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

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

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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)

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

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