Find the minimum value of $\cos e{c^2}\theta + {\sin ^2}\theta $.
A. 0
B. -1
C. 1
D. 2
Answer
633.3k+ views
Hint: Here, we will use the property $A \cdot M \geqslant G \cdot M$ to find the minimum value, where A.M is the Arithmetic mean and G.M is the Geometric mean.
Complete step-by-step answer:
Given equation is $\cos e{c^2}\theta + {\sin ^2}\theta $.
As we know that $A \cdot M \geqslant G \cdot M$, so let us use this property to find the minimum value of $\cos e{c^2}\theta + {\sin ^2}\theta $.
We know that the A.M i.e.., Arithmetic mean of two terms ‘a’ and ‘b’ will be $\dfrac{{a + b}}{2}$ i.e..,
$A \cdot M = \dfrac{{a + b}}{2} \to (1)$.
Similarly G.M i.e.., Geometric mean of two terms ‘a’ and ‘b’ will be $\sqrt {a \cdot b} $ i.e..,
$G \cdot M = \sqrt {a \cdot b} \to (2)$
Now, let us consider $\cos e{c^2}\theta $ and ${\sin ^2}\theta $ be two terms, then the A.M and G.M of these terms can be written as
$A \cdot M = \dfrac{{\cos e{c^2}\theta + {{\sin }^2}\theta }}{2}\left[ {\because {\text{ from eq(1)}}} \right]$
$G \cdot M = \sqrt {\cos e{c^2}\theta \cdot {{\sin }^2}\theta } \left[ {\because {\text{ from eq (2)}}} \right]$
Now, let us substitute the above values in the property $A \cdot M \geqslant G \cdot M$, we get
$
\Rightarrow \dfrac{{\cos e{c^2}\theta + {{\sin }^2}\theta }}{2} \geqslant \sqrt {\cos e{c^2}\theta \cdot {{\sin }^2}\theta } \\
\Rightarrow \dfrac{{\cos e{c^2}\theta + {{\sin }^2}\theta }}{2} \geqslant \sqrt 1 \left[ {\because \cos e{c^2}\theta \cdot {{\sin }^2}\theta = \dfrac{1}{{{{\sin }^2}\theta }}.{{\sin }^2}\theta = 1} \right] \\
\Rightarrow \cos e{c^2}\theta + {\sin ^2}\theta \geqslant 2 \\
$
So, from the above equation, we can say that the value of $\cos e{c^2}\theta + {\sin ^2}\theta $ is always greater than or equal to 2.
Hence, the minimum value of $\cos e{c^2}\theta + {\sin ^2}\theta $ is 2.
So, Option D is the required answer.
Note: While solving these types of problems, you must know the property $A \cdot M \geqslant G \cdot M$ such that it will give you the solution easily without performing any other operations to simplify the given terms. The definitions of Arithmetic mean and Geometric mean have to be known perfectly to solve.
Complete step-by-step answer:
Given equation is $\cos e{c^2}\theta + {\sin ^2}\theta $.
As we know that $A \cdot M \geqslant G \cdot M$, so let us use this property to find the minimum value of $\cos e{c^2}\theta + {\sin ^2}\theta $.
We know that the A.M i.e.., Arithmetic mean of two terms ‘a’ and ‘b’ will be $\dfrac{{a + b}}{2}$ i.e..,
$A \cdot M = \dfrac{{a + b}}{2} \to (1)$.
Similarly G.M i.e.., Geometric mean of two terms ‘a’ and ‘b’ will be $\sqrt {a \cdot b} $ i.e..,
$G \cdot M = \sqrt {a \cdot b} \to (2)$
Now, let us consider $\cos e{c^2}\theta $ and ${\sin ^2}\theta $ be two terms, then the A.M and G.M of these terms can be written as
$A \cdot M = \dfrac{{\cos e{c^2}\theta + {{\sin }^2}\theta }}{2}\left[ {\because {\text{ from eq(1)}}} \right]$
$G \cdot M = \sqrt {\cos e{c^2}\theta \cdot {{\sin }^2}\theta } \left[ {\because {\text{ from eq (2)}}} \right]$
Now, let us substitute the above values in the property $A \cdot M \geqslant G \cdot M$, we get
$
\Rightarrow \dfrac{{\cos e{c^2}\theta + {{\sin }^2}\theta }}{2} \geqslant \sqrt {\cos e{c^2}\theta \cdot {{\sin }^2}\theta } \\
\Rightarrow \dfrac{{\cos e{c^2}\theta + {{\sin }^2}\theta }}{2} \geqslant \sqrt 1 \left[ {\because \cos e{c^2}\theta \cdot {{\sin }^2}\theta = \dfrac{1}{{{{\sin }^2}\theta }}.{{\sin }^2}\theta = 1} \right] \\
\Rightarrow \cos e{c^2}\theta + {\sin ^2}\theta \geqslant 2 \\
$
So, from the above equation, we can say that the value of $\cos e{c^2}\theta + {\sin ^2}\theta $ is always greater than or equal to 2.
Hence, the minimum value of $\cos e{c^2}\theta + {\sin ^2}\theta $ is 2.
So, Option D is the required answer.
Note: While solving these types of problems, you must know the property $A \cdot M \geqslant G \cdot M$ such that it will give you the solution easily without performing any other operations to simplify the given terms. The definitions of Arithmetic mean and Geometric mean have to be known perfectly to solve.
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