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# What is the minimum of $\sin x+\cos x$? Verified
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Hint: To find the minimum value of the given trigonometric expression, we need to rearrange the form of the given expression into a trigonometric sine function. To calculate the minimum value, we have to minimise the trigonometric sine function.

The given trigonometric expression = $\sin x+\cos x$

Let the expression be equal to $f(x)$

Multiply and divide the function $f(x)$ with the value of $\dfrac{1}{\sqrt{2}}$, we get

$\Rightarrow$ $f(x)=\dfrac{\sqrt{2}\sin x+\sqrt{2}\cos x}{\sqrt{2}}$

By the numerator common value $\sqrt{2}$, we get

$\Rightarrow$ $f(x)=\sqrt{2}\left( \dfrac{\sin x+\cos x}{\sqrt{2}} \right)$

By reframing the arrangement, we get

$\Rightarrow$ $f(x)=\sqrt{2}\left( \dfrac{\sin x}{\sqrt{2}}+\dfrac{\cos x}{\sqrt{2}} \right)$

We know that the trigonometric values of $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}},\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$

By substituting the above values in the given trigonometric expression, we get

$f(x)=\sqrt{2}\left( \cos \left( \dfrac{\pi }{4} \right)\sin x+\cos x\sin \left( \dfrac{\pi }{4} \right) \right)$

We know that the formula for sine angle of summation of two angles is given by $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$

By comparing the above formula and the trigonometric expression, we get

The values of $\alpha ,\beta$ are equal to the value of $x,\dfrac{\pi }{4}$ respectively.

Thus, we obtain the above expression as

$f(x)=\sqrt{2}\left( \sin \left( x+\dfrac{\pi }{4} \right) \right)$
To calculate the minimum value of the expression we have to minimise the trigonometric sine function.

We know that the value of the trigonometric function $\sin \theta$ lies between -1 and 1 and $\theta$ is an angle.

Thus, we can say that the minimum of any angle of sine is equal to -1 which implies that the minimum value of $\sin \left( x+\dfrac{\pi }{4} \right)$ is equal to -1.

By substituting the above value in the given trigonometric expression, we get
$f(x)=\sqrt{2}\left( -1 \right)$

Thus, the minimum value of the given trigonometric expression is equal to $-\sqrt{2}$.

Note: The possibility of mistake can be adding the minimum value of both terms of the expression, which is wrong since both the functions are not minimum at the same point. The alternative procedure of solving the question is to use the formula $\cos \left( \alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta$ and then substitute the minimum value of $\cos \theta$ which is equal to -1, to calculate the minimum value.
Last updated date: 22nd Sep 2023
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