.
If 3sinA + 5cosA = 5 then 5sinA - 3cosA equals
~~~~~~~~~~~~~~~
I will give ABSOLUTELY UNUSUAL and UNEXPECTED solution that you never saw/heard before.
1. You are given that
3*sin(A) + 5*cos(A) = 5.
Divide both sides by = . You will get
+ = . (1)
2. Consider vector U = (,) and vector B = (sin(A),cos(A)).
The equality (1) means that the scalar product of these vectors is equal to .
In other words,
|U|*|B|*cos(a) = , (2)
where |U| is the modulus (= the magnitude, the length) of the vector U, |B| is the modulus of the vector B and
"a" is the angle between these two vectors.
But (!!!) the length of the vector U is equal to 1, as well as the length of the vector B.
(Everybody who knows how to calculate the length of the vector, can easily check it).
Therefore, the equality (2) means that the cosine of the angle "a" is equal to :
cos(a) = . (3)
3. Very good. Now let us consider the expression y = 5*sin(A) - 3*cos(A) which is under the question.
Again, divide both sides by . You will get
= . (4)
The right side of (4) is NOTHING ELSE AS the scalar product of the vectors V = (,) and vector B introduced above.
Notice that the vector V has the length of 1 and is orthogonal to vector U.
So, V is the unit vector orthogonal to U.
Therefore, (4) simply is
= cos(b), (5)
where "b" is the angle between the vectors V and B.
4. BUT (!!!), since the vectors U and V are orthogonal, the angles "a" and "b" are complementary:
b = - a.
Then
cos(b) = sin(a) = = = = = = +/-. (6)
Thus = cos(b) = +/- .
It implies that y = +/- 3.
Answer. If 3*sin(A) + 5*cos(A) = 5 then 5*sin(A) - 3*cos(A) = +/- 3.
Solved.
Plots y1 = 3*sin(x) + 5*cos(x) (red), y2 = 5 (green), y3 = 5*sin(x) - 3*cos(x) (blue) and y4 = 3 (purple).
The plot shows very clearly that there are two points in the segment [,] where 3*sin(x) + 5*cos(x) = 5.
One of these points is x = A = 0. But there is the other point, too.
The plot shows very clearly, also, that at the points where 3*sin(x) + 5*cos(x) = 5 we have 5*sin(x) - 3*cos(x) = +/- 3.
It serves as the CHECK.