by Walter VanniniIntroductionTrianglesUnit CircleRoot typical SquareRotation MatrixDeterminantOrthonormal BasisCombinatorial IdentitiesExponential FunctionAlgebraFactorizationParametrized Curve DerivativesIntegralConservation the EnergyHyperbolic AnaloguesFeedback
We will check out various interpretations the the over identity. Later on parts ofthis essay don"t often depend on previously parts, so please feel totally free to skimpast incomprehensible text until you find something you"re comfortable with.
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The usual method the identity is interpreted is via Pythagoras"theorem. In a right angled triangle through sidesa,b,c, and also angle t in ~ the vertexwhere a and c meet, cos(t) is by definitiona/c, sin(t) is by definition b/c , and also socos2(t) + sin2(t) is(a/c)2+(b/c)2, i m sorry by simplealgebraic manipulation is(a2+b2)/c2. Pythagoras"theorem says that a2+b2 isc2, and also so this simplifies come 1.
Traditionally (ie. The method I learned it), the following step is torealize that the unit circle focused at (0,0) in the(x,y) plane is characterized byx2+y2=1. The over identity can then beinterpreted together saying the the point (cos(t), sin(t)) is onthe unit circle. Furthermore, this method leads to a definition ofcos(t) and sin(t) for all genuine t.
Root typical Square
Another way to understand this identity is via the trigonometric identities
Incidentally, these identities pretty lot leap the end at you fromlooking at the graphs the y = cos2(x) and y =sin2(x).
Another way to look at it is to remember the the root median square ofcos is 1/√2, as is the root mean square ofsin. This way that cos2 is 1/2 +variation, and sin2 is 1/2 +variation, and also amazingly the variation ofcos2 is specifically the an adverse of the sport ofsin2. Actually, this does fit in v the factthat sin and cos are just phase-shifted versionsof every other, so that sin2andcos2 space phase-shifted version of each other, sotheir variations have to be somehow related.Anyway, we now have thatcos2(t)+sin2(t) is of the form (1/2+ deviation) + (1/2 - deviation), and also so this is 1.
Rotation MatrixYet another means to know the identification is via 2 by 2 matrices. Thelinear procession that represents an anticlockwise rotation by anglet is
The rotation matrix sends out the suggest (1,0) to(cos(x),sin(x)) and the allude (0,1) to (-sin(x),cos(x)). The traditional basis e1=(1,0),e2=(0,1) is one orthonormal basis, meaning thate1.e1=1 (check: 1.1+0.0=1)e1.e2=0 (check: 0.1+1.0=0)e2.e2=1 (check: 0.0+1.1=1)
Since a rotation is a "rigid motion", it will certainly transform an orthonormal communication toanother orthonormal basis. This means that the over three equations will betrue when e1 is (cos(t), sin(t)), ande2 is(-sin(t), cos(t)). The two equationse1.e1 = 1and e2.e2 = 1are each a restatement that cos2(t) + sin2(t) =1, if e1.e2 = 0is a explain of the identitycos(t)(-sin(t))+sin(t)cos(t) = 0.Strangely enough, also the e1.e2 statement is amanifestation the the identity, as is displayed below.
Combinatorial IdentitiesAnother method to look in ~ the identity is via the power series expansions ofcos(x) and also sin(x).
|cos2(x)||=||1||-||(1/2! + 1/2!)||x2||+||(1/4! + 1/2!2! + 1/4!)||x4/4!||-||…|
|sin2(x)||=||x2||-||(1/3! + 1/3!)||x4/4!||+||…|
|1/2! + 1/2!||=||1|
|1/4! + 1/2!2! + 1/4!||=||1/3! + 1/3!|
|1/6! + 1/4!2! + 1/2!4! + 1/6!||=||1/5! + 1/3!3! + 1/5!|
|2C0 + 2C2||=||2C1|
|4C0 + 4C2+ 4C4||=||4C1 + 4C3|
|6C0 + 6C2+ 6C4 + 6C6||=||6C1 + 6C3+ 6C5|
|Continuing follow me these lines, the truth that|
|ex||=||1 + x + x2/2! + x3/3! + x4/4! + …|
|immediately tells us that|
|eix||=||1 + ix - x2/2! - ix3/3! + x4/4! + …|
|which is, dividing into the even and also odd powers|
|eix||=||cos(x) + isin(x)|
|At this allude there room two ways to get to the identity.|
The first is come realize the the solitary equation eix =cos(x)+isin(x) provides us the equation e-ix =cos(x)-isin(x) and from these two equations we have the right to solve forcos(x) and also sin(x) to getcos(x) = (eix+e-ix)/2sin(x) = (eix-e-ix)/2i.Letting w represent eix, for this reason that1/w is e-ix, we have thatcos2(x)+sin2(x) is(w+1/w)2/4 - (w-1/w)2/4. Thissimplifies to 1. Therefore the trigonometric identity have the right to be viewedas the algebraic identification (w+1/w)2 -(w-1/w)2 = 4.
The 2nd is to be clever (in a different way), and also realize us canfactorize a2 + b2 as(a-ib)(a+ib). If you haven"t seen this before, it"s arestatement the a2-b2 = (a-b)(a+b).Using complex numbers, we have that -b2 is(ib)2, which leads toa2+b2 = a2-(-b2) =a2 - (ib)2 = (a-ib)(a+ib). Incidentally,it"s the course likewise true the a2+b2 =(b-ia)(b+ia). Anyway, using the administer tocos2(x)+sin2(x), us get(cos(x)+isin(x))(cos(x)-isin(x)), i beg your pardon iseixe-ix. Using w to representeix, the reality thatcos2(x)+sin2(x)=1 is just aconvoluted means of speak w(1/w)=1.
Parametrized Curve Derivatives
Consider the parametrized curve c(t) = (cos(t), sin(t)). Theidentity tells united state that this parametrized curve is constantly on the unitcircle about the origin. Differentiating, we have actually c"(t) =(-sin(t), cos(t)) . C"(t).c"(t) is (-sin(t))(-sin(t))+ cos(t)cos(t), i beg your pardon is justcos2(t)+sin2(t). The reality that this is1 tells united state that the parametrized curve is actuallyparametrized by arc length. So, the trigonometric identity deserve to beviewed as simply stating the reality that radians traverse the unitcircle in ~ unit speed. Finally, separating again, we getc""(t) = (-cos(t), -sin(t)). Obviously, c""(t).c""(t) =cos2(t) + sin2(t), so the the identitytells us that moving uniformly along the circle results in anacceleration of consistent magnitude.
Finally, an additional approach to creating thatcos2(x)+sin2(x) is 1 is torealize the it is really stating two different properties, namely thatcos2(x)+sin2(x) is a constant, and thatthe constant happens to be 1, ie. For some value of x,cos2(x)+sin2(x) is 1. Thelatter residential property is easily established: taking x=0, cos(x)=1 and sin(x)= 0. Clearly12+02 =1.
Given a role f, to create that f is aconstant function, it suffices to establish that the derivative off is zero. Using this technique tof=cos2+sin2, the very first step is usingthe chain dominance to obtain that the derivative ofcos2(x)+sin2(x) is 2cos(x)cos"x +2sin(x)sin"(x). Via the derivative identities cos"(x) =-sin(x), and also sin"(x) = cos(x), we have actually that2cos(x)cos"x + 2sin(x)sin"(x)simplifies to2cos(x)(-sin(x)) + 2sin(x)cos(x) , which is zero. For this reason theidentity deserve to be viewed as the integrated version of the trivialidentitycos(x)(-sin(x)) + sin(x)(cos(x)) = 0, which us saw previously ina totally different context.
Conservation the Energy
Yet another way to view that the cos2+sin2is constant is to realize that it represents the amount of the potentialand kinetic energies the the solution x=cos(t) come the equationfor straightforward harmonic activity x""(t)+x(t)=0.
For a fragment of fixed 1, the kinetic power is(1/2)x"(t)2, and the potential power is(1/2)x(t)2 (upto an additive constant).Conservation of energy tells united state that(1/2)x"(t)2+(1/2)x(t)2is a constant, and sox"(t)2+x(t)2is additionally a constant.Taking the solution x(t)=cos(t) (or x(t)=sin(t))we gain the identity.
I should cite that none of the over mathematics is dependence onthe underlying physics. If x(t) is a systems to the second orderequation x""(t)+F(x(t)) = 0,then (1/2)x"(t)2 + V(x(t)),where V is one antiderivative that F, is constant.Simple differentiation extablishes this.
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Incidentally, due to the fact that x(t)=acos(t)+bsin(t) is the basic solutionto the 2nd order equation, we have that(acos(t)+bsin(t))2+(-asin(t)+bcos(t))2is the consistent a2+b2
The next step to expertise the identification is come compare and also contrastit to the identity for hyperbolic cosine and hyperbolic sine, namelycosh2(t) - sinh2(t) = 1. I"ll deferthis to a later essay.