Using the T-formula (1 Viewer)

[Prazo1994]

Hey guys,
Just needed help with these questions

6a) Given that t = 112.5 show that 2t/2-t^2 = 1 (i was able to do this bit)

b) i hence show that tan 112.5 = - (2)^1/2 -1

7 Using the method from the previous Q:
a) show that tan 15 = 2-(3)^1/2
b) show that 7pi/8 = 1-(2)^1/2

 

[viraj30]

howd u do first bit???

 

[fishy89]

Double Angle Formula:
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{2tanx}{1-tan^2x} = tan2x" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{2tanx}{1-tan^2x} = tan2x" title="\frac{2tanx}{1-tan^2x} = tan2x" /></a>

So if <a href="http://www.codecogs.com/eqnedit.php?latex=t = tan112.5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?t = tan112.5" title="t = tan112.5" /></a>),
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{2tan112.5}{1-tan^2(112.5)} = tan2(112.5) = tan225 = 1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{2tan112.5}{1-tan^2(112.5)} = tan2(112.5) = tan225 = 1" title="\frac{2tan112.5}{1-tan^2(112.5)} = tan2(112.5) = tan225 = 1" /></a>

now if 2t/1-t^2 = 1,
rearranging gives
2t = 1 - t^2
t^2 + 2t - 1 = 0

using the quadratic formula:
<a href="http://www.codecogs.com/eqnedit.php?latex=t=\frac{-2\pm \sqrt{(-2)^2-4(1)(-1))))}}{2(1))}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?t=\frac{-2\pm \sqrt{(-2)^2-4(1)(-1))))}}{2(1))}" title="t=\frac{-2\pm \sqrt{(-2)^2-4(1)(-1))))}}{2(1))}" /></a>
= [-2 +- sqrt8]/2 = -1 +- rt2 = -1 - rt2 in our case.

Use the same method by replacing 112.5 with 15 and 7pi/8. Hope this helps.

 

[such_such]

how do you know t=tan112.5? I thought t=tan(1/2)x...

 

[Timske]

Hey guys,
Just needed help with these questions

6a) Given that t = 112.5 show that 2t/2-t^2 = 1 (i was able to do this bit)

b) i hence show that tan 112.5 = - (2)^1/2 -1

7 Using the method from the previous Q:
a) show that tan 15 = 2-(3)^1/2
b) show that 7pi/8 = 1-(2)^1/2

?

 

[karnbmx]

Double Angle Formula:


So if ),


now if 2t/1-t^2 = 1,
rearranging gives
2t = 1 - t^2
t^2 + 2t - 1 = 0

using the quadratic formula:

= [-2 +- sqrt8]/2 = -1 +- rt2 = -1 - rt2 in our case.

Use the same method by replacing 112.5 with 15 and 7pi/8. Hope this helps.

The real reason why we neglect the positive case is because the angle is obtuse, and therefore, tan is negative as it lies in the second quadrant.

Hence, it is -1 - rt(2) not -1 + rt(2).

Something to keep in mind when answering these questions.