Question

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

Answer 1

Hint: 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 \[\tan \left( {\dfrac{\theta }{2}} \right) = \dfrac{{\sin \theta }}{{1 + \cos \theta }}\] .