Question

How do you find the exact values of $\tan 112.5^\circ $ using the half angle formula?

Answer 1

Hint: Here, in this question, we need to find the exact value of \[\tan 112.5^\circ \] using the half angle identity. In order to use the half angle identity, we will first rewrite the given value as the double of the given angle and then, simplify in such a way that we get to use half angle identity and find the answer.

Formula used: $\tan x = \dfrac{{\sin x}}{{\cos x}}$
\[{\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{2}(1 - \cos x)\]
\[{\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{2}(1 + \cos x)\]
\[\tan 2t = \dfrac{{2t}}{{1 - {t^2}}}\;\]

Complete step-by-step solution:
The given value is \[\tan 165^\circ \] , we need to find its exact value by using the half angle identity.
This angle lies in the 2nd Quadrant.
First, we can also write \[\tan \left( {112.5} \right) = \tan (\dfrac{{225}}{2})\]
Expanding the function using $\tan x = \dfrac{{\sin x}}{{\cos x}}$ ,
\[\tan (\dfrac{{225}}{2}) = \dfrac{{\sin (\dfrac{{225}}{2})}}{{\cos (\dfrac{{225}}{2})}}\]
Taking square root and squaring the terms at the same time,
\[ = \sqrt {{\text{ }}{{\left[ {\dfrac{{\sin \left( {\dfrac{{225}}{2}} \right)}}{{\cos \left( {\dfrac{{225}}{2}} \right)}}} \right]}^2}} \]
Square is been multiplied inside,
\[ = \sqrt {{\text{ }}\dfrac{{{{\sin }^2}\left( {\dfrac{{225}}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{{225}}{2}} \right)}}} \]
We say it's negative because the value of tan is always negative in the second quadrant!
\[ = - \sqrt {{\text{ }}\dfrac{{{{\sin }^2}\left( {\dfrac{{225}}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{{225}}{2}} \right)}}} \]
Next, we will use the half angle formula:
Since we know that,
\[{\sin ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{2}(1 - \cos x)\]
\[{\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{2}(1 + \cos x)\]
We will use these trigonometric functions in here,
$\tan (112.5) = - \sqrt {{\text{ }}\dfrac{{{{\sin }^2}\left( {\dfrac{{225}}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{{225}}{2}} \right)}}} $
\[ - \sqrt {{\text{ }}\dfrac{{{{\sin }^2}\left( {\dfrac{{225}}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{{225}}{2}} \right)}}} = - \sqrt {{\text{ }}\dfrac{{\dfrac{1}{2}\left( {1 - \cos 225} \right)}}{{\dfrac{1}{2}\left( {1 + \cos 225} \right)}}} \]
Cancelling the $2$ on both sides,
\[ = - \sqrt {{\text{ }}\dfrac{{\left( {1 - \cos 225} \right)}}{{\left( {1 + \cos 225} \right)}}} \]
Now, we can rewrite \[225 = 180 + 45\] which changes into \[\cos (225) = \cos (180 + 45)\]
We also know $\cos (180 + x) = - \cos x$
\[\cos (180 + 45) = - \cos \left( {45} \right)\]
$ - \cos 45$ \[ = - \sqrt {{\text{ }}\dfrac{{1 - ( - \cos 45)}}{{1 + ( - \cos 45)}}} \]
\[ = - \sqrt {{\text{ }}\dfrac{{1 + \cos 45}}{{1 - \cos 45}}} \]
Substituting the value of $\cos 45$ that is $1$ ,
\[ = - \sqrt {{\text{ }}\dfrac{{1 + \dfrac{{\sqrt 2 }}{2}}}{{1 - \dfrac{{\sqrt 2 }}{2}}}} \]
Now, we will take LCM
\[ = - \sqrt {{\text{ }}\dfrac{{\dfrac{{2 + \sqrt 2 }}{2}}}{{\dfrac{{2 - \sqrt 2 }}{2}}}} \]
\[ = - \sqrt {{\text{ }}\dfrac{{2 + \sqrt 2 }}{{2 - \sqrt 2 }}} \]
Now we have to Rationalize this,
\[ = - \sqrt {{\text{ }}\dfrac{{2 + \sqrt 2 }}{{2 - \sqrt 2 }} \times \dfrac{{2 + \sqrt 2 }}{{2 + \sqrt 2 }}} \]
We know that ${a^2} - {b^2} = (a + b)(a - b)$ using this formula,
\[ = - \sqrt {{\text{ }}\dfrac{{{{\left( {2 + \sqrt 2 } \right)}^2}}}{{{2^2} - {{\left( {\sqrt 2 } \right)}^2}}}} \]
\[ = - \sqrt {{\text{ }}\dfrac{{{{\left( {2 + \sqrt 2 } \right)}^2}}}{{4 - 2}}} \]
\[ = - \dfrac{{\sqrt {{{\left( {2 + \sqrt 2 } \right)}^2}} }}{{\sqrt 2 }}\]
\[ = - \dfrac{{2 + \sqrt 2 }}{{\sqrt 2 }}\]
Taking the common number out, we get
\[ = - \dfrac{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 }}\]
\[ = - \left( {\sqrt 2 + 1} \right)\]
$ - (1 + \sqrt 2 )$

Therefore, the exact value of $\tan 112.5^\circ $ is $ - (1 + \sqrt 2 )$.

Note: Alternative method:
There is another method to find the exact value of $\tan 112.5^\circ $.
First, let \[\tan {\text{ }}112.5{\text{ }} = {\text{ }}\tan {\text{ }}t\]
Now, we can rewrite it as,
\[\tan {\text{ }}2t{\text{ }} = {\text{ }}\tan {\text{ }}225\]
Now, we can rewrite \[225 = 45 + 180\] which changes into \[\tan (225) = \tan (45 + 180)\]
 \[\tan \left( {45{\text{ }} + {\text{ }}180} \right) = \tan 45 = 1\]
Now, we will use this trigonometric identity, \[\tan 2t = \dfrac{{2t}}{{1 - {t^2}}}\;\]
\[\tan {\text{225}} = \tan {\text{2}}t = \tan 45 = 1\]
\[1 = \dfrac{{2t}}{{1 - {t^2}}}\;\]
Cross multiplying this,
$1 - {t^2} = 2t$
${t^2} + 2t - 1 = 0$
Now we get this quadratic equation,
Applying ${t^2} + 2t - 1 = 0$ in $\dfrac{{{\text{ - b \pm }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ - 4ac}}} }}{{{\text{2a}}}}$ ,
  $x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4(1)( - 1)} }}{{2(1)}}$
Simplifying the numerator,
${\text{x = }}\dfrac{{ - 2 \pm \sqrt {4 + 4} }}{2}$
${\text{x = }}\dfrac{{ - 2 \pm \sqrt 8 }}{2}$
Simplifying the numerator, we get,
${\text{x = }}\dfrac{{ - 2 \pm 2\sqrt 2 }}{2}$
${\text{x = }}\dfrac{{ - 2 + 2\sqrt 2 }}{2},\dfrac{{ - 2 - 2\sqrt 2 }}{2}$
${\text{x = }} - 1 + \sqrt 2 , - 1 - \sqrt 2 $
Since \[\tan (112.5{\text{ }}^\circ )\] is in Quadrant II, the $\tan $ value is negative.
Therefore, the value will be negative.
Therefore, the value of \[\tan (112.5{\text{ }}^\circ )\] is $ - (1 + \sqrt 2 )$ .