One reply on “Trig Identities are just really hard to prove”
I think the student doesn’t understand the concept that an identity is proved by constructing a chain of things that are equal to each other. Rather, the student thinks you prove an identity by performing a series of transformations of expressions that gradually lead from one expression to another. A transformation, according to this student, does not need to produce an expression equivalent to the given one. Rather, it needs to produce an expression that looks more like the desired goal.
To me, it seems this student recognized a difference of fourth powers on the left and a sum of squares on the right. Knowing that a difference of fourth powers factors into a sum of squares and a difference of squares, the student tries to make the left side more like the right by dividing it by a difference of squares.Unfortunately, his/her exponent notation on the secant terms is poor. And he/she incorrectly “cancels” the secant terms while simplifying the quotient. The result looks almost right — just missing a factor of 2. So the student quickly multiplies by a factor of 2 (without distributing), carelessly throwing a tangent squared in there as well, and declares both sides equal.
One reply on “Trig Identities are just really hard to prove”
I think the student doesn’t understand the concept that an identity is proved by constructing a chain of things that are equal to each other. Rather, the student thinks you prove an identity by performing a series of transformations of expressions that gradually lead from one expression to another. A transformation, according to this student, does not need to produce an expression equivalent to the given one. Rather, it needs to produce an expression that looks more like the desired goal.
To me, it seems this student recognized a difference of fourth powers on the left and a sum of squares on the right. Knowing that a difference of fourth powers factors into a sum of squares and a difference of squares, the student tries to make the left side more like the right by dividing it by a difference of squares.Unfortunately, his/her exponent notation on the secant terms is poor. And he/she incorrectly “cancels” the secant terms while simplifying the quotient. The result looks almost right — just missing a factor of 2. So the student quickly multiplies by a factor of 2 (without distributing), carelessly throwing a tangent squared in there as well, and declares both sides equal.