Euler s Formula and Trig Identities
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1 Euler s Formula and Trig Identities Steven W. Nydick May 5, 0 Introduction Many of the identities from trigonometry can be demonstrated relatively easily using Euler s formula, rules of exponents, basic complex analysis, multiplication rules, and one elementary rule from trigonometry. There are two things to consider during the remaining derivation. First, Euler s formula says exp[ix] (x (x Equation ( is due to the consequences of rotating a line in the complex plane. See Wikipedia for a more thorough explanation. Second, the real and imaginary part of a complex number remain the same before and after factoring. No manipulation can turn the imaginary part of a complex number into a real part. Pythagorean Identities Before using Euler s formula to demonstrate complex trigonometric identities, one might not remember the most basic trigonometric identity of them all: (x + sin (x. Due to the Pythagorian theorem h o + a. ( And from the definitions of sine (sin(x o, where o is the length of the horizontal or vertical h line opposite of the angle, x, and h is the hypotenuse formed from a right triangle and ine ((x a, where a is the length of the horizontal or vertical line adjacent to the angle, x, we h have that ( o ( a sin (x + (x + h h o + a h o + a o + a (3
2 Even/Odd Identities Using Equation (3, we can figure out some relatively basic identities. For instance, ( x (x and sin( x sin(x are both solved in one step. ( x ( x exp[ ix] (exp[ix] (4 exp(ix (by (x (x ( (x i sin(x (5 (x (x (x i sin(x (x i sin(x (x + sin (using i (x (x i sin(x (by 3 (x + i[ sin(x And because the real/imaginary part of ( x ( x is equal to the real/imaginary part of (x + i[ sin(x, we end up with ( x (x sin( x sin(x6 Many trig identities use Equation (6 in their derivation. Sum/Difference Formulas A straightforward derivation involves sums or differences inside sines and ines. The deriviation proceeds by noting the relationship among angles and Euler s formula Using properties of exponents (x ± y (x ± y exp[i(x ± y. exp[i(x ± y exp[ix] exp[i(±y [(x (x[(±y (±y [(x (x[(y ± i sin(y (by 6 (x (y sin(x sin(y + i[sin(x (y ± (x sin(y And because the real/imaginary part of (x ± y (x ± y is equal to the real/imaginary part of (x (y sin(x sin(y + i[sin(x (y ± (x sin(y, we end up with (x ± y (x (y sin(x sin(y sin(x ± y sin(x (y ± (x sin(y7
3 Double Angle Formulas (x and sin(x Another useful identity entails doubling an angle inside of a sine or consine operator. Even though this identity might not seem important, a consequence of the double angle formula is very useful. This derivation is similar to the previous one and starts by considering that (x (x exp[i(x. And the derivation follows a similar path to the previous one. exp[i(x exp[i(x + x exp[ix] exp[ix] (exp[ix] [(x (x (x + i (x sin(x sin (x (x sin (x + i[ (x sin(x And because the real/imaginary part of (x (x is equal to the real/imaginary part of (x sin (x + i[ (x sin(x, we end up with (x (x sin (x sin(x (x sin(x8 (x and sin (x Immediate consequences of Equations (3 and (8 include indentites for squares of sines and ines. For instance, we know that (x + (x sin (x + (by 8 (x sin (x + [sin (x + (x (by 3 (x and that (x (x sin (x (by 8 (x sin (x [sin (x + (x (by 3 sin (x9 so that we immediately end up with (x + (x sin (x (x (0 3
4 Sum to Product Formulas (x (y and sin(x sin(y We can also examine identities involving the differences outside of sines and ines. instance, we can put (x (y and sin(x sin(y into one step by noting that For (x (y + i[sin(x sin(y (x (x [(y (y exp(ix exp(iy. And therefore exp(ix exp(iy exp ( ix + i y i ( y exp iy + i x i x exp ( i + i exp i + i exp ( i exp i exp i exp i exp ( [ ( i exp i exp i [ ( [ ( Using Equation (6, we have [ ( [ ( [ ( sin ( ( [ ( i sin [ ( [ ( ( i sin sin [ ( + i sin i sin And because the real/imaginary part of (x (y+i[sin(x sin(y is equal to the real/imaginary part of sin ( sin [ ( + i sin, we end up with (x (y sin ( sin (x + (y and sin(x + sin(y sin(x sin(y sin ( ( A similar derivation indicates involving the sums of sines and ines. As before, note that (x + (y + i[sin(x + sin(y (x (x + [(y (y exp(ix + exp(iy And because the negative versus positive does not affect any of the steps inside the exponent, we end up with exp(ix exp(iy exp ( [ ( i exp i + exp i [ ( [ ( + [ ( [ ( + i sin [ ( [ ( ( [ ( + i sin 4
5 And because the real/imaginary part of (x+(y+i[sin(x+sin(y is equal to the real/imaginary part of ( [ ( + i sin, we end up with (x + (y ( sin(x + sin(y sin ( ( Extensions of the sum-to-product formulas result in product-to-sum formulas. However, because those formulas are not easily derived using Euler s formula, I leave them as exercises. 5
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