Question

For an acute angle theta sintheta + costheta take greatest value when theta is

Answer 1

Suppose we have to find the maximum value of the following.

$a\sin\theta+b\cos\theta$

For this we may assume a right triangle with perpendicular sides $a$ and $b$ Then by Pythagoras' Theorem, the hypotenuse is given by $c=\sqrt{a^2+b^2}.$

$\setlength{\unitlength}{1mm}\begin {picture}(5,5)\put(0,0){\line(1,0){30}}\put(0,0){\line(0,1){40}}\put(30,0){\line(-3,4){30}}\put(14,-4){ a }\put(-4,19){ b }\put(17.5,20.5){ c=\sqrt{a^2+b^2} }\put(25,1){ \alpha }\end {picture}$

Let $\alpha$ be the angle inside the triangle opposite to $b.$ Then,

$\sin\alpha=\dfrac {b}{c}\quad;\quad\cos\alpha=\dfrac {a}{c}$

Then come back to the function,

$a\sin\theta+b\cos\theta$

Just multiply and divide both the terms by $c$ as follows:

$c\left(\dfrac {a\sin\theta+b\cos\theta}{c}\right)\\\\\\=c\left (\dfrac {a}{c}\sin\theta+\dfrac {b}{c}\cos\theta\right)\\\\\\=c\left (\sin\theta\cos\alpha+\cos\theta\sin\alpha\right)\\\\\\=c\sin (\theta+\alpha)$

We know that the sine value of an arbitrary angle is in the closed interval [-1, 1]. Then we can say that,

$\sin (\theta+\alpha)\in [-1,\ 1]\\\\\implies\ \boxed {c\sin (\theta+\alpha)\in [-c,\ c]}$

From this we get that,

$\boxed{\begin {minipage}{5.5cm} \max(a\sin\theta+b\cos\theta)=\sqrt{a^2+b^2} \\\\ \min(a\sin\theta+b\cos\theta)=-\sqrt{a^2+b^2} \end {minipage}}$

So the maximum value of $\sin\theta+\cos\theta$ is $(a=b=1),$

$\sqrt {1^2+1^2}=\sqrt 2$

But we have to find the value of $\theta$ satisfying the given condition,

$\sin\theta+\cos\theta=\sqrt 2,\quad\theta\in\left (0,\ \dfrac {\pi}{2}\right)$

So,

$(\sin\theta+\cos\theta)^2=2\\\\\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=2\\\\\sin\theta\cos\theta=\dfrac {1}{2}\quad\quad [\because\ \sin^2\theta+\cos^2\theta=1]$

Again,

$\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=2\\\\\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta -4\sin\theta\cos\theta=2-2\quad\quad \left[\because\ \sin\theta\cos\theta=\dfrac {1}{2}\right]\\\\\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta=0\\\\(\sin\theta-\cos\theta)^2=0\\\\\sin\theta=\cos\theta$

Then,

$\tan\theta=1\\\\\implies\ \boxed {\boxed {\theta=\dfrac {\pi}{4}}}$

#answerwithquality

#BAL