Question

The initial phase angle for $ i = 10\sin \omega t + 8\cos \omega t $ is
(A) $ {\tan ^{ - 1}}\left( {\dfrac{4}{5}} \right) $
(B) $ {\tan ^{ - 1}}\left( {\dfrac{5}{4}} \right) $
(C) $ {\sin ^{ - 1}}\left( {\dfrac{4}{5}} \right) $
(D) $ {90^0} $

Answer 1

Hint: We are given with an equation and are asked to find the initial phase angle for the same. Thus, we will firstly evaluate the equation at time $ t = 0 $ . Then, we will use some basic trigonometric ideas to manipulate the evaluated value and then come up with an answer.

Complete Step By Step Solution
Here, The given equation is,
 $ i = 10\sin \omega t + 8\cos \omega t $
Now, For the initial value, we take time $ t = 0 $
Taking here, we get
 $ i = 10\sin \left( 0 \right) + 8\cos \left( 0 \right) $
We know,
 $ \sin \left( 0 \right) = 0 $ And $ \cos \left( 0 \right) = 1 $
Thus, we get
 $ i = 8\left( 1 \right) $
Further, we get
 $ i = 8 $
Now,
 $ {i_o} = \sqrt {{{\left( {10} \right)}^2} + {{\left( 8 \right)}^2}} $
Further, we get
 $ {i_o} = \sqrt {164} $
Where, $ {i_o} $ is the amplitude of the motion.
Now,
As per the generic equation of such motion,
 $ i = {i_o}\sin \left( {\omega t + \phi } \right) $
For time $ t = 0 $ ,
 $ i = {i_0}\sin \phi $
Then, we get
 $ \sin \phi = \dfrac{i}{{{i_o}}} $
Thus, we get
 $ \sin \phi = \dfrac{8}{{\sqrt {164} }} $
Thus,
 $ \tan \phi = \dfrac{8}{{\sqrt {164 - 64} }} $
Thus,
 $ \tan \phi = \dfrac{8}{{10}} $
Thus,
 $ \tan \phi = \dfrac{4}{5} $
Hence, we get
 $ \phi = {\tan ^{ - 1}}\left( {\dfrac{4}{5}} \right) $
Hence, the correct option is (A).

Note
We have converted the sine function to a tangent one as all the given options are in the same format. We used basic trigonometry for conversion. One should not confuse it to be a given parameter.