The greatest value of the function -5 sin the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The expression is of the form $a \sin 	heta + b \cos 	heta$.The maximum value of this expression is given by the formula $\sqrt{a^2 + b^2}$.Here

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) The expression is of the form $a \sin 	heta + b \cos 	heta$.The maximum value of this expression is given by the formula $\sqrt{a^2 + b^2}$.Here,$a = -5$ and $b = 12$.Substituting these values into the formula:Maximum value $= \sqrt{(-5)^2 + (12)^2}$$= \sqrt{25 + 144}$$= \sqrt{169}$$= 13$.Therefore,the greatest value is $13$.

The greatest value of the function $-5 \sin \theta + 12 \cos \theta$ is

Similar Questions

${d \over {dx}}({x^2}{e^x}\sin x) = $

$A$ particular straight line passes through the origin and a point whose abscissa is double the ordinate of the point. The equation of such a straight line is:

If $log_{10} (xy) = 2$,then the value of $xy$ is

Given that $y = \frac{10}{\sin x + \sqrt{3} \cos x}$. The minimum value of $y$ is

$\frac{d}{dx} {\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)}^2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module