Question 1009849
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If 3cosx - 4sinx = 5, then 3sinx + 4cosx =?
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3*cosx - 4*sinx = 5.            (1)

Let us rewrite this equation in the form

{{{3/5}}}.cos(x) - {{{4/5}}}.sin(x) = 1.     (2)

Now consider the right-angled triangle with the sides (3,4,5), and let {{{alpha}}} be its smallest of two acute angles.

Then {{{sin(alpha)}}} = {{{3/5}}} and {{{cos(alpha)}}} = {{{4/5}}},

so we can rewrite the left side of the equation (2) in the form 

{{{sin(alpha)*cos(x) - cos(alpha)*sin(x)}}} = 1,   or

{{{sin(alpha - x)}}} = 1.

It implies that 

{{{alpha - x}}} = {{{pi/2}}}   and

{{{cos(alpha - x)}}} = 0.                  (3)

But, from the other side, 

{{{cos(alpha-x)}}} = {{{cos(alpha)*cos(x)}}} + {{{sin(alpha)*sin(x)}}}.

Now substitute here {{{cos(alpha)}}} = {{{4/5}}} and {{{sin(alpha)}}} = {{{3/5}}}, and you will get

{{{(4/5)*cos(x)}}} + {{{(3/5)*sin(x)}}} = {{{0}}}.   (4)

As a last step, multiply both sides of (4) by 5, and you will get

4*cos(x) + 3*sin(x) = 0.

The problem is solved.

<U>Answer</U>.  If 3cosx - 4sinx = 5, then 3sinx + 4cosx = 0.
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