The discussion revolves around the equation 2Arctan(1/3) + Arctan(1/7) = π/4, focusing on the mathematical reasoning and potential errors in the approach to solving it.
Some participants have identified a mistake in the application of the sine function, noting that sin(A+B) does not equal sinA + sinB. There is a suggestion to follow a specific line of reasoning regarding the expansion of tangent double angles and sums.
Participants are working under the constraints of the original problem and are questioning the assumptions made in their approaches, particularly regarding the manipulation of trigonometric identities.
Alexx1 said:2Arctan(1/3) + arctan(1/7) = π/4
I try to solve it like this:
= sin(2arctan(1/3)) + sin(arctan(1/7) = sin(π/4)
But this isn't correct
What have I done wrong?
Alexx1 said:2Arctan(1/3) + arctan(1/7) = π/4
I try to solve it like this:
= sin(2arctan(1/3)) + sin(arctan(1/7) = sin(π/4)
But this isn't correct
What have I done wrong?
Mentallic said:The problem is that [tex]sin(A+B)\neq sinA+sinB[/tex] for all A and B, which is what you've done on the left side of the equation.
Since you took the sine of both sides, it should be [tex]sin\left(2arctan(1/3)+arctan(1/7)\right)=sin(\pi/4)[/tex]
and from the mistake above that I showed you, you can't separate each part in the sine to make two sine functions as you've done.
Try follow on and see where rockfreak is leading you. You need to know how to expand tan double angles and sums.