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The initial phase angle for i=10sinωt+8cosωt is
(A) tan1(45)
(B) tan1(54)
(C) sin1(45)
(D) 900

Answer
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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=10sinωt+8cosωt
Now, For the initial value, we take time t=0
Taking here, we get
 i=10sin(0)+8cos(0)
We know,
 sin(0)=0 And cos(0)=1
Thus, we get
 i=8(1)
Further, we get
 i=8
Now,
 io=(10)2+(8)2
Further, we get
 io=164
Where, io is the amplitude of the motion.
Now,
As per the generic equation of such motion,
 i=iosin(ωt+ϕ)
For time t=0 ,
 i=i0sinϕ
Then, we get
 sinϕ=iio
Thus, we get
 sinϕ=8164
Thus,
 tanϕ=816464
Thus,
 tanϕ=810
Thus,
 tanϕ=45
Hence, we get
 ϕ=tan1(45)
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.