Question

How do you find the exact values of $\tan {67.5^ \circ }$ using the half angle formula?

Answer 1

We use the half angle formulas for solving this problem. Using this formula, we can solve many other problems. The formulas are, $\sin \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}}$ and $\cos \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 + \cos \theta }}{2}}$ . And we will also discuss trigonometric ratios of angles like $\left( {{{90}^ \circ } \pm \theta } \right)$ in this problem. We will also use some known trigonometric ratios like sine and cosine values of ${45^ \circ }$ .

Complete step by step answer:
Tangent value is positive in the first quadrant, so, $\tan {67.5^ \circ }$ is a positive value. So we should get a positive answer.
Firstly, the angle ${67.5^ \circ }$ is half of the angle ${135^ \circ }$ .
And, we can write as, $\tan {67.5^ \circ } = \tan \left( {\dfrac{{{{135}^ \circ }}}{2}} \right)$
Now, we also know that, $\sin \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}}$ and also $\cos \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 + \cos \theta }}{2}}$
So, dividing these both, we get,
$\dfrac{{\sin \left( {\dfrac{\theta }{2}} \right)}}{{\cos \left( {\dfrac{\theta }{2}} \right)}} = \tan \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}}$
To rationalize the denominator, we multiply both numerator and denominator by its conjugate
(For $\sqrt {a + b}$ the conjugate is $\sqrt {a - b}$ )
On rationalizing, we get, $\tan \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}} \dfrac{{\sqrt {1 - \cos \theta } }}{{\sqrt {1 - \cos \theta } }}$
$\Rightarrow \tan \left( {\dfrac{\theta }{2}} \right) = \dfrac{{1 - \cos \theta }}{{\sin \theta }}$
Here, we get both positive and positive values for this. But for our convenience, we take only positive values.
(Because, $1 - {\cos ^2}\theta = {\sin ^2}\theta$ )
So, substituting $\theta = {135^ \circ }$
$\Rightarrow \tan \left( {\dfrac{{{{135}^ \circ }}}{2}} \right) = \dfrac{{1 - \cos {{135}^ \circ }}}{{\sin {{135}^ \circ }}}$
And, ${135^ \circ } = {90^ \circ } + {45^ \circ }$
So, $\cos {135^ \circ } = \cos ({90^ \circ } + {45^ \circ })$
$\Rightarrow \cos {135^ \circ } = - \sin {45^ \circ }$
(Because $\cos ({90^ \circ } + \theta ) = - \sin \theta$ ; as cosine is negative in second quadrant and $\left( {{{90}^ \circ } + \theta } \right)$ belongs to second quadrant)
So, $\cos {135^ \circ } = - \dfrac{1}{{\sqrt 2 }}$ (as $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ )
And \sin {135^ \circ } = \sin ({90^ \circ } + {45^ \circ }) = \cos {45^ \circ }$$$$ = \dfrac{1}{{\sqrt 2 }} (as $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ )
(Because $\sin ({90^ \circ } + \theta ) = \cos \theta$ ; as sine is positive in second quadrant and $\left( {{{90}^ \circ } + \theta } \right)$ belongs to second quadrant)
So,
$\Rightarrow \tan \left( {\dfrac{{{{135}^ \circ }}}{2}} \right) = \dfrac{{1 - \left( { - \dfrac{1}{{\sqrt 2 }}} \right)}}{{\dfrac{1}{{\sqrt 2 }}}}$
$\Rightarrow \tan \left( {\dfrac{{{{135}^ \circ }}}{2}} \right) = \dfrac{{1 + \dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}}$
So finally, we get, $\tan \left( {\dfrac{{{{135}^ \circ }}}{2}} \right) = \sqrt 2 + 1$
$\Rightarrow \tan \left( {{{67.5}^ \circ }} \right) = \sqrt 2 + 1$
We know that the value of $\sqrt 2$ is equal to the $1.414$ .
So, $\tan ({67.5^ \circ }) = 1.414 + 1 = 2.414$
And this is the required value.

Note: To rationalize the denominator, we need to multiply both numerator and denominator by its conjugate and here the conjugate is $\sqrt {1 - \cos \theta }$ . But instead, we can also multiply both numerator and denominator by $\sqrt {1 + \cos \theta }$ and we can get another value which is also equal to the first.
So, $\tan \left( {\dfrac{\theta }{2}} \right) = \sqrt {\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}} \dfrac{{\sqrt {1 + \cos \theta } }}{{\sqrt {1 + \cos \theta } }}$
So, that implies as $\tan \left( {\dfrac{\theta }{2}} \right) = \dfrac{{\sin \theta }}{{1 + \cos \theta }}$ .