Question

How do we find the maximum and the minimum values of $\dfrac{1}{3\sin x-4\cos x+7}$ ?

Answer 1

We can solve this question by first converting the denominator to a from such that you can write the values in terms sin and cos. And then we have to use the formula of sin(a-b) = cosa sinb - sina cosb to simplify the denominator. Then we can use the properties of sine to find the maximum and the minimum values of the given term.

Complete step by step solution:
According to the problem, we are asked to find the maximum and the minimum values of $\dfrac{1}{3\sin x-4\cos x+7}$.
For this, we first convert the denominator to a from such that you can write the values in terms sin and cos. And then we have to use the formula of sin(a-b) = cosa sinb - sina cosb to simplify the denominator. Then we can use the properties of sine to find the maximum and the minimum values of the given term.
Here we take the given term as equation 1.
$\Rightarrow f=\dfrac{1}{3\sin x-4\cos x+7}$------- (1)
Now we will take 5 as a common from the denominator. Therefore, we get:
$\Rightarrow f=\dfrac{1}{5\left( \dfrac{3}{5}\sin x-\dfrac{4}{5}\cos x \right)+7}$ ------- (2)
Now we take $\cos \theta =\dfrac{3}{5}$ and $\sin \theta =\dfrac{4}{5}$ . So substituting this in the equation 2, we get :
$\Rightarrow f=\dfrac{1}{5\left( \cos \theta \sin x-\sin \theta \cos x \right)+7}$-
$\Rightarrow f=\dfrac{1}{5\sin \left( x-\theta \right)+7}$ -
Now we know that $-1<\sin \theta <1$. Therefore using this property, we get:
$\Rightarrow \dfrac{1}{5\left( 1 \right)+7}< f <\dfrac{1}{5\left( -1 \right)+7}$
$\Rightarrow \dfrac{1}{12}< f <\dfrac{1}{2}$
Therefore, after all the solving, we get the final answer of the question as – the maximum value is $\dfrac{1}{2}$ and the minimum value as $\dfrac{1}{12}$.

Note: It is important that you should know the formulas of trigonometry to solve these kinds of questions. Otherwise you will not be able to solve them. You also should know the basic properties of sine, cos , tan , cot and others.