If the maximum value of 7 cos x + 6 sin x is expressed as a , what is the value of a ?
This problem is posed by Minimario M .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
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.
good solution.
Nice..did the same. But can you tell me what exactly is R method?
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 ⟹ tan x = 7 6 . Using this, sin x = 8 5 6 and cos x = 8 5 7 . 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 = 8 5 4 9 + 8 5 3 6 = 8 5 8 5 = 8 5 , so our answer is a = 8 5 .
Trop bien, too well!
Did the same!
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
Then making the derivative equal to 0 we have : 0 = 6 cos x − 7 sin x tan x = 7 6 x = arctan ( 7 6 ) + n π
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 ( 7 6 ) + 6 sin arctan ( 7 6 ) ≈ 9 . 2 2 f ( x 2 ) = 7 cos ( arctan ( 7 6 ) + π ) + 6 sin ( arctan ( 7 6 ) + π ) ≈ − 9 . 2 2
The first value is the maximum, now we must find what is its value squared. Using these formulas :
cos ( arctan x ) = 1 + x 2 1 sin ( arctan x ) = 1 + x 2 x
We have
f ( x 1 ) 2 = ( 8 5 4 9 + 3 6 ) 2 = ( 8 5 ) 2 = 8 5
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 . Therefore, we can choose y such that sin y cos y = 6 7 . Since cos 2 y + sin 2 y = 1 , we have cos y = 8 5 7 and sin y = 8 5 6 , for some angle y. For angle y, we have 5 cos x + 6 sin x = 8 5 ( 8 5 7 cos x + 8 5 6 sin x ) = 8 5 ( cos x cos y + sin x sin y ) = 8 5 cos ( y − x ) . Since the maximum value of cos ( y − x ) is 1, the desired maximum is 8 5
Log in to reply
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.
I think my solution is quite similar to yours.
f ( x ) = 7 c o s x + 6 s i n x
f ′ ( x ) = − 7 s i n x + 6 c o s x
For max/min, f ′ ( x ) = 0
Either 7 s i n x = 6 c o s x or − 7 s i n x , 6 c o s x = 0
Latter is impossible so we have t a n x = 7 6
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
We can clearly see that for this triangle A C = 8 5
So 6 s i n x = 6 ∗ 8 5 6 = 8 5 3 6 and 7 c o s x = 7 ∗ 8 5 7 = 8 5 4 9
6 s i n x + 7 c o s x = 8 5 3 6 + 4 9 = 8 5 so a = 8 5
Note that if we used the alternate value of x for which t a n x = 7 6 with 0 ∘ ≤ x ≤ 3 6 0 ∘ then s i n x , c o s x < 0 thus making the total a minimum.
I have only just realised there is a typo here.
The last line should read 0 ∘ ≤ x ≤ 3 6 0 ∘
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 , d x d y = − 7 sin x + 6 cos x So when y is at its maximum, − 7 sin x + 6 cos x = 0 ⇒ 7 sin x = 6 cos x ⇒ tan x = 7 6 ⇒ x = tan − 1 7 6 = β We can show that β is the value of x for which y attains its maximum by finding the second derivative, d x 2 d 2 y = − 7 cos x − 6 sin x − 7 cos β − 6 sin β < 0 which implies that β is indeed the value of x for which y is maximum.
Thus we know the value of x for which a attains its maximum. Using this value, we find that the maximum value of a is, a = 7 cos β + 6 sin β = 8 5 ⇒ a = 8 5
Let 7 cos x + 6 sin x ≡ R cos ( x − α )
7 cos x + 6 sin x ≡ R cos x cos α + R sin x sin α
When x = 0 ∘ , 7 = R cos α ......(1)
When x = 9 0 ∘ , 6 = R sin α ......(2)
( 1 ) 2 + ( 2 ) 2 :
7 2 + 6 2 = R 2 ( cos 2 α + sin 2 α )
7 2 + 6 2 = R 2
R = 7 2 + 6 2
R cos ( x − α ) = a
Since we need to find the maximum value of a , we let R = a , i.e. cos ( x − α ) = 1 , which is the maximum value of cos ( x − α ) .
a = 7 2 + 6 2
Hence, maximum value of a = 8 5 .
Using Cauchy-Schwartz inequality:
∣ 7 cos x + 6 sin x ∣ ≤ ( 7 2 + 6 2 ) ( cos 2 x + sin 2 x ) = 8 5 ∴ a = 8 5
What a coincidence!
By Cauchy-Schwarz inequality, we obtain: ∣ 7 cos x + 6 sin x ∣ ≤ cos 2 x + sin 2 x ⋅ 7 2 + 6 2 Since cos 2 x + sin 2 x = 1 we have: 7 cos x + 6 sin x ≤ 8 5 So the result is 8 5
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
Problem Loading...
Note Loading...
Set Loading...
In general k cos x + ℓ sin x = k 2 + ℓ 2 ( k 2 + ℓ 2 k cos x + k 2 + ℓ 2 ℓ sin x ) = k 2 + ℓ 2 ( sin y cos x + cos y sin x ) = k 2 + ℓ 2 sin ( x + y ) ≤ k 2 + ℓ 2 Hence $$a = 7^2 + 6^2 = \boxed{85}$$