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Question: Range of \({\sin ^2}\theta + {\cos ^4}\theta = \) A) \(1 \leqslant A \leqslant 2\) B) \(\dfrac{3...

Range of sin⁡2θ+cos⁡4θ={\sin ^2}\theta + {\cos ^4}\theta =
A) 1⩽A⩽21 \leqslant A \leqslant 2
B) 34⩽A⩽1\dfrac{3}{4} \leqslant A \leqslant 1
C) 136⩽A⩽1\dfrac{{13}}{6} \leqslant A \leqslant 1
D) None of these

Explanation

Solution

It is known that range of −1⩽cos⁡θ⩽1 - 1 \leqslant \cos \theta \leqslant 1. 1st we will convert sin⁡θ\sin \theta into cos⁡θ\cos \theta by using sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1then sin⁡2θ+cos⁡4θ=1−cos⁡2θ+cos⁡4θ{\sin ^2}\theta + {\cos ^4}\theta = 1 - {\cos ^2}\theta + {\cos ^4}\theta . After this we will convert this square form by complete square method. after that using range of cos⁡θ\cos \theta we will proceed further and convert it into sin⁡2θ+cos⁡4θ{\sin ^2}\theta + {\cos ^4}\theta

Complete step by step solution:
sin⁡2θ+cos⁡4θ={\sin ^2}\theta + {\cos ^4}\theta =
We know that sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, substituting this in equation we get
⇒1−cos⁡2θ+cos⁡4θ\Rightarrow 1 - {\cos ^2}\theta + {\cos ^4}\theta
We can also write it as
⇒1−2×12×cos⁡2θ+cos⁡4θ+14−14\Rightarrow 1 - 2 \times \dfrac{1}{2} \times {\cos ^2}\theta + {\cos ^4}\theta + \dfrac{1}{4} - \dfrac{1}{4}
⇒(cos⁡2θ−12)2+34\Rightarrow {\left( {{{\cos }^2}\theta - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} [∵(a−b)2=a2+b2−2ab]\left[ {\because {{\left( {a - b} \right)}^2} = {a^2} + {b^2} - 2ab} \right]
It is known that
−1⩽cos⁡θ⩽1- 1 \leqslant \cos \theta \leqslant 1
On squaring we get
0⩽cos⁡2θ⩽10 \leqslant {\cos ^2}\theta \leqslant 1
Subtracting 12\dfrac{1}{2} we get
−12⩽cos⁡2θ−12⩽12- \dfrac{1}{2} \leqslant {\cos ^2}\theta - \dfrac{1}{2} \leqslant \dfrac{1}{2}
Again, on squaring we get
0⩽(cos⁡2θ−12)2⩽140 \leqslant {\left( {{{\cos }^2}\theta - \dfrac{1}{2}} \right)^2} \leqslant \dfrac{1}{4}
Then adding 34\dfrac{3}{4} we get
34⩽(cos⁡2θ−12)2+34⩽1\dfrac{3}{4} \leqslant {\left( {{{\cos }^2}\theta - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \leqslant 1
There for range of sin⁡2θ+cos⁡4θ={\sin ^2}\theta + {\cos ^4}\theta = 34⩽A⩽1\dfrac{3}{4} \leqslant A \leqslant 1

Hence, option B is the correct option.

Note:
Point to remember are, in these types of questions we will convert all trigonometric ratios in a single form. Range of −1⩽cos⁡θ⩽1 - 1 \leqslant \cos \theta \leqslant 1, −1⩽sin⁡θ⩽1 - 1 \leqslant \sin \theta \leqslant 1. While converting division should be avoided as far as possible it will reduce the possible value.