Question

The minimum value of the function $f(x)=(3\sin x-4\cos x-10)(3\sin x+4\cos x-10)$ is
 $A-$ $49$
 $B-$ ${\displaystyle \frac{190-60\sqrt{2}}{2}}$
 $C-$ $84$
 $D-$ $48$

Answer 1

Hint: If we carefully look at the expression it should strike that it is of the form of algebraic expression given by $a^2-b^2=(a+b)(a-b)$ . It is with the properties of trigonometric functions we will be using the algebraic expression. Properties of trigonometric functions we would be using is ${\sin }^2\theta +{\cos }^2\theta =1$ and its restructured form. It should be noted that before even starting the sum , students should immediately notice that this identity i.e. $a^2-b^2=(a+b)(a-b)$ is being used in the given numerical.

Complete step-by-step answer:
We are going to mainly use the following properties in the given numerical
 $a^2-b^2=(a+b)(a-b).........(1)$
 ${\sin }^2\theta +{\cos }^2\theta =1.............(2)$
First step in solving this numerical is to rearrange the given numerical in the form of Equation $1$
The given question after re-arrangement becomes as follows
 $f(x)=((3\sin x-10)-4\cos x)((3\sin x-10)+4\cos x)$
Simplifying the RHS further in the form of $a^2-b^2$ we get
 $\Rightarrow f(x)=(3\sin x-10{)}^2-(4\cos x{)}^2$
Opening the brackets of RHS to further simplify the problem
 $\Rightarrow f(x)=9{\sin }^{2}x+100-60\sin x-16{\cos }^{2}x........(3)$
Since most of the terms in Equation $3$ are having $\sin$ , we will convert the $\cos$ function in Equation $3$ to $\sin$ by using the property Number $2$ as listed above. New expression would now become
 $\Rightarrow f(x)=9{\sin }^{2}x+100-60\sin x-16(1-{\sin }^{2}x)........(4)$
Opening the brackets and on rearranging we get the following equation
$$\Rightarrow f(x)=25{\sin }^{2}x-60\sin x+84........(5)$$
Now since this is an equation involving only $\sin$ , we will have to use a property $(a+b{)}^2=a^2+b^2+2ab$ to further solve the sum. Thus we will have to rearrange equation $5$ in such a way that it is of the form of the above expression.
 $\therefore$ $$\Rightarrow f(x)=(5\sin x-6{)}^2+48..........(6)$$
We have to find the minimum value of the given question. Thus the value would be minimum only if we try to bring the value of the bracket in equation $6$ as small as possible. This will happen when only $\sin$ is maximum . Thus we know that the maximum value of $\sin$ is $1$ . Now the expressions becomes as follows
$$\Rightarrow f(x)=(5-6{)}^2+48..........(7)$$
 $\therefore$ The minimum value of the expression after solving equation $7$ is $49$ .
Thus, the answer is option $A$ .
So, the correct answer is “Option A”.

Note: Though this sum seems a bit easy , it wouldn’t have been possible to solve the sum if the student was unaware of the identities used in the numerical. Also it is extremely important to first carefully look at the sum and think for a minute as to which identities can be used in a particular sum before blindly starting to solve .This would help the student to complete the given question in less time. Always remember that sinx value will always lie between +1 to -1 for every x.