Question

Answer this

i will mark as brainliest​

Answer 1

We are given a function:

$\displaystyle f=\frac{2}{\sin \theta + \cos \theta}$

The denominator consists of $\sin\theta + \cos\theta$

We will first bring them in a single trigonometric function form.

$\displaystyle f=\frac{2}{\sin\theta+\cos\theta}\\\\\\\implies\text{In the denominator, multiply and divide by }\sqrt{2}\\\\\\\implies f=\frac{2}{\sqrt{2}\times\frac{1}{\sqrt{2}}(\sin\theta+\cos\theta)}\\\\\\\implies f=\frac{2}{\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin\theta+\frac{1}{\sqrt{2}}\cos\theta}\right)}\\\\\\\implies f=\frac{\sqrt{2}}{\sin\theta\cos\frac{\pi}{4}+\cos\theta\sin\frac{\pi}{4}}\\\\\\\text{Use the identity } \sin A\cos B+\cos A\sin B=\sin(A+B)$

$\displaystyle \implies \boxed{f=\frac{\sqrt{2}}{\sin \left( \theta+\frac{\pi}{4}\right)}}$

Now, Minimum Value can have two meanings. We will take both.

1) Minimum Numerical Value

We can also call this the Minimum Absolute Value.

For example, -1 and 1 have the same numerical or absolute value.

We have a sine function in the denominator. If we want the minimum value of f, then we need to get the maximum numerical value of denominator.

And, we know that the Maximum Value of Sine Function is 1.

So, we have:

$\displaystyle \left(\sin \left( \theta+\frac{\pi}{4}\right) \right)_{max} = 1$

So, we have:

$\displaystyle f_{min} = \frac{\sqrt{2}}{\left(\sin \left( \theta+\frac{\pi}{4}\right) \right)_{max}}\\\\\\\implies f_{min}=\frac{\sqrt{2}}{1}\\\\\\\implies \boxed{\boxed{\bold{f_{min}=\sqrt{2}}}}$

2) Minimum Mathematical Value

By this, we mean that, if we have two numbers, for example -1 and 1, then we consider -1 < 1.

In this way. the minimum mathematical value possible can be obtained here as follows:

$\displaystyle f_{min} \text{ when limit } \sin \left( \theta+\frac{\pi}{4}\right) \to 0^{-}\\\\\\\implies f_{min} = \lim_{\sin \left( \theta+\frac{\pi}{4}\right) \to 0^{-}} \, \, \frac{\sqrt{2}}{\sin \left( \theta+\frac{\pi}{4}\right)}\\\\\\\implies f_{min} = \frac{\sqrt{2}}{0^{-}}\\\\\\\implies \boxed{\boxed{\bold{f_{min} = -\infty}}}$

A Graph is also attached for visual understanding.

The red curve represents the original function.

The green line represents $y=\sqrt{2}$, which is the minimum numerical value.

As we can see, the graph also goes all the way down to $-\infty$. That is our minimum Mathematical Value.

Answer 2

Please mark me as the brainliest because I need 3 more brainliest answers to get Virtuoso please help me