trigonometric identities formulas double angle

T a n s i n c o s c o s s i n c o s s i n c o s s i n c o s c o s s i n c o s s i n s i n 6 𝜃 = 6 𝜃 6 𝜃 = 6 𝜃 𝜃 − 2 0 𝜃 𝜃 + 6 𝜃 𝜃 𝜃 − 1 5 𝜃 𝜃 + 1 5 𝜃 𝜃 − 𝜃. Using the identities from part 1, we can write          īy equating the real and imaginary parts, we arrive at the followingĬ o s c o s c o s s i n c o s s i n s i n s i n c o s s i n c o s s i n c o s s i n 6 𝜃 = 𝜃 − 1 5 𝜃 𝜃 + 1 5 𝜃 𝜃 − 𝜃, 6 𝜃 = 6 𝜃 𝜃 − 2 0 𝜃 𝜃 + 6 𝜃 𝜃. Using the binomial theorem, we can expand the parentheses as follows:Ĭ o s s i n c o s c o s s i n c o s s i n c o s s i n c o s s i n c o s s i n s i n 6 𝜃 + 𝑖 6 𝜃 = 𝜃 + 𝐶 𝜃 ( 𝑖 𝜃 ) + 𝐶 𝜃  𝑖 𝜃  + 𝐶 𝜃  𝑖 𝜃  + 𝐶 𝜃  𝑖 𝜃  + 𝐶 𝜃  𝑖 𝜃  + 𝑖 𝜃. Using Euler’s formula, we can rewrite this asĬ o s s i n c o s s i n 6 𝜃 + 𝑖 6 𝜃 = ( 𝜃 + 𝑖 𝜃 ).

Using the properties of the exponential function, we can write Use Euler’s formula to derive a formula for.To simplify the derivation, we can use the binomial theorem.Įxample 4: Multiple Angle Formulas from Euler’s Formula Using a similar method, we can derive multiple angle formulas forĪnd c o s 𝑛 𝜃 for larger values of 𝑛. Get the following two trigonometric identities:Ĭ o s c o s s i n s i n c o s s i n 2 𝜃 = 𝜃 − 𝜃, 2 𝜃 = 2 𝜃 𝜃.     įinally, we can equate the real and imaginary parts to In both of these cases, we will get the same result:Īpplying Euler’s formula to both of these expression, we haveĬ o s s i n c o s s i n 2 𝜃 + 𝑖 2 𝜃 = ( 𝜃 + 𝑖 𝜃 ). Obvious choices for properties of the exponential function to use, either the power Since we are lookingįor identities in 2 𝜃, it makes sense that we would consider 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 ( 𝜃 + 𝜑 ) = 𝜃 𝜑 − 𝜃 𝜑, ( 𝜃 + 𝜑 ) = 𝜃 𝜑 + 𝜃 𝜑. 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 ( 𝜃 + 𝜑 ) + 𝑖 ( 𝜃 + 𝜑 ) = ( 𝜃 𝜑 − 𝜃 𝜑 ) + 𝑖 ( 𝜃 𝜑 + 𝜃 𝜑 ). 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 ( 𝜃 + 𝜑 ) + 𝑖 ( 𝜃 + 𝜑 ) = 𝜃 𝜑 + 𝑖 𝜃 𝜑 + 𝑖 𝜃 𝜑 + 𝑖 𝜃 𝜑. 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 ( 𝜃 + 𝜑 ) + 𝑖 ( 𝜃 + 𝜑 ) = ( 𝜃 + 𝑖 𝜃 ) ( 𝜑 + 𝑖 𝜑 ). 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 Use Euler’s formula to express 𝑒    in.Example 1: Using Euler’s Formula to Derive Trigonometric Identities

Trigonometric identities formulas double angle