SOLUTION: Prove the formula tan3@ = 3tan@ - tan^3@ / 1 - 3tan^2@. Hence solve the equation tan3@ = 0 If 0≤@≤360

Algebra ->  Finance -> SOLUTION: Prove the formula tan3@ = 3tan@ - tan^3@ / 1 - 3tan^2@. Hence solve the equation tan3@ = 0 If 0≤@≤360      Log On


   

Question 1202935: Prove the formula tan3@ = 3tan@ - tan^3@ / 1 - 3tan^2@.
Hence solve the equation tan3@ = 0
If 0≤@≤360
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

One of the many trig identities is
tan%28A%2BB%29+=+%28tan%28A%29%2Btan%28B%29%29%2F%281-tan%28A%29%2Atan%28B%29%29
Let's plug in A = x and B = x

tan%28A%2BB%29+=+%28tan%28A%29%2Btan%28B%29%29%2F%281-tan%28A%29%2Atan%28B%29%29

tan%28x%2Bx%29+=+%28tan%28x%29%2Btan%28x%29%29%2F%281-tan%28x%29%2Atan%28x%29%29

tan%282x%29+=+%282%2Atan%28x%29%29%2F%281+-+tan%5E2%28x%29%29
this will be useful in substitution steps later on.


Go back to
tan%28A%2BB%29+=+%28tan%28A%29%2Btan%28B%29%29%2F%281-tan%28A%29%2Atan%28B%29%29

Let's plug in
A = x
B = 2x
to get the following
tan%28A%2BB%29+=+%28tan%28A%29%2Btan%28B%29%29%2F%281-tan%28A%29%2Atan%28B%29%29

tan%28x%2B2x%29+=+%28tan%28x%29%2Btan%282x%29%29%2F%281-tan%28x%29%2Atan%282x%29%29

tan%283x%29+=+%28tan%28x%29%2Btan%282x%29%29%2F%281-tan%28x%29%2Atan%282x%29%29

This is one large fraction with
numerator = tan(x)+tan(2x)
denominator = 1-tan(x)*tan(2x)

In other words,
tan%283x%29+=+P%2FQ
where
P+=+tan%28x%29%2Btan%282x%29
and
Q+=+1-tan%28x%29%2Atan%282x%29

Let's apply a substitution in the numerator
P+=+tan%28x%29%2Btan%282x%29

P+=+tan%28x%29%2B%282%2Atan%28x%29%29%2F%281+-+tan%5E2%28x%29%29





P+=+%28tan%28x%29+-+tan%5E3%28x%29%2B2%2Atan%28x%29%29%2F%281+-+tan%5E2%28x%29%29

P+=+%283tan%28x%29+-+tan%5E3%28x%29%29%2F%281+-+tan%5E2%28x%29%29

Apply that same substitution with the denominator.

Q+=+1-tan%28x%29%2Atan%282x%29

Q+=+1-tan%28x%29%2A%28%282%2Atan%28x%29%29%2F%281+-+tan%5E2%28x%29%29%29

Q+=+1-%282%2Atan%5E2%28x%29%29%2F%281+-+tan%5E2%28x%29%29



Q+=+%281+-+tan%5E2%28x%29-2%2Atan%5E2%28x%29%29%2F%281+-+tan%5E2%28x%29%29

Q+=+%281+-+3%2Atan%5E2%28x%29%29%2F%281+-+tan%5E2%28x%29%29

------------------------------------

Recall that this identity
tan%283x%29+=+%28tan%28x%29%2Btan%282x%29%29%2F%281-tan%28x%29%2Atan%282x%29%29
is of the form P%2FQ

where
P+=+tan%28x%29%2Btan%282x%29
and
Q+=+1-tan%28x%29%2Atan%282x%29

but we found these equivalent forms
P+=+%283tan%28x%29+-+tan%5E3%28x%29%29%2F%281+-+tan%5E2%28x%29%29
and
Q+=+%281+-+3%2Atan%5E2%28x%29%29%2F%281+-+tan%5E2%28x%29%29

The 2nd versions of P and Q involve the denominator 1-tan%5E2%28x%29.
Those expressions cancel when dividing.

Therefore,
tan%283x%29+=+%283tan%28x%29+-+tan%5E3%28x%29%29%2F%281+-+3%2Atan%5E2%28x%29%29

------------------------------------

For the next part of the problem, here are a few hints:
tan%283x%29+=+0

%283tan%28x%29+-+tan%5E3%28x%29%29%2F%281+-+3%2Atan%5E2%28x%29%29+=+0

3tan%28x%29+-+tan%5E3%28x%29+=+0

tan%28x%29%283+-+tan%5E2%28x%29%29+=+0

I'll let you take over from here.