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Expert: - 1/9/2009


Question
Hi,
I'm stuck on the following question:

If 2x + y = pi/4 show that

tan y = (1-2tanx- tan^2x)/(1+2tanx - tan^2x)

Hence deduce that tan(pi/8) is a root of t^2 + 2t -1 =0, and that its value is( [sqrt 2] -1).

Thank you very much!!

Answer
OK.  That wasn't hard - just took awhile, once I thought about it.
The key to the problem is to know that

[1] tan(a+b) = {tan(a)+tan(b)}/{1 - tan(a)tan(b)}
note that if the problem is tan(a-b), you have
tan(a-b) = {tan(a) - tan(b)}/{1 + tan(a)tan(b)} since
tan(-b)= -tan(b).

This implies (by letting a=b) that [2] tan(2x) = 2tan(x)/{1-tanē(x)}.

You may need to reverse this process if you are suppose to turn the right side into the left, but here's how to do it.

To start with, we know that y = π/4 - 2x, so
tan(y) = tan(π/4 - 2x).

From [1], we know that this is
{tan(π/4) - tan(2x)}/{1 - tan(π/4)tan(2x)}.

Since it is known that tan(π/4) = 1, this reduces to
{1 - tan(2x)}/{1 + tan(2x)}.

Now from [2], this converts to
{1 - 2tan(x)/[1-tanē(x)]}/{1 + 2tan(x)/[1-tanē(x)]}.

Now if we multiply by {1-tanē(x)]}/{1-tanē(x)]}, we get
{1 - tanē(x) - 2tan(x)}/{1 - tanē(x) + 2tan(x)}.

By merely rearranging the terms on the bottom and not putting in the parenthesis on tanx, this looks like what we're looking for, nameley (1-2tanx- tanēx)/(1+2tanx - tanēx).

You know, once you get these, they don't seem that hard ... all you've got to know is [1], which will give you [2].

Thanks for making me think about such a challenging problem that seems like a piece of cake now.  It took me four pages, but the final solution on paper was only four simple two line steps.

Can I give you a rating of 10 for such a wonderful problem?

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