Question

Find value of A when $3\sin \theta + 4\cos \theta = A\sin \left( {\theta + \phi } \right)$.

Answer 1

We will find the solution of this type of problem by substitution method. In this type of problem $\left( {a\sin \theta + b\cos \theta } \right)$ we substitute ‘a’ as $r\cos \phi \,\,and\,\,\,b\,\,as\,\,r\sin \phi$ and then on squaring and adding finding value of ‘r’ and on dividing them we will get$\phi$. Using these in the given equation on comparing we can get the value of A and hence the required solution.

Formula used: $\sin (A + B) = \sin A\cos B + \cos A\sin B$

Complete Answer:
We have
$3\sin \theta + 4\cos \theta = A\sin \left( {\theta + \phi } \right)$
Taking,
$\Rightarrow 3 = r\cos \phi \,..................(i) \\\ \,and\, \\\ \Rightarrow \,4 = r\sin \phi ..................(ii) \\\$
In the above equation. We have
$\Rightarrow r\cos \phi \sin \theta + r\sin \phi \cos \theta = A\sin \left( {\theta + \phi } \right)$
taking ’r’ common from left side
$\Rightarrow r\left( {\cos \phi \sin \theta + \sin \phi \cos \theta } \right) = A\sin \left( {\theta + \phi } \right) \\\ \Rightarrow r\sin \left( {\theta + \phi } \right) = A\sin \left( {\theta + \phi } \right).............................(iii) \\\$
From (i) and (ii), on squaring and adding. We have
$\Rightarrow {(3)^2} = {\left( {r\cos \phi } \right)^2}, \\\ \Rightarrow {(4)^2} = {\left( {r\sin \phi } \right)^2}, \\\ \\\ \Rightarrow {(3)^2} + {(4)^2} = {\left( {r\cos \phi } \right)^2} + {\left( {r\sin \phi } \right)^2} \\\ \Rightarrow 9 + 16 = {r^2}{\cos ^2}\phi + {r^2}{\sin ^2}\phi \\\ or \\\ 25 = {r^2}\left( {{{\cos }^2}\phi + {{\sin }^2}\phi } \right) \\\ \Rightarrow 25 = {r^2} \\\ or\, \\\ {r^2} = 25 \\\ \Rightarrow r = 5 \\\$
Also, on dividing (ii) by (i) we have,
$\Rightarrow \dfrac{{r\sin \phi }}{{r\cos \phi }} = \dfrac{4}{3} \\\ \Rightarrow \tan \phi = \dfrac{4}{3} \\\ \Rightarrow \phi = {\tan ^{ - 1}}\left( {\dfrac{4}{3}} \right) \\\$
Using ‘r’ and $\phi$in (iii) we have
$\Rightarrow 5\sin \left( {\theta + \phi } \right) = A\sin \left( {\theta + \phi } \right) \\\ \Rightarrow A = 5 \\\$
Hence, from above we see that required value of A is $5.$

Note: In this problem we can take substitution as either $a = r\cos \phi \,\,and\,\,\,b = r\sin \phi$ or $a = r\sin \phi \,\,and\,\,b = r\cos \phi$ and on solving along either way, results remains same in both cases. Hence, there are two different substitutions but results at the end will be the same for both.