We can apply to get two different but equal expressions. The first thing we need to consider is what property of the exponential function Use Euler’s formula to derive a formula for Įxample 3: Double Angle Formulas from Euler’s Formula įinally, we can equate the real and imaginary parts to get theĬ o s c o s c o s s i n s i n s i n c o s s i n s i n c o s ( □ + □ ) = □ □ − □ □, ( □ + □ ) = □ □ + □ □. Gathering together the like terms on the right-hand side, we find thatĬ o s s i n c o s c o s s i n s i n c o s s i n s i n c o s ( □ + □ ) + □ ( □ + □ ) = ( □ □ − □ □ ) + □ ( □ □ + □ □ ). īy expanding the parentheses on the right-hand side, we haveĬ o s s i n c o s c o s c o s s i n s i n c o s s i n s i n ( □ + □ ) + □ ( □ + □ ) = □ □ + □ □ □ + □ □ □ + □ □ □. We can apply Euler’s formula to each one to getĬ o s s i n c o s s i n c o s s i n ( □ + □ ) + □ ( □ + □ ) = ( □ + □ □ ) ( □ + □ □ ). Now that we have two different but equal expressions, The obvious choice here is the multiplicative property: The first thing we need to do is apply one of the properties of the exponentialįunction. What trigonometric identities can be derived by applying Euler’s The next example willĭemonstrate the versatility of this method.Įxample 2: Trigonometric Identities from Euler’s Formula We have proven the trigonometric identityĪs we saw in the last example, by using the properties of the exponential c o s c o s s i n c o s s i n s i n Using Euler’s identity and our answer from part 1, we can rewrite Using the odd/even identities for sine and cosine, Identity can be derived by expanding the exponentials in terms
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