I've been in the trouble to deduce the right term

regatamin2 2022-01-03 Answered
I've been in the trouble to deduce the right term from the left term.
\(\displaystyle{a}{\cos{{\left(\theta\right)}}}+{b}{\sin{{\left(\theta\right)}}}=\sqrt{{{a}^{{2}}+{b}^{{2}}}}\cdot{\sin{{\left(\theta+\phi\right)}}}\)
\(\displaystyle{\tan{{\left(\phi\right)}}}={\frac{{a}}{{b}}}\)
The textbook says that the right term can be gained using the trigonometric addition theorem.However I have no idea what should I do for next.

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Expert Answer

John Koga
Answered 2022-01-04 Author has 4537 answers
Hint:
\(\displaystyle{\tan{\phi}}={\frac{{a}}{{b}}}\)
\(\displaystyle{\frac{{{\sin{\phi}}}}{{{a}}}}={\frac{{{\cos{\phi}}}}{{{b}}}}=\pm\sqrt{{{\frac{{?}}{{{a}^{{2}}+{b}^{{2}}}}}}}\)
Now replace the values of a,b
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censoratojk
Answered 2022-01-05 Author has 5464 answers
HINT
Begin by assuming that
\(\displaystyle{a}{\cos{\theta}}+{b}{\sin{\theta}}\)
can be written in the form
\(\displaystyle{R}{\sin{{\left({x}+\phi\right)}}}\)
for some constants R and \(\displaystyle\phi\). Expand the second expression using the angle addition formula, form some equations, and you should have it.
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Vasquez
Answered 2022-01-08 Author has 9499 answers

\(\begin{array}{}\tan \phi=\frac ab \\Thus, \\\cos \phi=\frac{b}{\sqrt{a^2+b^2}} \\\sin \phi=\frac{a}{\sqrt{a^2+b^2}} \\L.H.S. \\a \cos \theta+ b \sin \theta=\sqrt{a^2+b^2} \frac{(a \cos \theta+b \sin \theta)}{\sqrt{a^2+b^2}} \\=\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\cos \theta+\frac{b}{\sqrt{a^2+b^2}}\sin \theta) \end{array}\)

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