Suppose # s(x) # and # c(x) # are 2 functions where: 1) # s'(x) = c(x) # and # c'(x) = -s(x); # 2) # s(0) = 0 # and # c(0) = 1. # What can you say about the quantity: # \qquad [ s(x) ]^2 + [ c(x) ]^2 # ?

3 Answers
Cesareo R.
Feb 6, 2018

#c^2+s^2=1#

Explanation:

We have

#s'=c rArr s cdot s' = s cdot c# and
#s=-c' rArr c cdot s = -c cdot c'# then

#s cdot s'+c cdot c' = 0# or

#c^2+s^2=C_0#

now putting the initial conditions

#c^2(0)+s^2(0) = 0+1=C_0 rArr C_0 = 1#

and finally

#c^2+s^2=1#

Steve M
Feb 6, 2018

# [ s(x) ]^2 + [ c(x) ]^2 =1 #

Explanation:

We are given that:

# s'(x) = c(x) \ \ \ \ \ \ \ \ \ \ \ # ..... [A]
# c'(x) = -s(x) \ \ \ \ \ \ # ..... [B]

Differentiating the second equation [B] wrt #x# we get:

#c''(x) = -s'(x) #

And then incorporating the first equation [A]:

# -c''(x) = c(x) #

Or:

# c''(x) + c(x) = 0 #

Which is a Second Order ODE with constant coefficients, so we consider the associated Auxiliary equation:

# m^2 + 1 = 0 => m= +- i #

So as we have two pure imaginary roots, the solution is of the

# c(x) =Acosx + Bsinx #

And then using [B] we have:

# s(x) = -c'(x) #
# :. s(x) = -{-Asinx+Bcosx} #
# \ \ \ \ \ \ \ \ \ \ \ = Asinx-Bcosx #

Using the given condition, #c(0)=1# and #s(0)=0# we have:

# A + 0 = 1 => A = 1#
# 0-B = 0 => B=0 #

Thus we have:

# c(x) = cosx #
# s(x) = sinx #

And so we infer that:

# [ s(x) ]^2 + [ c(x) ]^2= sin^2x+cos^2 =1 #

Ultrilliam
Apr 29, 2018

If you're going to submit an inefficient answer, it may as well be interesting !

Explanation:

We have:

#((s'),(c')) = ((0, 1),(-1,0)) ((s),(c)) #

# implies mathbf s' = M mathbf s #

That solves trivially as:

  • #mathbf s = e^(x M) mathbf s_o#

Now, "what can we say":

#s^2 + c^2 = mathbf s^T mathbf s#

#= (e^(x M) mathbf s_o)^T (e^(x M) mathbf s_o)#

#= mathbf s_o ^T (e^(x M))^T * e^(x M) mathbf s_o #

#= mathbf s_o ^T e^(x M^T) * e^(x M) mathbf s_o #

Looking the matrices:

#M M^T = ((0, 1),(-1,0)) ((0, -1),(1,0)) = mathbb I = M^T M #

Symmetry....and Commutation

#implies e^(x M^T) * e^(x M)= e^(x (M^T+ M)) #

#= mathbf s_o ^T e^(x (((0, 1),(-1,0)) + ((0, -1),(1,0)))) mathbf s_o #

#= mathbf s_o ^T e^(x( 0)) mathbf s_o = s_o^2 = 1#

So:

#s^2 + c^2 = 1#

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