
The minimum value of $3\cos x + 4\sin x + 5$ is:
(A) $5$
(B) $9$
(C) $7$
(D) $0$
Answer
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Hint: In the given question, we are provided with a trigonometric expression in sine and cosine and we are required to find the minimum value of the expression. To solve the problem, we must know the technique of calculating the range of the trigonometric expression of type $\left( {a\sin x + b\cos x} \right)$. We first calculate the range of the expression $3\cos x + 4\sin x$. Then, we add the constant $5$ to the maximum and minimum values of the range in order to get to the required answer.
Complete answer: So, the given trigonometric expression is $\left( {3\cos x + 4\sin x + 5} \right)$.
Now, to calculate the minimum value of the trigonometric expression $\left( {3\cos x + 4\sin x + 5} \right)$, we first substitute the range of the sum of the terms $\left( {3\cos x + 4\sin x} \right)$ and then add five to the resultant minimum and maximum value of the expression.
Now, we know that the range of the trigonometric expression $\left( {a\sin x + b\cos x} \right)$ is \[\left[ { - \sqrt {{a^2} + {b^2}} ,\sqrt {{a^2} + {b^2}} } \right]\].
So, the range of the trigonometric expression $\left( {3\cos x + 4\sin x} \right)$ will be \[\left[ { - \sqrt {{3^2} + {4^2}} ,\sqrt {{3^2} + {4^2}} } \right]\].
So, we get the minimum value of the expression $\left( {3\cos x + 4\sin x} \right)$ as \[\left( { - \sqrt {{3^2} + {4^2}} } \right)\].
Now, we know that the square of $3$ is $9$ and the square of $4$ is $16$. So, we get the minimum value of expression as,
\[ \Rightarrow \left( { - \sqrt {9 + 16} } \right)\]
Adding the terms, we get,
\[ \Rightarrow \left( { - \sqrt {25} } \right)\]
We know that the square root of $25$ is $5$.
\[ \Rightarrow - 5\]
So, the minimum of the trigonometric expression $\left( {3\cos x + 4\sin x} \right)$ is \[\left( { - 5} \right)\].
Now, we have to find the minimum value of the expression $\left( {3\cos x + 4\sin x + 5} \right)$ by adding $5$ to the minimum value of $\left( {3\cos x + 4\sin x} \right)$.
So, we get the minimum value of the expression $\left( {3\cos x + 4\sin x + 5} \right)$as $ - 5 + 5 = 0$.
Hence, the minimum value of $\left( {3\cos x + 4\sin x + 5} \right)$is zero.
So, option (D) is the correct answer.
Note:
Such questions require grip over the concepts of trigonometry and inequalities. One must know the methodology to calculate the range of trigonometric expressions of the form $\left( {a\sin x + b\cos x} \right)$ in order to solve the given problem. We also must know the technique of calculating the range of a composite complex number in order to tackle such problems.
Complete answer: So, the given trigonometric expression is $\left( {3\cos x + 4\sin x + 5} \right)$.
Now, to calculate the minimum value of the trigonometric expression $\left( {3\cos x + 4\sin x + 5} \right)$, we first substitute the range of the sum of the terms $\left( {3\cos x + 4\sin x} \right)$ and then add five to the resultant minimum and maximum value of the expression.
Now, we know that the range of the trigonometric expression $\left( {a\sin x + b\cos x} \right)$ is \[\left[ { - \sqrt {{a^2} + {b^2}} ,\sqrt {{a^2} + {b^2}} } \right]\].
So, the range of the trigonometric expression $\left( {3\cos x + 4\sin x} \right)$ will be \[\left[ { - \sqrt {{3^2} + {4^2}} ,\sqrt {{3^2} + {4^2}} } \right]\].
So, we get the minimum value of the expression $\left( {3\cos x + 4\sin x} \right)$ as \[\left( { - \sqrt {{3^2} + {4^2}} } \right)\].
Now, we know that the square of $3$ is $9$ and the square of $4$ is $16$. So, we get the minimum value of expression as,
\[ \Rightarrow \left( { - \sqrt {9 + 16} } \right)\]
Adding the terms, we get,
\[ \Rightarrow \left( { - \sqrt {25} } \right)\]
We know that the square root of $25$ is $5$.
\[ \Rightarrow - 5\]
So, the minimum of the trigonometric expression $\left( {3\cos x + 4\sin x} \right)$ is \[\left( { - 5} \right)\].
Now, we have to find the minimum value of the expression $\left( {3\cos x + 4\sin x + 5} \right)$ by adding $5$ to the minimum value of $\left( {3\cos x + 4\sin x} \right)$.
So, we get the minimum value of the expression $\left( {3\cos x + 4\sin x + 5} \right)$as $ - 5 + 5 = 0$.
Hence, the minimum value of $\left( {3\cos x + 4\sin x + 5} \right)$is zero.
So, option (D) is the correct answer.
Note:
Such questions require grip over the concepts of trigonometry and inequalities. One must know the methodology to calculate the range of trigonometric expressions of the form $\left( {a\sin x + b\cos x} \right)$ in order to solve the given problem. We also must know the technique of calculating the range of a composite complex number in order to tackle such problems.
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