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, $$\sin \left({\displaystyle \frac{\theta }{2}}\right)=\pm \sqrt{{\displaystyle \frac{1-\cos \theta }{2}}}$$ and $$\cos \left({\displaystyle \frac{\theta }{2}}\right)=\pm \sqrt{{\displaystyle \frac{1+\cos \theta }{2}}}$$ . 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, $$\tan {67.5}^{\circ }=\tan \left({\displaystyle \frac{{135}^{\circ }}{2}}\right)$$
Now, we also know that, $$\sin \left({\displaystyle \frac{\theta }{2}}\right)=\pm \sqrt{{\displaystyle \frac{1-\cos \theta }{2}}}$$ and also $$\cos \left({\displaystyle \frac{\theta }{2}}\right)=\pm \sqrt{{\displaystyle \frac{1+\cos \theta }{2}}}$$
So, dividing these both, we get,
$${\displaystyle \frac{\sin \left({\displaystyle \frac{\theta }{2}}\right)}{\cos \left({\displaystyle \frac{\theta }{2}}\right)}}=\tan \left({\displaystyle \frac{\theta }{2}}\right)=\pm \sqrt{{\displaystyle \frac{1-\cos \theta }{1+\cos \theta }}}$$
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, $$\tan \left({\displaystyle \frac{\theta }{2}}\right)=\pm \sqrt{{\displaystyle \frac{1-\cos \theta }{1+\cos \theta }}}{\displaystyle \frac{\sqrt{1-\cos \theta }}{\sqrt{1-\cos \theta }}}$$
$$\Rightarrow \tan \left({\displaystyle \frac{\theta }{2}}\right)={\displaystyle \frac{1-\cos \theta }{\sin \theta }}$$
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 $$\theta ={135}^{\circ }$$
$$\Rightarrow \tan \left({\displaystyle \frac{{135}^{\circ }}{2}}\right)={\displaystyle \frac{1-\cos {135}^{\circ }}{\sin {135}^{\circ }}}$$
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, $$\cos {135}^{\circ }=-{\displaystyle \frac{1}{\sqrt{2}}}$$ (as $$\sin {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}}$$ )
And $$\sin {135}^{\circ }=\sin ({90}^{\circ }+{45}^{\circ })=\cos {45}^{\circ }$$$$={\displaystyle \frac{1}{\sqrt{2}}}$$ (as $$\cos {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}}$$ )
(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,
$$\Rightarrow \tan \left({\displaystyle \frac{{135}^{\circ }}{2}}\right)={\displaystyle \frac{1-\left(-{\displaystyle \frac{1}{\sqrt{2}}}\right)}{{\displaystyle \frac{1}{\sqrt{2}}}}}$$
$$\Rightarrow \tan \left({\displaystyle \frac{{135}^{\circ }}{2}}\right)={\displaystyle \frac{1+{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}}}$$
So finally, we get, $$\tan \left({\displaystyle \frac{{135}^{\circ }}{2}}\right)=\sqrt{2}+1$$
$$\Rightarrow \tan \left({67.5}^{\circ }\right)=\sqrt{2}+1$$
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, $$\tan \left({\displaystyle \frac{\theta }{2}}\right)=\sqrt{{\displaystyle \frac{1-\cos \theta }{1+\cos \theta }}}{\displaystyle \frac{\sqrt{1+\cos \theta }}{\sqrt{1+\cos \theta }}}$$
So, that implies as $$\tan \left({\displaystyle \frac{\theta }{2}}\right)={\displaystyle \frac{\sin \theta }{1+\cos \theta }}$$ .