Question #8977f

Question

Question #8977f

1 Answer

Impact of this question

1793 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-16T02:02:57

$1 + \sqrt{2}$ $1 + sqrt2$

Explanation:

Call tan ( 67.5) = tan t.
tan (2t) = tan (135) = - 1
Use trig identity:
$tan 2 t = \frac{2 tan t}{1 - {tan}^{2} t}$ $tan 2t = (2tan t)/(1 - tan^2 t$
In this case:
$- 1 = \frac{2 tan t}{1 - {tan}^{2} t}$ $-1 = (2tan t)/(1 - tan^2 t)$
Cross multiply and bring quadratic equation to standard form:
${tan}^{2} t - 2 tan t - 1 = 0$ $tan^2 t - 2tan t - 1 = 0$
$D = d^{2} = b^{2} - 4 a c = 4 + 4 = 8$ $D = d^2 = b^2 - 4ac = 4 + 4 = 8$ --> $d = \pm 2 \sqrt{2}$ $d = +- 2sqrt2$
There are 2 real roots:
$tan t = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{2}{2} \pm \frac{2 \sqrt{2}}{2} = 1 \pm \sqrt{2}$ $tan t = -b/(2a) +- d/(2a) = 2/2 +- (2sqrt2)/2 = 1 +- sqrt2$
Since tan (67.5) > 0, there for:
$tan t = tan (67.5) = 1 + \sqrt{2}$ $tan t = tan (67.5) = 1 + sqrt2$

Science

Math

Humanities

... and beyond

Question #8977f

1 Answer

Explanation:

Related questions

Impact of this question