Video: MATH-ALG+GEO-2018-S1-Q02

Which of the following is equal to 𝑒^πœƒπ‘– + 𝑒^βˆ’πœƒπ‘–? [A] 𝑒^2πœƒπ‘– [B] 2cos πœƒ [C] 2sin πœƒ [D] 𝑒^βˆ’2πœƒπ‘–

02:53

Video Transcript

Which of the following is equal to 𝑒 to the power of πœƒπ‘– plus 𝑒 to the power of negative πœƒπ‘–? Is it option a) 𝑒 to the power of two πœƒπ‘–, option b) two cos πœƒ, option c) two sin πœƒ, or option d) 𝑒 to the power of negative two πœƒπ‘–.

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 πœƒ times 𝑖, 𝑖 being the square root of negative one and negative πœƒ 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 𝑖π‘₯ equals cos π‘₯ plus 𝑖 sin π‘₯.

We can apply this to 𝑒 to the πœƒπ‘–. πœƒπ‘– is just 𝑖 times πœƒ. And so, we substitute πœƒ for π‘₯ in Euler’s formula to find that 𝑒 to the πœƒπ‘– is cos πœƒ plus 𝑖 sin πœƒ. Now, for 𝑒 to the power of negative πœƒπ‘–, we can see that π‘₯ is negative πœƒ. And so, we substitute negative πœƒ for π‘₯ in Euler’s formula. Making this substitution then, we get cos of negative πœƒ plus 𝑖 times sin of negative πœƒ. Now, we can rewrite cos of negative πœƒ and sin of negative πœƒ using the odd even identities for trigonometric functions.

Sin is an odd function. And so, sin of negative π‘₯ is negative sin of π‘₯, whereas cos is an even function. And so, cos of negative π‘₯ is just cos of π‘₯. cos of negative πœƒ is therefore just cos πœƒ. And similarly, sin of negative πœƒ is negative sin πœƒ. Making the substitution then, we get cos πœƒ plus 𝑖 sin πœƒ plus cos πœƒ minus 𝑖 sin πœƒ. You might in fact really have known that 𝑒 to the power of negative 𝑖π‘₯ is cos π‘₯ minus 𝑖 sin π‘₯. Now, we see that there is some cancelation: a plus 𝑖 sin πœƒ cancels with a minus 𝑖 sin πœƒ. And we’re left with cos πœƒ plus cos πœƒ which is of course two cos πœƒ. So our answer is option b. 𝑒 to the power of πœƒπ‘– plus 𝑒 to the power of negative πœƒπ‘– is two cos πœƒ.

Just as a remark, you might notice that if you divide both sides by two, you get the very useful identity cos πœƒ equals 𝑒 to the πœƒπ‘– plus 𝑒 to the negative πœƒπ‘– over two. The equivalent results for sin is that sin πœƒ equals 𝑒 to the πœƒπ‘– minus 𝑒 to the negative πœƒπ‘– all over two 𝑖. This can be proved in a similar way.

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