Question 1204328
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Find the largest possible domain and largest possible range of the function
𝑔(𝑥) = 4 cos(3𝑥) − 3 sin(3𝑥).
Give your answers in set/interval notations.
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<pre>
The domain is, OBVIOUSLY, the set of all real numbers, since this function (this expression) 
is defined over all this set.


To find the range, let's make this identical transformation


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


Next, notice that  {{{(4/5)^2}}} + {{{(3/5)^2}}} = {{{16/25 + 9/25}}} = {{{25/25}}} = 1.


THEREFORE, there is such angle {{{theta}}} that  {{{sin(theta)}}} = {{{4/5}}},  {{{cos(theta)}}} = {{{3/5}}}.

This  {{{theta}}}  is simply the angle in QI, which satisfies this equation  {{{tan(theta)}}} = {{{4/3}}},  or  {{{theta}}} = {{{arctan(4/3)}}}.


Then we can continue the equality (1) this way


      4*cos(3x) - 3*sin(3x) = {{{5*((4/5)*cos(3x) - (3/5)*sin(3x))}}} = {{{5*(sin(theta)*cos(3x) - cos(theta)*sin(3x))}}} = 

          now apply the formula for sine of the sum of arguments   

    = {{{5*sin(theta-3x)}}}.


Thus we presented the original expression as the sine function with amplitude 5 of argument  {{{theta-3x}}}

    4*cos(3x) - 3*sin(3x) = {{{5*sin(theta-3x)}}}.


It tells you that the range of  4*cos(3x) - 3*sin(3x)  is the interval from -5 to 5, or, in the interval form, [-5,5].


<U>ANSWER</U>.  The domain of the given function is the entire number line (-oo,oo).

         The range of the given function is the interval [-5,5].
</pre>

Solved.


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<pre>
This transformation and the logic, which I used, may seem as a focus - pocus.


But actually, it is a general transformation of the expression a*cos(x) - b*sin(x) with real coefficients "a" and "b"
into single harmonic function


    a*sin(x) - b*cos(x) = {{{sqrt(a^2+b^2)*((a/sqrt(a^2+b^2))*cos(x) - (b/sqrt(a^2+b^2))*sin(x)))}}} = 

                        = {{{sqrt(a^2+b^2)*((a/sqrt(a^2+b^2))*cos(x) - (b/sqrt(a^2+b^2))*sin(x))}}} = 

                        = {{{sqrt(a^2+b^2)*(sin(theta)*cos(x) - cos(theta)*sin(x))}}} = {{{sqrt(a^2+b^2)*sin(theta-x)}}}.


where  {{{theta}}} = {{{arctan(a/b)}}}.


It works always for any real coefficients "a" and "b" and transforms any linear combination a*cos(x) + b*sin(x) 

into single harmonic function  {{{sin(theta-x)}}}  with the shift {{{theta)}}} = {{{arctan(a/b)}}}  and the amplitude  {{{sqrt(a^2+b^2)}}}.


It is very useful classic trigonometric transformation and the identity to know and to use in different 
trigonometric problems.


So, it makes sense to learn and to memorize it.
</pre>