How do you find the exact values of tan 112.5 degrees using the half angle formula?

Question

How do you find the exact values of tan 112.5 degrees using the half angle formula?

2 Answers

Impact of this question

20612 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-22T13:24:43

$tan (112.5) = - (1 + \sqrt{2})$ $tan(112.5)=-(1+sqrt(2))$

Explanation:

$112.5 = 112 \frac{1}{2} = \frac{225}{2}$ $112.5=112 1/2=225/2$

**NB : ** This angle lies in the 2nd Quadrant.

$\Rightarrow tan (112.5) = tan (\frac{225}{5}) = \frac{sin (\frac{225}{2})}{cos (\frac{225}{2})} = - \sqrt{{⎡ ⎢ ⎣ \frac{sin (\frac{225}{2})}{cos (\frac{225}{2})} ⎤ ⎥ ⎦}^{2}} = - \sqrt{\frac{{sin}^{2} (\frac{225}{2})}{{cos}^{2} (\frac{225}{2})}}$ $=>tan(112.5)=tan(225/5)=sin(225/2)/cos(225/2)=-sqrt([sin(225/2)/cos(225/2)]^2)=-sqrt(sin^2(225/2)/cos^2(225/2))$

We say it's negative because the value of $tan$ $tan$ is always negative in the second quadrant!

Next, we use the half angle formula below :

${sin}^{2} (\frac{x}{2}) = \frac{1}{2} (1 - cos x)$ $sin^2(x/2)=1/2(1-cosx)$

${cos}^{2} (\frac{x}{2}) = \frac{1}{2} (1 + cos x)$ $cos^2(x/2)=1/2(1+cosx)$

$\Rightarrow tan (112.5) = - \sqrt{\frac{{sin}^{2} (\frac{225}{2})}{{cos}^{2} (\frac{225}{2})}} = - \sqrt{\frac{\frac{1}{2} (1 - cos (225))}{\frac{1}{2} (1 + cos (225))}} = - \sqrt{\frac{1 - cos (225)}{1 + cos (225)}}$ $=>tan(112.5)= -sqrt(sin^2(225/2)/cos^2(225/2))= -sqrt((1/2(1-cos(225)))/(1/2(1+cos(225))))= -sqrt((1-cos(225))/(1+cos(225)))$

Notice that : $225 = 180 + 45 \Rightarrow cos (225) = - cos (45)$ $225=180+45 => cos(225)=-cos(45)$

$\Rightarrow tan (112.5) = - \sqrt{\frac{1 - (- cos 45)}{1 + (- cos 45)}} = - \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}}} = \sqrt{\frac{2 + \sqrt{2}}{2 - \sqrt{2}}}$ $=>tan(112.5)=-sqrt((1-(-cos45))/(1+(-cos45)))=-sqrt((1+sqrt(2)/2)/(1-sqrt(2)/2))=sqrt((2+sqrt(2))/(2-sqrt(2)))$

Now you want to Rationalize;

$\Rightarrow - \sqrt{\frac{(2 + \sqrt{2}) \times (2 + \sqrt{2})}{(2 - \sqrt{2}) \times (2 + \sqrt{2})}} = - \sqrt{\frac{{(2 + \sqrt{2})}^{2}}{4 - 2}} = - \frac{2 + \sqrt{2}}{\sqrt{2}} = - \frac{\sqrt{2} \times (2 + \sqrt{2})}{\sqrt{2} \times \sqrt{2}} = - \frac{2 \sqrt{2} + 2}{2} = - (1 + \sqrt{2})$ $=>-sqrt(((2+sqrt(2))xx(2+sqrt(2)))/((2-sqrt(2))xx(2+sqrt(2))))= -sqrt(((2+sqrt(2))^2)/(4-2))=-(2+sqrt(2))/sqrt(2)=-(sqrt(2)xx(2+sqrt(2)))/(sqrt2xxsqrt2)=-(2sqrt2+2)/2=color(blue)(-(1+sqrt(2)))$

Answer 2 · 2015-07-22T15:31:42

Find tan 112.5

Ans: (-1 - sqrt2)

Explanation:

Call tan 112.5 = tan t
tan 2t = tan 225 = tan (45 + 180) = tan 45 = 1
Use trig identity: $tan 2 t = \frac{2 t}{1 - t^{2}}$ $tan 2t = (2t)/(1 - t^2)$ -->

$1 = \frac{2 t}{1 - t^{2}}$ $1 = (2t)/(1 - t^2)$ --> $t^{2} + 2 t - 1 = 0$ $t^2 + 2t - 1 = 0$
$D = d^{2} = b^{2} - 4 a c = 4 + 4 = 8 - \to d = \pm 2 \sqrt{2}$ $D = d^2 = b^2 - 4ac = 4 + 4 = 8 --> d = +- 2sqrt2$

$t = tan 112.5 = - \frac{2}{2} \pm \frac{2 \sqrt{2}}{2} = - 1 \pm \sqrt{2}$ $t = tan 112.5 = -2/2 +- (2sqrt2)/2 = - 1 +- sqrt2$

Since t = 112.5 deg is in Quadrant II, its tan is negative, then only the negative answer is accepted : (-1 - sqrt2)

Science

Math

Humanities

... and beyond

How do you find the exact values of tan 112.5 degrees using the half angle formula?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question