Which of the following is equal to 𝑒 to the power of 𝜃𝑖 plus 𝑒 to the power of negative 𝜃𝑖? Is it option a) 𝑒 to the power of two 𝜃𝑖, option b) two cos 𝜃, option c) two sin 𝜃, or option d) 𝑒 to the power of negative two 𝜃𝑖.
Now we have 𝑒 to the power of some things. So perhaps, index or exponent laws will help us simplify this expression. But we noticed that the index or exponent in each case is imaginary. We have 𝜃 times 𝑖, 𝑖 being the square root of negative one and negative 𝜃 times 𝑖. And whenever we see a number raised to the power of an imaginary number or a complex number, we should think of Euler’s formula: 𝑒 to the power of 𝑖𝑥 equals cos 𝑥 plus 𝑖 sin 𝑥.
We can apply this to 𝑒 to the 𝜃𝑖. 𝜃𝑖 is just 𝑖 times 𝜃. And so, we substitute 𝜃 for 𝑥 in Euler’s formula to find that 𝑒 to the 𝜃𝑖 is cos 𝜃 plus 𝑖 sin 𝜃. Now, for 𝑒 to the power of negative 𝜃𝑖, we can see that 𝑥 is negative 𝜃. And so, we substitute negative 𝜃 for 𝑥 in Euler’s formula. Making this substitution then, we get cos of negative 𝜃 plus 𝑖 times sin of negative 𝜃. Now, we can rewrite cos of negative 𝜃 and sin of negative 𝜃 using the odd even identities for trigonometric functions.
Sin is an odd function. And so, sin of negative 𝑥 is negative sin of 𝑥, whereas cos is an even function. And so, cos of negative 𝑥 is just cos of 𝑥. cos of negative 𝜃 is therefore just cos 𝜃. And similarly, sin of negative 𝜃 is negative sin 𝜃. Making the substitution then, we get cos 𝜃 plus 𝑖 sin 𝜃 plus cos 𝜃 minus 𝑖 sin 𝜃. You might in fact really have known that 𝑒 to the power of negative 𝑖𝑥 is cos 𝑥 minus 𝑖 sin 𝑥. Now, we see that there is some cancelation: a plus 𝑖 sin 𝜃 cancels with a minus 𝑖 sin 𝜃. And we’re left with cos 𝜃 plus cos 𝜃 which is of course two cos 𝜃. So our answer is option b. 𝑒 to the power of 𝜃𝑖 plus 𝑒 to the power of negative 𝜃𝑖 is two cos 𝜃.
Just as a remark, you might notice that if you divide both sides by two, you get the very useful identity cos 𝜃 equals 𝑒 to the 𝜃𝑖 plus 𝑒 to the negative 𝜃𝑖 over two. The equivalent results for sin is that sin 𝜃 equals 𝑒 to the 𝜃𝑖 minus 𝑒 to the negative 𝜃𝑖 all over two 𝑖. This can be proved in a similar way.