Maximize Minimario's trigonometric expression

Geometry Level 1

If the maximum value of 7 cos x + 6 sin x 7 \cos x+6 \sin x is expressed as a \sqrt{a} , what is the value of a a ?


This problem is posed by Minimario M .


The answer is 85.

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11 solutions

Jan J.
Oct 28, 2013

In general k cos x + sin x = k 2 + 2 ( k k 2 + 2 cos x + k 2 + 2 sin x ) = k 2 + 2 ( sin y cos x + cos y sin x ) = k 2 + 2 sin ( x + y ) k 2 + 2 \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 $$a = 7^2 + 6^2 = \boxed{85}$$

Moderator note:

The R-method provides a simple explanation which doesn't require the use of calculus.

Also in case anyone wonders, this is known as the R-method.

Jan J. - 7 years, 7 months ago

good solution.

Møhsyne Kâhøuâ - 7 years, 7 months ago

Nice..did the same. But can you tell me what exactly is R method?

Krishna Ar - 6 years, 11 months ago

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See this .

Prasun Biswas - 6 years, 3 months ago
Ahaan Rungta
Oct 28, 2013

Let f ( x ) = 7 cos x + 6 sin x 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 0 , and we get f ( x ) = 7 sin x + 6 cos x = 0 tan x = 6 7 . f'(x) = -7 \sin x + 6 \cos x = 0 \implies \tan x = \dfrac {6}{7}. Using this, sin x = 6 85 \sin x = \dfrac {6}{\sqrt{85}} and cos x = 7 85 \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 7 cos x + 6 sin x = 49 85 + 36 85 = 85 85 = 85 , \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 = 85 a = \boxed {85} .

Trop bien, too well!

Kevin H - 7 years, 7 months ago

Did the same!

Kartik Sharma - 6 years, 8 months ago
Samuel Hatin
Oct 27, 2013

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

f ( x ) = 7 cos x + 6 sin x f ( x ) = 6 cos x 7 sin x \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 : 0 = 6 cos x 7 sin x tan x = 6 7 x = arctan ( 6 7 ) + n π \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 : f ( x 1 ) = 7 cos arctan ( 6 7 ) + 6 sin arctan ( 6 7 ) 9.22 f ( x 2 ) = 7 cos ( arctan ( 6 7 ) + π ) + 6 sin ( arctan ( 6 7 ) + π ) 9.22 \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 :

cos ( arctan x ) = 1 1 + x 2 sin ( arctan x ) = x 1 + x 2 \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 ( x 1 ) 2 = ( 49 + 36 85 ) 2 = ( 85 ) 2 = 85 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.

Here's my intended solution:

Note that cos ( y x ) = cos y cos x + sin y sin x \cos(y-x)=\cos y \cos x+\sin y \sin x . Therefore, we can choose y y such that cos y sin y = 7 6 \frac{\cos y}{\sin y}=\frac{7}{6} . Since cos 2 y + sin 2 y = 1 \cos^2 y+\sin^2 y=1 , we have cos y = 7 85 \cos y=\frac{7}{\sqrt{85}} and sin y = 6 85 \sin y=\frac{6}{\sqrt{85}} , for some angle y. For angle y, we have 5 cos x + 6 sin x = 85 ( 7 85 cos x + 6 85 sin x ) = 85 ( cos x cos y + sin x sin y ) = 85 cos ( y x ) 5 \cos x+6 \sin x=\sqrt{85}\left(\frac{7}{\sqrt{85}}\cos x+\frac{6}{\sqrt{85}}\sin x\right)=\sqrt{85} (\cos x \cos y+\sin x\sin y)=\sqrt{85} \cos (y-x) . Since the maximum value of cos ( y x ) \cos(y-x) is 1, the desired maximum is 85 \boxed{\sqrt{85}}

minimario minimario - 7 years, 7 months ago

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This is quite interesting, I wouldn't have thought about that. It is way more easy to deal with it like this then the way I did it.

Samuel Hatin - 7 years, 7 months ago

I think my solution is quite similar to yours.

Siao Chi Mok - 7 years, 7 months ago
Danny He
Oct 28, 2013

f ( x ) = 7 c o s x + 6 s i n x f\left(x\right) = 7cosx + 6sinx

f ( x ) = 7 s i n x + 6 c o s x f'\left(x\right) = -7sinx + 6cosx

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

Either 7 s i n x = 6 c o s x 7sinx = 6cosx or 7 s i n x , 6 c o s x = 0 -7sinx, 6cosx = 0

Latter is impossible so we have t a n x = 6 7 tanx = \frac{6}{7}

We can now construct a right-angled triangle A B C , A B C = 9 0 , B C A = x , A B = 6 , B C = 7 ABC, \: \angle ABC =90^ \circ, \: \angle BCA = x^\circ, \: AB = 6, BC =7

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

So 6 s i n x = 6 6 85 = 36 85 6sinx = 6*\frac{6}{\sqrt{85}} = \frac{36}{\sqrt{85}} and 7 c o s x = 7 7 85 = 49 85 7cosx = 7*\frac{7}{\sqrt{85}} = \frac{49}{\sqrt{85}}

6 s i n x + 7 c o s x = 36 + 49 85 = 85 6sinx + 7 cosx = \frac{36+49}{\sqrt{85}} = \sqrt{85} so a = 85 a= 85

Note that if we used the alternate value of x x for which t a n x = 6 7 tanx = \frac{6}{7} with 0 x 36 0 0^\circ \leq x \leq 360^\circ then s i n x , c o s x < 0 sinx, cosx < 0 thus making the total a minimum.

I have only just realised there is a typo here.

The last line should read 0 x 36 0 0^\circ \leq x \leq 360^\circ

Danny He - 7 years, 7 months ago
Oliver Welsh
Oct 28, 2013

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

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

Siao Chi Mok
Oct 30, 2013

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

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

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

When x = 9 0 x = 90^ \circ , 6 = R sin α 6 = R \sin \alpha ......(2)

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

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

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

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

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

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

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

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

Vu Vincent
Dec 24, 2017

Using Cauchy-Schwartz inequality:

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

What a coincidence!

Emmanuel Lasker
Dec 27, 2013

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

Lutful Kabir Chy
Dec 23, 2013

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

Nirmit Tripathii
Oct 31, 2013

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

Kim Chan
Oct 29, 2013

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

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