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Trigonometric Identities with Euler’s Formula

April 29th, 2009 |

Author: Merrin

Euler’s formula is a great way to derive many trigonometric identities and relationships. Instead of drawing a reference triangle and surmising different geometric properties from triangles you can solve most everything with algebra. Euler’s formula is the glue between algebra, geometry, trigonometry, and complex numbers. You may recall Euler’s formula from complex variables.

$e^{i\theta } =\cos \theta +i\sin \theta$

We can prove this formula later. But for now we can form a second Euler equation with the negative of the angle.

$e^{-i\theta } =\cos (-\theta )+i\sin (-\theta )=\cos \theta -i\sin \theta$

The cosine is an even function and the sine function is odd. Adding or subtracting these two Euler equations give the algebraic forms of sine and cosine.

$\cos \theta =\cfrac{e^{i\theta } +e^{-i\theta } }{2}$

$\sin \theta =\cfrac{e^{i\theta } -e^{-i\theta } }{2i}$

You should be able to verify that cosine and sine are even and odd from the formulas.

Example 1. Derive the following equation using Euler’s formula.

$\cos ^{2} \theta +\sin ^{2} \theta =1$

Solution 1. The following identity is just the Pythagorean theorem, the base of the unit reference triangle is cosine and the altitude is sine. The proof with Euler’s formula is also straightforward

${e^{i\theta } e^{-i\theta } =1}$

${(\cos \theta +i\sin \theta )(\cos (-\theta )+i\sin (-\theta ))=1}$

$(\cos \theta +i\sin \theta )(\cos \theta -i\sin \theta )=1$

${\cos ^2\theta +\sin^2 \theta =1}$

Example 2. Find simple equations for the sum or difference of two angles as the argument of sine and cosine.

Solution 2. We can use Euler’s formula and separate the real and imaginary parts out.

$\cos (A+B)= \text{Re}[e^{iA} e^{iB} ]=\text{Re}[(\cos A+i\sin A)(\cos B+i\sin B)]$

$\sin (A+B)=\text{Im}[e^{iA} e^{iB}]=\text{Im}[ (\cos A+i\sin A)(\cos B+i\sin B)]$

The sum formulas are then just

${\cos (A+B)} = {\cos A\cos B-\sin A\sin B }$

${\sin (A+B)} = {\cos A\sin B+\sin A\cos B}$

Under the transformation $latex B \to -B&s=1$ the difference equations are

${\cos (A-B)} = {\cos A\cos B+\sin A\sin B }$

${\sin (A-B)} = {\sin A\cos B-\cos A\sin B}$

Example 3. Derive the four product formulas for sine and cosine.

Solution 3. Using Euler’s formula we just calculate the products.

${\sin (A)\sin (B)} = {\cfrac{1}{2i} \cfrac{1}{2i} (e^{iA} -e^{-iB} )(e^{iB} -e^{-iB} )} \\ = {-\cfrac{1}{4} (e^{i(A+B)} +e^{-i(A+B)} -e^{i(B-A)} -e^{i(A-B)} )} \\ = {\cfrac{1}{2}\cos [(A-B)]-\cfrac{1}{2} \cos [(A+B)]}$

All the products can be derived in a similar fashion.

${\sin A\sin B} = {\cfrac{1}{2} [\cos (A-B)-\cos (A+B)] }$

${\cos A\cos B} = {\cfrac{1}{2} [\cos (A-B)+\cos (A+B)]}$

${\sin A\cos B} = {\cfrac{1}{2} [\sin (A+B)+\sin (A-B]}$

${\cos B\sin A} = {\cfrac{1}{2} [\sin (A+B)-\sin (A-B)]}$

Example 4. Find the double angle identities for sine and cosine.

Solution 4. From the previous identities we have the double angle formulas $latex A=B=\theta &s=1 $.

$\cos (2\theta) = \cos ^{2} \theta -\sin ^{2} \theta = \cos ^{2} \theta -(1-\cos ^{2} \theta)=2\cos ^{2} \theta -1$

$\sin (2\theta )=\sin \theta \cos \theta +\cos \theta \sin \theta = 2\sin \theta \cos \theta$

Example 5. Express $latex \sin^2\theta &s=1$ and $latex \cos^2\theta&s=1$ in terms of the double angle of cosine, $latex \cos(2\theta)&s=1$.

Solution 5. We already have one answer from rearranging the previous problem.

$\cos^2 \theta = \cfrac{1+\cos (2\theta )}{2}$

And now since

$\sin^2 (2\theta) = 1-\cos^2 (2\theta)$

We have the result that

$\sin^2 (2\theta) = \cfrac{1-\cos (2\theta )}{2}$

Example 6. Find a formula for tangent in terms of sines and cosines as well as the double angle formula for tangent.

Solution 6. The tangent of a sum can be found from the sine and cosine of the same sum. That result can be used to find the double angle formula for tangent.

$\tan (A+B)=\cfrac{\sin (A+B)}{\cos (A+B)} =\cfrac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}$

Divide by $latex \cos A \cos B&s=1$.

$\tan (A+B)=\cfrac{\tan A+\tan B}{1-\tan A\tan B}$

From these relations you can set A=B and get the double angle formula for tangents.

$\tan (2A)=\cfrac{2\tan A}{1-(\tan A)^{2} }$

Example 7. Calculate an expansion for $latex \sin ^7 x&s=1$ in terms of single powers of sines.

Solution 7. The simplest way to proceed is to write the function in terms of complex variables using Euler’s formula. After this expand the expression using the binomial identity.

$(\sin \theta )^{7} = \left( \cfrac{1}{2i} \right)^{7} (e^{i\theta } -e^{-i\theta } )^{7}$