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Welcome to NRICH.

tan(q) = 1 ± 2^0.5

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By Anonymous on Saturday, May 5, 2001 - 06:50 pm:

How would you find the values of q if tan(q) = 1 ± 2^0.5?

Thanks

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By Brad Rodgers (P1930) on Sunday, May 6, 2001 - 03:14 am:

The answer to the 'plus one' is 3p/8, and I haven't yet figured any sort of a solution to the second one in terms of pi. Here's a relatively easy way to prove that tan(3p/8)=1+2^1/2
See if you can figure out why the angles are as they are. I'll for now assume you're okay with the way the angles are labeled. First, use the angle bisector theorem to deduce that

1/a=2^1/2/b

or

b=2^1/2a

Remembering that

a+b=1, you should be able to substitute and see that a=1/(1+2^1/2) Thus the tangent of 3p/8=1/[1/(1+2^1/2)]=1+2^1/2 as required. I'm still working on an angle that has tangent of 1-2^1/2, it seems to be -p/8, but I'm having trouble proving this.

Brad

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By Brad Rodgers (P1930) on Sunday, May 6, 2001 - 03:27 am:

The proof is actually easier for the negative solution. The answer is -p/8 and you get there by noticing that

in the preceding triangle, we get, as before,

b=2^1/2a

and

a+b=1

So a=1/(1+2^1/2) or 2^1/2-1. Therefore

tan(p/8)=2^1/2-1

Which, recalling that tan(-x)=-tan(x), gives us the answer. To get the full answer then, just remember that tan oscillates every pi.

Brad

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By Kerwin Hui (Kwkh2) on Sunday, May 6, 2001 - 03:14 pm:

Another way of solving:

tan q = 1 ± sqrt(2)

so (tan q - 1)² = 2

i.e. tan² q - 2 tan q -1 = 0

Now rearrange to get

2 tan q / (1 - tan² q) = -1

But we recognise the left hand side is tan 2q, and proceed from there.

Kerwin

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By Anonymous on Sunday, May 6, 2001 - 08:22 pm:

Thanks Brad and Kerwin

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