Happy Labor Day Weekend! Today, I’ll explain how to add an infinite number of trigonometric functions. In particular, I’ll explain why it is not as intimidating as it seems. This post makes use of geometric series and differences of squares, which are useful in Calculus.
$$ 1 + \sin(\pi/4) + \sin^2(\pi/4) + \sin^3(\pi/4) + \cdots$$ $$ = 2 + \sqrt{2} $$
Breaking the Problem Down
Notice that the left-hand side can be written as
$$ 1 + u + u^2 + u^3 + \cdots $$
where \( u = \sin(\pi/4) \). The above expression is an example of a geometric series. In turns out that
$$ 1 + u + u^2 + u^3 + \cdots = \frac{1}{1 \, – \, u} $$
See my post on geometric series for more on how that works.
Because \( u = \sin(\pi/4) \), we conclude that
$$ 1 + \sin(\pi/4) + \sin^2(\pi/4) + \sin^3(\pi/4) + \cdots $$ $$ = \frac{1}{1 \, – \, \sin(\pi/4)} $$
Using Trigonometry
It turns out that
$$ \sin(\pi/4) = \frac{1}{\sqrt{2}} $$
See this Wikipedia link for more on that. Using the above formula, we conclude that
$$ 1 + \sin(\pi/4) + \sin^2(\pi/4) + \sin^3(\pi/4) + \cdots $$ $$ = \frac{1}{1 \, – \, 1 / \sqrt{2}} \,. $$
Using Difference of Squares
Let’s analyze the number
$$ \frac{1}{1 \, – \, 1 / \sqrt{2}} $$
in more depth. Note that
$$ \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} $$
So, using a technique known as difference of squares, we calculate
$$ \bigg{(}1 \, – \, \frac{1}{\sqrt{2}}\bigg{)}\bigg{(}1 \, + \, \frac{1}{\sqrt{2}}\bigg{)} = 1 \; – \; \frac{1}{2} $$
See some of my older posts, such as this one, for more on how this is used. Now, multiply both the numerator and the denominator by \( 1 + 1/\sqrt{2} \):
$$ \frac{1}{1 \, – \, 1 / \sqrt{2}} = \frac{1 + 1/\sqrt{2}}{(1 – 1/\sqrt{2})(1 + 1/\sqrt{2})} $$
By the above, \( (1 – 1/\sqrt{2})(1 + 1/\sqrt{2}) = 1 – 1/2 \). So,
$$ \frac{1}{1 \, – \, 1/\sqrt{2}} = \frac{1 + 1/\sqrt{2}}{1 \; – \, 1/2} $$
From the above, we conclude that
$$ 1 + \sin(\pi/4) + \sin^2(\pi/4) + \sin^3(\pi/4) + \cdots $$ $$ = \frac{1 + 1/\sqrt{2}}{1 \; – \, 1/2} \,. $$
The Final Touches
The number
$$ \frac{1 + 1/\sqrt{2}}{1 \; – \, 1/2} $$
can be further simplified:
$$ \frac{1 + 1/\sqrt{2}}{1 \; – \, 1/2} = \frac{1 + 1/\sqrt{2}}{1/2} $$
Dividing any number by \( 1/2 \) has the same effect as multiplying it by \( 2 \). So,
$$ \frac{1 + 1/\sqrt{2}}{1/2} = 2 \cdot \bigg{(}1 + \frac{1}{\sqrt{2}} \bigg{)} $$
Additionally,
$$ 2 \cdot \bigg{(}1 + \frac{1}{\sqrt{2}} \bigg{)} = 2 + \frac{2}{\sqrt{2}} $$
Recall that \( \sqrt{2} \cdot \sqrt{2} = 2 \). So, \( 2/\sqrt{2} = \sqrt{2}\) and we have
$$ 2 \cdot \bigg{(}1 + \frac{1}{\sqrt{2}} \bigg{)} = 2 + \sqrt{2} $$
At last, we have:
$$ 1 + \sin(\pi/4) + \sin^2(\pi/4) + \sin^3(\pi/4) + \cdots $$ $$ = 2 + \sqrt{2} $$
