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Question: Find the maximum and minimum value of \(12\sin \theta -9{{\sin }^{2}}\theta \)....

Find the maximum and minimum value of 12sin⁡θ−9sin⁡2θ12\sin \theta -9{{\sin }^{2}}\theta .

Explanation

Solution

First, before proceeding for this, we must see that the given function be in the form of the perfect square. Then, to get the maximum or minimum value from the question, we want the term of the perfect square to be zero. Then, by substituting the value of sin⁡θ\sin \theta as 23\dfrac{2}{3} in the function given in the question to get the maximum value. Then, to get the minimum value of the function, we need to get a sine function minimum which is only possible when θ=0∘\theta ={{0}^{\circ }}.

Complete step-by-step answer :
In this question, we are supposed to find the maximum and minimum value of 12sin⁡θ−9sin⁡2θ12\sin \theta -9{{\sin }^{2}}\theta .
So, before proceeding for this, we must see that the given function be in the form of the perfect square.
Now, by seeing the equation we get to know that by adding and subtracting 4 from the function gives the perfect square as:
12sin⁡θ−9sin⁡2θ−4+4 ⇒4−(4−12sin⁡θ+9sin⁡2θ) ⇒4−(2−3sin⁡θ)2 \begin{aligned} & 12\sin \theta -9{{\sin }^{2}}\theta -4+4 \\\ & \Rightarrow 4-\left( 4-12\sin \theta +9{{\sin }^{2}}\theta \right) \\\ & \Rightarrow 4-{{\left( 2-3\sin \theta \right)}^{2}} \\\ \end{aligned}
Now, to get the maximum or minimum value from the question, we want the term of the perfect square to be zero.
So, to fulfil this condition, we proceed as:
(2−3sin⁡θ)2=0{{\left( 2-3\sin \theta \right)}^{2}}=0
Now, solve the above expression to get the maximum value of the function as:
2−3sin⁡θ=0 ⇒2=3sinθ ⇒sin⁡θ=23 \begin{aligned} & 2-3\sin \theta =0 \\\ & \Rightarrow 2=3sin\theta \\\ & \Rightarrow \sin \theta =\dfrac{2}{3} \\\ \end{aligned}
Now, by substituting the value of sin⁡θ\sin \theta as 23\dfrac{2}{3} in the function given in the question to get the maximum value.
So, the maximum value is as:
12(23)−9(23)212\left( \dfrac{2}{3} \right)-9{{\left( \dfrac{2}{3} \right)}^{2}}
Now, by solving the above expression to get the maximum value as:
12(23)−9(49)=8−4 ⇒4 \begin{aligned} & 12\left( \dfrac{2}{3} \right)-9\left( \dfrac{4}{9} \right)=8-4 \\\ & \Rightarrow 4 \\\ \end{aligned}
So, the maximum value of the function is 4.
Now, to get the minimum value of the function, we need to get sine function minimum which is only possible when θ=−90∘\theta =-{{90}^{\circ }} or 270∘{{270}^{\circ }} which gives the value assin⁡θ=−1\sin \theta =-1.
So, we get the minimum function value at sin⁡θ=−1\sin \theta =-1as:
4−(2−3(−1))2=4−(2+3)2 ⇒4−(5)2 ⇒4−25 ⇒−21 \begin{aligned} & 4-{{\left( 2-3\left( -1 \right) \right)}^{2}}=4-{{\left( 2+3 \right)}^{2}} \\\ & \Rightarrow 4-{{\left( 5 \right)}^{2}} \\\ & \Rightarrow 4-25 \\\ & \Rightarrow -21 \\\ \end{aligned}
So, the minimum value of the function is -21.
Hence, the maximum and minimum value of the function 12sin⁡θ−9sin⁡2θ12\sin \theta -9{{\sin }^{2}}\theta is 4 and -21 respectively.

Note : To solve these types of the questions we need to know some of the basics of the trigonometric functions and some rules of the maximum and minimum value of the functions. So, some facts of the maximum and minimum value of the function is that the double differentiation will give the values as:
d2ydx2>0\dfrac{{{d}^{2}}y}{d{{x}^{2}}}>0 gives minimum value and d2ydx2<0\dfrac{{{d}^{2}}y}{d{{x}^{2}}}<0 gives maximum value.