}+{\cos{{x}}}={\sin{{x}}}\pm\sqrt{{{1}-{{\sin}^{{2}}{x}}}}$$
Using complement / cofunction identity
$$\displaystyle{\cos{{x}}}={\sin{{\left({\frac{{\pi}}{{{2}}}}-{x}\right)}}}$$
$$\displaystyle{\sin{{x}}}+{\cos{{x}}}={\sin{{x}}}+{\sin{{\left({\frac{{\pi}}{{{2}}}}-{2}\right)}}}$$
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lovagwb
Explanation
Suppose that $$\displaystyle{\sin{{x}}}+{\cos{{x}}}={R}{\sin{{\left({x}+\alpha\right)}}}$$
Then
$$\displaystyle{\sin{{x}}}+{\cos{{x}}}={R}{\sin{{x}}}{\cos{\alpha}}+{R}{\cos{{x}}}{\sin{\alpha}}$$
$$\displaystyle={\left({R}{\cos{\alpha}}\right)}{\sin{{x}}}+{\left({R}{\sin{\alpha}}\right)}{\cos{{x}}}$$
The coefficients of $$\displaystyle{\sin{{x}}}$$ and of $$\displaystyle{\cos{{x}}}$$ must be equal so
$$\displaystyle{R}{\cos{\alpha}}={1}$$
$$\displaystyle{R}{\sin{\alpha}}={1}$$
$$\displaystyle{R}^{{2}}{{\cos}^{{2}}\alpha}+{R}^{{2}}{{\sin}^{{2}}\alpha}={2}\ \text{ so }\ {R}^{{2}}{\left({{\cos}^{{2}}\alpha}+{{\sin}^{{2}}\alpha}\right)}={2}$$
$$\displaystyle{R}=\sqrt{{{2}}}$$
And now
$$\displaystyle{\cos{\alpha}}={\frac{{{1}}}{{\sqrt{{2}}}}}$$
$$\displaystyle{\sin{\alpha}}={\frac{{{1}}}{{\sqrt{{2}}}}}$$
$$\displaystyle\text{So }\ \alpha={{\cos}^{{-{1}}}{\left({\frac{{{1}}}{{\sqrt{{2}}}}}\right)}}={\frac{{\pi}}{{{4}}}}$$
$$\displaystyle{\sin{{x}}}+{\cos{{x}}}=\sqrt{{2}}{\sin{{\left({x}+{\frac{{\pi}}{{{4}}}}\right)}}}$$
Vasquez

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