Maximize Minimario's trigonometric expression

In general $\begin{aligned}k\cos x + \ell \sin x &= \sqrt{k^2 + \ell^2}\left(\frac{k}{\sqrt{k^2 + \ell^2}} \cos x + \frac{\ell}{\sqrt{k^2 + \ell^2}} \sin x\right) \\ &= \sqrt{k^2 + \ell^2} (\sin y \cos x + \cos y \sin x) \\ &= \sqrt{k^2 + \ell^2} \sin (x + y) \leq \sqrt{k^2 + \ell^2}\end{aligned}$ Hence $$a = 7^2 + 6^2 = \boxed{85}$$

Let $f(x) = 7 \cos x + 6 \sin x$ . Then, if we want to find a global extremum, we find the derivative and set it equal to $0$ , and we get $f'(x) = -7 \sin x + 6 \cos x = 0 \implies \tan x = \dfrac {6}{7}.$ Using this, $\sin x = \dfrac {6}{\sqrt{85}}$ and $\cos x = \dfrac {7}{\sqrt{85}}$ . Using the Second Derivative Test, we find that this is indeed a maximum, not a minimum. Thus, our answer is $\begin{aligned} 7 \cos x + 6 \sin x &= \dfrac {49}{\sqrt{85}} + \dfrac {36}{\sqrt{85}} \\&= \dfrac {85}{\sqrt{85}} \\&= \sqrt {85}, \end{aligned}$ so our answer is $a = \boxed {85}$ .

Let's take the derivative to find all the points where we have local maximum and minimum.

$\begin{array}{l} f(x) = 7\cos x + 6\sin x\\ f'(x) = 6\cos x - 7\sin x \end{array}$

Then making the derivative equal to 0 we have : $\begin{array}{l} 0 = 6\cos x - 7\sin x\\ \tan x = \frac{6}{7}\\ x = \arctan \left( {\frac{6}{7}} \right) + n\pi \end{array}$

The function given is periodic within 2pi because both member are periodic within 2 pi so the max and min must also be periodic. If there is a maximum, it must be included in that period. Let's try the first 2 values within the first period. We have : $\begin{array}{l} f({x_1}) = 7\cos \arctan \left( {\frac{6}{7}} \right) + 6\sin \arctan \left( {\frac{6}{7}} \right) \approx 9.22\\ f({x_2}) = 7\cos \left( {\arctan \left( {\frac{6}{7}} \right) + \pi } \right) + 6\sin \left( {\arctan \left( {\frac{6}{7}} \right) + \pi } \right) \approx - 9.22 \end{array}$

The first value is the maximum, now we must find what is its value squared. Using these formulas :

$\begin{array}{l} \cos (\arctan x) = \frac{1}{{\sqrt {1 + {x^2}} }}\\ \sin \left( {\arctan x} \right) = \frac{x}{{\sqrt {1 + {x^2}} }} \end{array}$

We have

$f{\left( {{x_1}} \right)^2} = {\left( {\frac{{49 + 36}}{{\sqrt {85} }}} \right)^2} = {\left( {\sqrt {85} } \right)^2} = 85$

So the answer for the value of a is 85.

$f\left(x\right) = 7cosx + 6sinx$

$f'\left(x\right) = -7sinx + 6cosx$

For max/min, $f'\left(x\right) = 0$

Either $7sinx = 6cosx$ or $-7sinx, 6cosx = 0$

Latter is impossible so we have $tanx = \frac{6}{7}$

We can now construct a right-angled triangle $ABC, \: \angle ABC =90^ \circ, \: \angle BCA = x^\circ, \: AB = 6, BC =7$

We can clearly see that for this triangle $AC= \sqrt{85}$

So $6sinx = 6*\frac{6}{\sqrt{85}} = \frac{36}{\sqrt{85}}$ and $7cosx = 7*\frac{7}{\sqrt{85}} = \frac{49}{\sqrt{85}}$

$6sinx + 7 cosx = \frac{36+49}{\sqrt{85}} = \sqrt{85}$ so $a= 85$

Note that if we used the alternate value of $x$ for which $tanx = \frac{6}{7}$ with $0^\circ \leq x \leq 360^\circ$ then $sinx, cosx < 0$ thus making the total a minimum.

Assume, $y=7\cos x + 6\sin x$ Then the maximum value of $y$ will occur when the derivative of $7\cos x + 6\sin x$ is equal to $0$ , $\frac{\mathrm dy}{\mathrm d x} = -7\sin x + 6 \cos x$ So when $y$ is at its maximum, $-7\sin x + 6 \cos x = 0$ $\Rightarrow 7 \sin x = 6\cos x$ $\Rightarrow \tan x = \frac{6}{7}$ $\Rightarrow x=\tan^{-1}\frac{6}{7} = \beta$ We can show that $\beta$ is the value of $x$ for which $y$ attains its maximum by finding the second derivative, $\frac{\mathrm d^2 y}{\mathrm d x^2} = -7 \cos x -6\sin x$ $-7 \cos \beta - 6\sin \beta < 0$ which implies that $\beta$ is indeed the value of $x$ for which $y$ is maximum.

Thus we know the value of $x$ for which $\sqrt{a}$ attains its maximum. Using this value, we find that the maximum value of $\sqrt{a}$ is, $\sqrt{a} = 7\cos \beta + 6\sin \beta = \sqrt{85}$ $\Rightarrow a = \fbox{85}$

Let $7 \cos x + 6 \sin x \equiv R \cos (x - \alpha)$

$7 \cos x+ 6 \sin x \equiv R \cos x \cos \alpha + R \sin x \sin \alpha$

When $x = 0^ \circ$ , $7 = R \cos \alpha$ ......(1)

When $x = 90^ \circ$ , $6 = R \sin \alpha$ ......(2)

$(1)^2 + (2)^2$ :

$7^2 + 6^2 = R^2(\cos^2 \alpha + \sin^2 \alpha)$

$7^2 + 6^2 = R^2$

$R = \sqrt{7^2 + 6^2}$

$R \cos (x - \alpha) = \sqrt{a}$

Since we need to find the maximum value of $a$ , we let $R = \sqrt{a}$ , i.e. $\cos (x - \alpha) = 1$ , which is the maximum value of $\cos (x - \alpha)$ .

$\sqrt{a} = \sqrt{7^2 + 6^2}$

Hence, maximum value of $a = \boxed {85}$ .

Using Cauchy-Schwartz inequality:

$|7\cos x + 6 \sin x| \le \sqrt{(7^2+6^2)(\cos^2 x + \sin^2 x)} = \sqrt{85}$ $\therefore a=85$

What a coincidence!

By Cauchy-Schwarz inequality, we obtain: $|7\cos{x}+6\sin{x}|\leq \sqrt{\cos^2{x}+\sin^2{x}}\cdot \sqrt{7^2+6^2}$ Since $\cos^2{x}+\sin^2{x}=1$ we have: $7\cos{x}+6\sin{x}\leq \sqrt{85}$ So the result is $\boxed{85}$

y=6sin(x)+7cos(x), dy/dx=0 tanx=6/7 :: x=tan-1(6/7) max y^2=85

This problem is of the type [a(cos(x)) + b(sin(x))]-> multiplying and dividing the equation by √(a^2+b^2) we get-> √(a^2+b^2)[a(acos(x))/√(a^2+b^2) + b(bsin(x))/√(a^2+b^2)] let a/√(a^2+b^2)=siny and b/√(a^2+b^2)=cosy there4 √(a^2+b^2)[a(acos(x))/√(a^2+b^2) + b(bsin(x))/√(a^2+b^2)]=√(a^2+b^2)[(sin(y))(cos(x)) + (cos(y))(sin(x))] =>√(a^2+b^2)[a(acos(x))/√(a^2+b^2) + b(bsin(x))/√(a^2+b^2)]=√(a^2+b^2)[sin(x+y)] The max. value of sin(x+y)=1 There4 max. value of-> √(a^2+b^2)[a(acos(x))/√(a^2+b^2) + b(bsin(x))/√(a^2+b^2)]=√(a^2+b^2) and √(a^2+b^2)[a(acos(x))/√(a^2+b^2) + b(bsin(x))/√(a^2+b^2)]=[a(cos(x)) + b(sin(x))]=√(a^2+b^2) Similarly, The maximum value of 7cosx+6sinx is √49+36 = √85

dy/dx=(7cosx+6sinx)/dx=\sqrt{a} =6cos x-7sin x=sqrt{a} =cosx(6-7tan x)=sqrt{a} find for angle x 6-7tanx=0 tanx=6/7 x=40.deg frm original equation.. {7cos(40.6)+6sin(40.6)}^2=a a=85