The equation 2Arctan(1/3) + Arctan(1/7) = π/4 is incorrectly approached by applying the sine function separately to each term. The correct method involves recognizing that sin(A+B) ≠ sinA + sinB. Instead, the equation should be treated as sin(2Arctan(1/3) + Arctan(1/7)) = sin(π/4). Understanding the expansion of tangent double angles and sums is crucial for solving this equation correctly.
PREREQUISITESMathematics students, educators, and anyone interested in solving trigonometric equations or enhancing their understanding of inverse trigonometric functions.
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 sin(A+B)\neq sinA+sinB 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 sin\left(2arctan(1/3)+arctan(1/7)\right)=sin(\pi/4)
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.