Question

If we have a trigonometric equation $A={{\sin }^{2}}\theta +{{\cos }^{4}}\theta$, then for all real values of $\theta$
A. $1\le A\le 2$
B. $\dfrac{3}{4}\le A\le 1$
C. $\dfrac{13}{16}\le A\le 1$
D. $\dfrac{3}{4}\le A\le \dfrac{13}{16}$

Answer 1

We are given a question with an expression given to us and we have to find the value from the options which is in line with the given expression. We will first solve the expression given to us and reduce it as much as possible, then we will find the minimum and the maximum value that ‘A’ can have depending upon the maximum and minimum values of terms in the reduced form of the given equation. Hence, we will have the range of values that ‘A’ can have.

Complete step by step solution:
According to the given question, we have an expression given to us and we have to find the range in which the given expression will give the value for all real values of $\theta$.
The expression given to us,
$A={{\sin }^{2}}\theta +{{\cos }^{4}}\theta$----(1)
We will now reduce this equation (1) and the new expression that we will have is,
$\Rightarrow A={{\sin }^{2}}\theta +{{\cos }^{2}}\theta .{{\cos }^{2}}\theta$
We know the formulae of cosine function, one of which is, ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$. Applying this formula, we get,
$\Rightarrow A={{\sin }^{2}}\theta +{{\cos }^{2}}\theta \left( 1-{{\sin }^{2}}\theta \right)$
Opening up the brackets in the above expression, we get,
$\Rightarrow A={{\sin }^{2}}\theta +{{\cos }^{2}}\theta -{{\cos }^{2}}\theta {{\sin }^{2}}\theta$
Multiplying and divided the composite term in the RHS, we get the new expression as,
$\Rightarrow A={{\sin }^{2}}\theta +{{\cos }^{2}}\theta -\dfrac{4}{4}{{\cos }^{2}}\theta {{\sin }^{2}}\theta$
Separating the components, we get,
$\Rightarrow A={{\sin }^{2}}\theta +{{\cos }^{2}}\theta -\dfrac{1}{4}{{\left( 2\cos \theta \sin \theta \right)}^{2}}$
Now, we know that, $\sin 2\theta =2\cos \theta \sin \theta$, applying this formula in the above expression we get,
$\Rightarrow A={{\sin }^{2}}\theta +{{\cos }^{2}}\theta -\dfrac{1}{4}{{\left( \sin 2\theta \right)}^{2}}$
Also, we know that, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. The expression we will get is,
$\Rightarrow A=1-\dfrac{1}{4}{{\left( \sin 2\theta \right)}^{2}}$
Here, the value of $\sin 2\theta$ ranges from 0 to 1, that is, the lowest possible value is 0 and the maximum value is 1.
So, the highest value that A can have is when $\sin 2\theta =0$, that is,
$\Rightarrow A=1-\dfrac{1}{4}{{\left( 0 \right)}^{2}}$
$\Rightarrow A=1$
Also, the lowest value that A can have is when $\sin 2\theta =1$, that is,
$\Rightarrow A=1-\dfrac{1}{4}{{\left( 1 \right)}^{2}}=1-\dfrac{1}{4}$
$\Rightarrow A=\dfrac{3}{4}$
So, ‘A’ ranges from $\dfrac{3}{4}$ to $1$.
Therefore, the option B. $\dfrac{3}{4}\le A\le 1$ is correct.

Note: The commonly used trigonometric identities should be known to reduce the given expression. In the above solution, while reducing the expression, the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ was used. Also, while finding the range of range, the maximum negative value will give the least value of the expression and the minimum negative value will give the maximum value of the expression.