A calculus problem by Sarthak Srivastava
Solve for real
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sin
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Image credit: Wikipedia Dirk Hünniger
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3 solutions
I agree that the maximum value of sin ( θ ) + cos ( θ ) is 2 , but this is only true if the arguments are both the same. In this case we have one argument being sin ( sin ( x ) ) and the other being cos ( cos ( x ) ) . So all we can say is that the maximum of sin ( sin ( sin ( x ) ) ) + cos ( cos ( cos ( x ) ) ) is 2 , which is greater than 2 π .
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2 > ( 2 π = 1 . 5 7 )
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Yes, that is what I have pointed out, which would as a result not be sufficient information to make the desired conclusion. As I have pointed out above, we can't automatically say that the maximum of sin ( sin ( sin ( x ) ) ) + cos ( cos ( cos ( x ) ) ) is 2 , since the arguments are not the same.
As Richard Levine has pointed out, the maximum of this sum turns out to be approximately 1 . 3 3 5 9 , which is in fact < 2 π , but it took WolframAlpha to get that result.
As Math Man indicates, we can solve this problem using calculus, analyzing relative maximums and minimums of the functions on the two sides of the original equation. This makes this question more of a Level 4 calculus problem than a level 2 geometry problem. I'm still looking for a way to avoid using calculus. :)
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@Brian Charlesworth – It there is a non-calculus approach, I think it will take quite a lot of machinery.
The reason why this is Level 2, is because the answer is somewhat easily guessed from the options. Given that it needs Calculus, I have moved this into Calculus instead.
@Brian Charlesworth – Yes there is a solution through calculus
@Brian Charlesworth – yes as richard said max(cos(cos(cos x)) + sin(sin(sin x)) is 1-3359. putting cos(cos(cosx))) as cos(sin(pi/2 -cosx))) and written all cos terms in terms of sin
According to WolframAlpha online, max(cos(cos(cos x)) + sin(sin(sin x))) is approx. 1.3359 when x is approx. .904984. But pi/2 is approx. 1.57, which is greater than 1.3359. So Samarth's answer would be correct if squareroot(2) was changed to approx. 1.3359 in the answer given.
calced it lol
Note that just because sin α = sin β doesn't immediately imply that α = β ( + 2 n π ) . We could have α + β = π ( + 2 n π ) instead.
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Sir i have edited the solution and submitted as per your note and using (in general) if s i n x = s i n θ t h e n x = n π + ( − 1 ) n θ n ∈ N
sir general solution for sinx is s i n x = s i n θ t h e n x = n π + ( − 1 ) n θ n ∈ N how can (as you mention) solution for tanx be a solution for sinx
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This looks much better now.
We are saying the same thing, just that I split it out into cases of n = 2 k and n = 2 k + 1 . Using your general solution, we have
α = 2 k π + β or α = ( 2 k + 1 ) π − β .
FYI - To type equations in Latex, you just need to add around your math code. I've edited the first line in your solution, so you can refer to it.
is my solution correct or wrong?
i m not satisfied with your solution
I'm not sure, but I think there is 1 solution!! , the equation can be written as tan(tan(tan(tan(x)))) = 1 , then x will be 89.35 degrees "and if you consider the first atan value is 225 instead of 45, it will converge to the same value of 89.35, correct me if I'm wrong..
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No. That does not work. It is not true that cos sin = tan , even thought that is a simplified way of remembering. You have to consider the argument of the function.
What is true is that cos θ sin θ = tan θ .
I just have a question. Would you tell me, please, is it correct expression "cos(cos(cos(x)))) or sin(sin(sin(x))))"? Is it correct to get cos, sin, tan, etc. of number that is not value of some angle in degrees of radians? In my opinion this problem does not have a sense, hense, do not have solution, and should not be posted here. Are you agree?
It does not have solution because x supposed to be some angle value in degrees or radians, but sin(x) or cos(x) or is just numbers. They are not degrees, not radians. it's impossible to get sin(sin(sin(x)))) or cos(cos(cos(cos(x)))) hence sin(sin(x)) and cos(cos(x)) are numeric values, but not degrees or radians value. Therefore, sorry, this problem is nonsense!
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This equation is in radians.
While I agree with you that "sin of a numeric value doesn't make sense", the convention is to implicitly assume that the numeric value is a radian measure. For example, we say sin 2 π , instead of saying sin ( 2 π radians ).
From another point of view, we say that the function sin : R → R , which is a function from the reals to the reals.
As such, repeated iterations of the sin (and cos) function still make sense.
cos[cos[cos[cos[X]]]] = sin[pi/2 -cos[cos[cos[X]]]] therefore , sin[pi/2 -cos[cos[cos[X]]]] = sin(sin(sin(sin(x)))] therfore , pi/2 -cos[cos[cos[X]]] = sin(sin(sin(x))) i.e. sin(sin(sin(x))) + cos[cos[cos[X]]] = pi/2 which is not possibe bcoz maX vaue of sin(x) + cos(x) is sqrt 2 pi/2 > sqrt 2
It'll be a fluctuating pulse over x axis throughout real axis without touching any point on x axis
c o s ( c o s ( c o s ( c o s x ) ) ) = s i n ( s i n ( s i n ( s i n x ) ) )
can be rewritten as
s i n ( 2 π − c o s ( c o s ( c o s x ) ) ) = s i n ( s i n ( s i n ( s i n x ) ) ) 2 π − c o s ( c o s ( c o s x ) ) = n π + ( − 1 ) n s i n ( s i n ( s i n x ) ) ( s i n x = s i n θ t h e n x = n π + ( − 1 ) n θ n ∈ N ) n π − 2 π = ( − 1 ) n s i n ( s i n ( s i n x ) ) + c o s ( c o s ( c o s x ) ) s i n c e ∣ ∣ n π − 2 π ∣ ∣ > 2 f o r a n y n ∈ N i f n i s o d d t h e n − s i n ( s i n ( s i n x ) ) + c o s ( c o s ( c o s x ) ) a n d i f n i s e v e n s i n ( s i n ( s i n x ) ) + c o s ( c o s ( c o s x ) ) b u t m a x i m u m v a l u e o f ( − 1 ) n s i n ( s i n ( s i n x ) ) + c o s ( c o s ( c o s x ) ) i s 1 . 3 3 5 9 b u t ∣ ∣ n π − 2 π ∣ ∣ > 1 . 3 3 5 9 t h e r e f o r e n π − 2 π > ( − 1 ) n s i n ( s i n ( s i n x ) ) + c o s ( c o s ( c o s x ) ) t h e r e i s n o s o l u t i o n f o r x
Edit: To find the maximum of sin sin sin x + cos cos cos x , Wolfram Alpha tells us that it is 1.3359, which is achieved when x ≈ 0 . 9 4 5 . It is hard to solve for this value explicitly, without powerful computational tools.