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

Salvatore Boone 2021-12-27 Answered
What is \(\displaystyle{\sin{{\left({x}\right)}}}+{\cos{{\left({x}\right)}}}\) in terms of sine?

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Expert Answer

Bubich13
Answered 2021-12-28 Author has 4611 answers
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
Answered 2021-12-29 Author has 4237 answers
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
\(\displaystyle{R}{\cos{\alpha}}={1}\)
\(\displaystyle{R}{\sin{\alpha}}={1}\)
Squaring and adding, we get
\(\displaystyle{R}^{{2}}{{\cos}^{{2}}\alpha}+{R}^{{2}}{{\sin}^{{2}}\alpha}={2}\ \text{ so }\ {R}^{{2}}{\left({{\cos}^{{2}}\alpha}+{{\sin}^{{2}}\alpha}\right)}={2}\)
\(\displaystyle{R}=\sqrt{{{2}}}\)
And now
\(\displaystyle{\cos{\alpha}}={\frac{{{1}}}{{\sqrt{{2}}}}}\)
\(\displaystyle{\sin{\alpha}}={\frac{{{1}}}{{\sqrt{{2}}}}}\)
\(\displaystyle\text{So }\ \alpha={{\cos}^{{-{1}}}{\left({\frac{{{1}}}{{\sqrt{{2}}}}}\right)}}={\frac{{\pi}}{{{4}}}}\)
\(\displaystyle{\sin{{x}}}+{\cos{{x}}}=\sqrt{{2}}{\sin{{\left({x}+{\frac{{\pi}}{{{4}}}}\right)}}}\)
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Vasquez
Answered 2022-01-08 Author has 9054 answers

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 \(\sin x + \sin ( \frac{\pi}{2} – x).\)

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