Question

The maximum value of 3/5sinx-12cosx+19

Answer 1

Answer:

Step-by-step explanation:

The given trigonometric expression is :

11 cos^2 x +3 sin^2 x+6sinx cosx +5

or, we can write it as,

(9 cos^2 x + 2 cos^2 x) + (2 sin^2 x + sin^2 x) + 6sinx cosx +5

Again, after rearranging the terms, we can write the whole expression as,

(9 cos^2 x + sin^2 x + 6sinx cosx) + (2 cos^2 x + 2 sin^2 x) + 5

Then if you factor the following underlined section as you would with a polynomial:

(9 cos^2 x + sin^2 x + 6sinx cosx) + (2 cos^2 x + 2 sin^2 x) + 5

You get:

(3 cos x + sin x)^2 + 2 (cos^2 x + sin ^2 x) + 5

Now, the term inside the second bracket (cos^2 x + sin ^2 x) is a very popular trigonometric identity and it's value is equal to one.

So, now the whole expression becomes,

(3 cos x + sin x)^2 +7

Now, the maximum and the minimum value of the whole expression depends upon the maximum and the minimum value of the term (3 cos x + sin x), which is of the form (a cosx + b sinx),

The maximum and minimum value of (a cosx + b sinx) is relatively easy to find.

So, I've attached a screenshot from a relevant document below:

Here, a=3 and b=1,

So, R= √10

As the value of cosine of any angle lies between -1 to 1, so the value of the value of expression cos(x − α) will lie between -1 to 1.

Hence, the maximum and the minimum value of (a cosx + b sinx) will be -R and R and all the values of the expression will lie between them.

i.e., in our case between (-√10) to √10.

Again, coming back to our original expression,

(3 cos x + sin x)^2 +7

The value of the term in bracket will lie between (-√10) and √10.

But, there is a catch here, as the squares of negative terms come out be positive, hence we can't take the negative term to find the minimum value of our expression. the minimum value of the expression will be at the minimum non-negative value in the range, which is zero.

So, the minimum value will be,

(0)^2 + 7=7

and the maximum value will be,

(√10)^2 +7 = 17