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Question: Find the maximum and minimum values of \({\cos ^6}\theta + {\sin ^6}\theta \) respectively. \(\lef...

Find the maximum and minimum values of cos⁡6θ+sin⁡6θ{\cos ^6}\theta + {\sin ^6}\theta respectively.
(a) 1 and 14\left( a \right){\text{ 1 and }}\dfrac{1}{4}
(b) 1 and 0\left( b \right){\text{ 1 and 0}}
(c) 2 and 0\left( c \right){\text{ 2 and 0}}
(d) 1 and 12(d){\text{ 1 and }}\dfrac{1}{2}

Explanation

Solution

Hint: First simplify the expression using various algebraic & trigonometric identities & then use the range of the trigonometric function in the simplified form.

Complete step-by-step answer:
We have to find the maximum and minimum value of cos⁡6θ+sin⁡6θ{\cos ^6}\theta + {\sin ^6}\theta .
So let’s simplify it first,
Let f(θ)=cos⁡6θ+sin⁡6θf(\theta ) = {\cos ^6}\theta + {\sin ^6}\theta .
So we can also write this as
f(θ)=(sin2θ)3+(cos⁡2θ)3f(\theta ) = {\left( {si{n^2}\theta } \right)^3} + {\left( {{{\cos }^2}\theta } \right)^3}
Now using a3+b3=(a+b)(a2+b2−ab){a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right), we have
f(θ)=(sin⁡2θ+cos⁡2θ)(sin⁡4θ+cos⁡4θ−sin⁡2θcos⁡2θ)f(\theta ) = \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right)
Using sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1…………………………… (1)
f(θ)=(sin⁡4θ+cos⁡4θ−sin⁡2θcos⁡2θ)f(\theta ) = \left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right)
Now we can write (sin⁡4θ+cos⁡4θ−sin⁡2θcos⁡2θ)=(sin⁡2θ+cos⁡2θ)2−3sin⁡2θcos⁡2θ\left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta {{\cos }^2}\theta } \right) = {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^2} - 3{\sin ^2}\theta {\cos ^2}\theta
So we can write f(θ)=(sin⁡2θ+cos⁡2θ)2−3sin⁡2θcos⁡2θf(\theta ) = {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^2} - 3{\sin ^2}\theta {\cos ^2}\theta
Now using equation 1 we have
f(θ)=1−3sin⁡2θcos⁡2θf(\theta ) = 1 - 3{\sin ^2}\theta {\cos ^2}\theta
We can write this as
f(θ)=1−34×4sin⁡2θcos⁡2θf(\theta ) = 1 - \dfrac{3}{4} \times 4{\sin ^2}\theta {\cos ^2}\theta
Now using 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta = \sin 2\theta
We have f(θ)=34(sin⁡2θ)2f(\theta ) = \dfrac{3}{4}{\left( {\sin 2\theta } \right)^2}
Using half angle formulae, (1−cos⁡4θ)=2sin⁡22θ\left( {1 - \cos 4\theta } \right) = 2{\sin ^2}2\theta
⇒1−38(1−cos⁡4θ)\Rightarrow 1 - \dfrac{3}{8}\left( {1 - \cos 4\theta } \right)
Let’s simplify it further we get 1−38+38cos⁡4θ1 - \dfrac{3}{8} + \dfrac{3}{8}\cos 4\theta
Hence f(θ)=58+38cos⁡4θf(\theta ) = \dfrac{5}{8} + \dfrac{3}{8}\cos 4\theta ……………………………. (2)
Now we know that −1⩽ cos4θ ⩽ 1 - 1 \leqslant {\text{ cos4}}\theta {\text{ }} \leqslant {\text{ 1}} (maximum and minimum inbound of cos x)
⇒−38⩽ 38cos4θ ⩽ 38\Rightarrow - \dfrac{3}{8} \leqslant {\text{ }}\dfrac{3}{8}{\text{cos4}}\theta {\text{ }} \leqslant {\text{ }}\dfrac{3}{8}
⇒58−38⩽ 58+38cos4θ ⩽58+38\Rightarrow \dfrac{5}{8} - \dfrac{3}{8} \leqslant {\text{ }}\dfrac{5}{8} + \dfrac{3}{8}{\text{cos4}}\theta {\text{ }} \leqslant \dfrac{5}{8} + \dfrac{3}{8} (Adding 58 \dfrac{5}{{8{\text{ }}}}to all sides of inequality)
Now using equation 2 we know that f(θ)=58+38cos⁡4θf(\theta ) = \dfrac{5}{8} + \dfrac{3}{8}\cos 4\theta
Hence
14⩽f(θ)⩽1\dfrac{1}{4} \leqslant f(\theta ) \leqslant 1
Thus the minimum value of the required quantity is 14\dfrac{1}{4}and maximum value is 1
So option (a) is correct.

Note: Whenever we have to solve such problems, try to simplify as much as possible in order to reach the simplest form of expression, then use the min and max inbounds of the simplified part to reach up to the solution.