Question

How do you find the exact value of the half angle of $\tan 157.5$?

Answer 1

In this question, we need to find the exact value of the given tangent function. Firstly, we will rewrite the given angle $157.5$ as $\dfrac{{315}}{2}$. Then we will apply the tangent half angle identity given by $\tan \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}}$. Then we find the reference angle and substitute in the formula. We then simplify and solve the problem to obtain the exact value of the given function.

Complete step-by-step answer:
Given the function $\tan 157.5$
We are asked to find the exact value of the half angle of the above function.
Firstly, rewrite $157.5$ as an angle, which we write as $\dfrac{{315}}{2}$.
So we need to find the exact value of $\tan \left( {\dfrac{{315}}{2}} \right)$.
Now we apply the tangent half angle identity which is given by,
$\tan \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}}$
Here $x = 315$ and we change $\pm$ sign to $-$, since tangent is negative in the second quadrant.
Hence we get,
$\tan \left( {\dfrac{{315}}{2}} \right) = - \sqrt {\dfrac{{1 - \cos 315}}{{1 + \cos 315}}}$ …… (1)
Now we simplify the expression inside the square root which is given by $\dfrac{{1 - \cos 315}}{{1 + \cos 315}}$
Now we will find the reference angle for 315 degrees. Since the angle is positive, we subtract it from 360 degrees. Hence we get,
Reference angle $= 360 - 315 = 45$.
So we need to simplify $\dfrac{{1 - \cos 45}}{{1 + \cos 45}}$.
We know the value $\cos 45 = \dfrac{1}{{\sqrt 2 }}$.
Substituting this we get,
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = \dfrac{{1 - \dfrac{1}{{\sqrt 2 }}}}{{1 + \dfrac{1}{{\sqrt 2 }}}}$
Taking LCM in the numerator and denominator on the right hand side, we get,
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = \dfrac{{\dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}}}{{\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 }}}}$
Now multiply the numerator by the reciprocal of the denominator, we get,
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 + 1}}$
Cancelling the common factor $\sqrt 2$, we get,
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}$
Now we simplify by the method of rationalization. So multiply and divide by $\sqrt 2 - 1$ we get,
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \times \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}$
This can be written as,
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = \dfrac{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}}$
In the denominator we have the form $(a + b)(a - b)$.
We have the identity given by $(a + b)(a - b) = {a^2} - {b^2}$
So here we have $a = \sqrt 2$ and $b = 1$.
So we get,
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = \dfrac{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 - 1} \right)}}{{{{\left( {\sqrt 2 } \right)}^2} - {1^2}}}$
Now we simplify the numerator and we square the terms in the denominator, we get,
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = \dfrac{{{{\left( {\sqrt 2 - 1} \right)}^2}}}{{2 - 1}}$
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = \dfrac{{{{\left( {\sqrt 2 - 1} \right)}^2}}}{1}$
This can be written as,
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = {\left( {\sqrt 2 - 1} \right)^2}$
We have the formula ${(a - b)^2} = {a^2} - 2ab + {b^2}$
Here $a = \sqrt 2$ and $b = 1$.
So we have,
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = {\left( {\sqrt 2 } \right)^2} - 2 \times \sqrt 2 \times 1 + {1^2}$
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = 2 - 2\sqrt 2 + 1$
$\Rightarrow \dfrac{{1 - \cos 45}}{{1 + \cos 45}} = 3 - 2\sqrt 2$
Substituting this in the equation (1), we get,
$\tan \left( {\dfrac{{315}}{2}} \right) = - \sqrt {3 - 2\sqrt 2 }$
In decimal form we get the value as $- 0.41421356 \approx - 0.41$
Hence the exact value of the half angle of $\tan 157.5$ is $- 0.41$.

Note:
Reference angles are the measure between a given angle and the x-axis.
To find the reference angle keep subtracting 360 degrees from the given angle until it lies between ${0^ \circ }$ and ${360^ \circ }$. For negative angles add 360 degrees instead. Students must know how to rationalize the given number and simplify such problems. The principle of rationalization is that we are going to multiply and divide the conjugate of the denominator and then simplify.
Also remember the formulas such as,
(1) $(a + b)(a - b) = {a^2} - {b^2}$
(2) ${(a - b)^2} = {a^2} - 2ab + {b^2}$
(3) ${(a + b)^2} = {a^2} + 2ab + {b^2}$