Plane Trigonometry

De Moivre's Theorem

756 (cos𝛼+𝑖sin𝛼)(cos𝛽+𝑖cos𝛽)⋯=cos(𝛼+𝛽+𝛾+⋯)+𝑖sin(𝛼+𝛽+𝛾+⋯), where 𝑖=−1 Proved by Induction 757 (cos𝜃+𝑖sin𝜃)^𝑛=cos𝑛𝜃+𝑖sin𝑛𝜃 Proof: By Induction, or by putting 𝛼, 𝛽, ⋯ each = 𝜃 in (756).

Expansion of cos𝑛𝜃, ⋯ in powers sin𝜃 and cos𝜃

758 cos𝑛𝜃=cos^𝑛𝜃−𝐶(𝑛,2)cos^𝑛−2𝜃sin²𝜃+𝐶(𝑛,4)cos^𝑛−4𝜃sin⁴𝜃−⋯ 759 sin𝑛𝜃=𝑛cos^𝑛−1𝜃sin𝜃−𝐶(𝑛,3)cos^𝑛−3𝜃sin³𝜃+⋯ Proof: Expand (757) by Bin. Th., and equate real and imaginary parts. 760 tan𝑛𝜃=𝑛tan𝜃−𝐶(𝑛,3)tan³𝜃+⋯1−𝐶(𝑛,2)tan²𝜃+𝐶(𝑛,4)tan⁴𝜃−⋯ In series (758, 759), stop at, and exclude, all terms with indices greater than 𝑛. Note, 𝑛 is here an integer. 761 Let 𝑠_𝑟=sum of the 𝐶(𝑛,𝑟) products of tan𝛼, tan𝛽, tan𝛾, ⋯ to 𝑛 terms. sin(𝛼+𝛽+𝛾+⋯)=cos𝛼cos𝛽⋯(𝑠₁−𝑠₃+𝑠₅−⋯) 762 cos(𝛼+𝛽+𝛾+⋯)=cos𝛼cos𝛽⋯(1−𝑠₂+𝑠₄−⋯) Proof: By equating real and imaginary parts in (756). 763 tan(𝛼+𝛽+𝛾+⋯)=(𝑠₁−𝑠₃+𝑠₅−𝑠₇+⋯)1−𝑠₂+𝑠₄−𝑠₆+⋯

Expansion of sine and cosine in powers the angle

764 sin𝜃=𝜃−𝜃³3!+𝜃⁵5!−⋯, cos𝜃=1−𝜃²2!+𝜃⁴4!−⋯ Proof: Put 𝜃𝑛 for 𝜃 in (757) and 𝑛=∞, employing (754) and (755). 766 𝑒^𝑖𝜃=cos𝜃+𝑖sin𝜃, 𝑒^−𝑖𝜃=cos𝜃−𝑖sin𝜃By 150 768 𝑒^𝑖𝜃+𝑒^−𝑖𝜃=2cos𝜃, 𝑒^𝑖𝜃−𝑒^−𝑖𝜃=2𝑖sin𝜃 770 𝑖tan𝜃=𝑒^𝑖𝜃−𝑒^−𝑖𝜃𝑒^𝑖𝜃+𝑒^−𝑖𝜃, 1+𝑖tan𝜃1−𝑖tan𝜃=𝑒^2𝑖𝜃

Expansion of cos^𝑛𝜃 and sin^𝑛𝜃 in cosines or sines of multiples of 𝜃

772 2^𝑛−1cos^𝑛𝜃=cos𝑛𝜃+𝑛cos(𝑛−2)𝜃+𝐶(𝑛,2)cos(𝑛−1)𝜃+𝐶(𝑛,3)cos(𝑛−6)𝜃+⋯ 773 When 𝑛 is even, 2^𝑛−1(−1)^12𝑛sin^𝑛𝜃=cos𝑛𝜃−𝑛cos(𝑛−2)𝜃+𝐶(𝑛,2)cos(𝑛−4)𝜃−𝐶(𝑛,3)cos(𝑛−6)𝜃+⋯ 774 And when 𝑛 is odd, 2^𝑛−1(−1)^𝑛−12sin^𝑛𝜃=sin𝑛𝜃−𝑛sin(𝑛−2)𝜃+𝐶(𝑛,2)sin(𝑛−4)𝜃−𝐶(𝑛,3)sin(𝑛−6)𝜃+⋯ Observe that in these series the coefficients are those of the Binomial Theorem, with this exception: If 𝑛 be even, the last term must be divided by 2.
The series are obtained by expanding (𝑒^𝑖𝜃±𝑒^−𝑖𝜃)^𝑛 by the Binomial Theorem, collecting the equidistant terms in pairs, and employing (768) and (769).

Expansion of cos𝑛𝜃 and sin𝑛𝜃 in powers of sin𝜃

775 When 𝑛 is even, cos𝑛𝜃=1−𝑛²2!sin²𝜃+𝑛²(𝑛²−2²)4!sin⁴𝜃−𝑛²(𝑛²−2²)(𝑛²−4²)6!sin⁶𝜃+⋯ 776 When 𝑛 is odd, cos𝑛𝜃=cos𝜃1−𝑛²−12!sin²𝜃+(𝑛²−1)(𝑛²−3²)4!sin⁴𝜃−(𝑛²−1)(𝑛²−3²)(𝑛²−5²)6!sin⁶𝜃+⋯ 777 When 𝑛 is even, sin𝑛𝜃=cos𝜃sin𝜃−𝑛²−2²3!sin³𝜃+(𝑛²−2²)(𝑛²−4²)5!sin⁵𝜃−(𝑛²−2²)(𝑛²−4²)(𝑛²−6²)7!sin⁷𝜃+⋯ 778 When 𝑛 is odd, sin𝑛𝜃=𝑛sin𝜃−𝑛(𝑛²−1)3!sin³𝜃+𝑛(𝑛²−1)(𝑛²−3²)5!sin⁵𝜃−𝑛(𝑛²−1)(𝑛²−3²)(𝑛²−5²)7!sin⁷𝜃+⋯ Proof: By (758), we may assume, when 𝑛 is an even integer cos𝑛𝜃=1+𝐴₂sin²𝜃+𝐴₄sin⁴𝜃+⋯+𝐴_2𝑟sin^2𝑟𝜃+⋯ Put 𝜃+𝑥 for 𝜃, and in cos𝑛𝜃cos𝑛𝑥−sin𝑛𝜃sin𝑛𝑥 substitute for cos𝑛𝑥 and sin𝑛𝑥 their values in powers of 𝑛𝑥 from (764). Each term on the right is of the type 𝐴_2𝑟(sin𝜃cos𝑥+cos𝜃sin𝑥)^2𝑟. Make similar substitutions for cos𝑥 and sin𝑥 in powers of 𝑥. Collect the two coefficients of 𝑥² in each term by the multinomial theorem (137) and equate them all to the coefficient of 𝑥² on the left. In this equation write cos²𝜃 for 1−sin²𝜃 everywhere, and then equate the coefficients of sin^2𝑟𝜃 to obtain the relation between the successive equatities 𝐴_2𝑟 and 𝐴_2𝑟+2 for the series (775).
When 𝑛 is an odd integer, begin by assuming, by (759) sin𝑛𝜃=𝐴₁sin𝜃+𝐴₃sin³𝜃+⋯ 779 The expansions of cos𝑛𝜃 and sin𝑛𝜃 in powers of cos𝜃 are obtained by changing 𝜃 into 12𝜋−𝜃 in (775) to (778).

Expansion of cos𝑛𝜃 in descending powers of cos𝜃

780 2cos𝑛𝜃=(2cos𝜃)^𝑛−𝑛(2cos𝜃)^𝑛−2+𝑛(𝑛−3)2!(2cos𝜃)^𝑛−4−⋯+(−1)^𝑟𝑛(𝑛−r−1)(𝑛−r−2)⋯(𝑛−2r+1)r!(2cos𝜃)^𝑛−2r+⋯ up to the last positive power of 2cos𝜃.
Proof: By expanding each term of the identity log(1−𝑥𝑧)+log1−𝑧𝑥=log1−𝑧𝑥+1𝑥−𝑧 By (156), equating coefficients of 𝑧^𝑛, and substituting from (768). 783 sin𝛼+𝑐sin(𝛼+𝛽)+𝑐²sin(𝛼+2𝛽)+⋯ to 𝑛 terms =sin𝛼−𝑐sin(𝛼−𝛽)−𝑐^𝑛sin(𝛼+𝑛𝛽)+𝑐^𝑛+1sin{𝛼+(𝑛−1)𝛽}1−2𝑐cos𝛽+𝑐² 784 If 𝑐 be < 1 and 𝑛 infinite, this becomes =sin𝛼−𝑐sin(𝛼−𝛽)1−2𝑐cos𝛽+𝑐² 785 cos𝛼+𝑐cos(𝛼+𝛽)+𝑐²cos(𝛼+2𝛽)+⋯ to 𝑛 terms = a similar result, changing sin into cos in the numerator. 786 similarly when 𝑐 is < 1 and 𝑛 infinite. 787 Method of summation: Substitute for the sines or cosines their exponential values (768). Sum the two resulting geometrical series, and substitute the sines or cosines again for the exponential values by (766). 788 𝑐sin(𝛼+𝛽)+𝑐²2!sin(𝛼+2𝛽)+𝑐³3!sin(𝛼+3𝛽)+⋯ to infinity =𝑒^𝑐cos𝛽sin(𝛼+𝑐sin𝛽)−sin𝛼 789 𝑐cos(𝛼+𝛽)+𝑐²2!cos(𝛼+2𝛽)+𝑐³3!cos(𝛼+3𝛽)+⋯ to infinity =𝑒^𝑐cos𝛽cos(𝛼+𝑐sin𝛽)−cos𝛼 Obtained by the rule in (787) 790 If, in the series (783) to (789), 𝛽 be changed into 𝛽+𝜋, the signs of the alternate terms will thereby be changed.

Expansion of 𝜃 in powers of tan𝜃 (Gregory's series)

791 𝜃=tan𝜃−tan³𝜃3+tan⁵𝜃5−⋯ The series converges if tan𝜃 be not >1. Proof: By expanding the logarithm of the value of 𝑒^2𝑖𝜃 in (771) by (158).

formula for the calculation of the value of 𝜋 by Gregor's series

792 𝜋4=tan⁻¹12+tan⁻¹13=tan⁻¹15−tan⁻¹1239791 794 𝜋4=4tan⁻¹15−tan⁻¹170+tan⁻¹199 Proof: By employing the formula for tan(𝐴±𝐵), (631)

To Prove that 𝜋 is Incommensurable

795 Convert the value of tan𝜃 in terms of 𝜃 from (764) and (765) into a continued fraction, thus tan𝜃=𝜃1₋𝜃²3₋𝜃²5₋𝜃²7₋⋯; or this result may be obtained by putting 𝑖𝜃 for 𝑦 in (294), and by (770). Hence 1−𝜃tan𝜃=𝜃²3₋𝜃²5₋𝜃²7₋⋯ Put 𝜋2 for , and assume that 𝜋, and therefore 𝜋²4, is commensurable. Let 𝜋²4=𝑚𝑛, 𝑚 and 𝑛 being integers. Then we shall have 1=𝑚2𝑛₋𝑚𝑛5𝑛₋𝑚𝑛7𝑛₋⋯
The continued fraction is incommensurable, by (177). But unity cannot be equal to an incommensurable quantity. Therefore 𝜋 is not commensurable. 796 If sin𝑥=𝑛sin(𝑥+𝛼), 𝑥=𝑛sin𝛼+𝑛²2sin2𝛼+𝑛³3sin3𝛼+⋯ 797 If tan𝑥=𝑛tan𝑦, 𝑥=𝑦−𝑚sin2𝑦+𝑚²2sin4𝑦−𝑚³3sin6𝑦+⋯, where 𝑚=1−𝑛1+𝑛
Proof:By substiuting the exponential values of the sine or tangent (769) and (770), and then eliminating 𝑥. 798 Coefficient of 𝑥^𝑛 in the expansion of 𝑒^𝑎𝑥cos𝑏𝑥=(𝑎²+𝑏²)^𝑛2𝑛!cos𝑛𝜃, where 𝑎=𝑟cos𝜃 and 𝑏=𝑟sin𝜃.
For proof, substitute for cos𝑏𝑥 from (768); expand by (150); put 𝑎=𝑟cos𝜃 and 𝑏=𝑟sin𝜃 in the coefficient of 𝑒^𝑥, employ (757). 799 When 𝑒<1, 1−𝑒²1−𝑒cos𝜃=1+2𝑏cos𝜃+2𝑏²cos2𝜃+2𝑏³cos3𝜃+⋯, where 𝑏=𝑒1+1−𝑒²
For proof, put 𝑒=2𝑏1+𝑏² and 2cos𝜃=𝑥+1𝑥, expand the fraction in two series of powers of 𝑥 by the method of (257), and substitute from (768). 800 sin𝛼+sin(𝛼+𝛽)+sin(𝛼+2𝛽)+⋯+sin{𝛼+(𝑛−1)𝛽}=sin𝛼+𝑛−12𝛽sin𝑛2𝛽sin𝛽2 801 cos𝛼+cos(𝛼+𝛽)+sin(𝛼+2𝛽)+⋯+cos{𝛼+(𝑛−1)𝛽}=cos𝛼+𝑛−12𝛽sin𝑛2𝛽sin𝛽2 802 If the terms in these series have the signs + and − alternately, change 𝛽 into 𝛽+𝜋 in the results.
Proof: Multiply the series by 2sin𝛽2, and apply (669) and (666). 803 If 𝛽=2𝜋𝑛 in (800) and (801), each series vanishes. 804 Generally, if 𝛽=2𝜋𝑛, and if 𝑟 be an integer not a multiple of 𝑛, the sum of the 𝑟^th powers of the sines or cosines in (800) or (801) is zero if 𝑟 be odd; and if 𝑟 be even it is =𝑛2^𝑟; by (772) to (774) 805 General Theorem: Denoting the sum of the series 𝑐+𝑐₁𝑥+𝑐₂𝑥²+⋯+𝑐_𝑛𝑥^𝑛 by 𝐹(𝑥); then 𝑐cos𝛼+𝑐₁cos(𝛼+𝛽)+⋯+𝑐_𝑛cos(𝛼+𝑛𝛽)=12{𝑒^𝑖𝛼𝐹(𝑒^𝑖𝛽)+𝑒^−𝑖𝛼𝐹(𝑒^−𝑖𝛽)} and 806 𝑐sin𝛼+𝑐₁sin(𝛼+𝛽)+⋯+𝑐_𝑛sin(𝛼+𝑛𝛽)=12𝑖{𝑒^𝑖𝛼𝐹(𝑒^𝑖𝛽)−𝑒^−𝑖𝛼𝐹(𝑒^−𝑖𝛽)} Provd by substituting for the sines and cosines their exponential values (766), ⋯.

Expansion of the sine and cosine in factors

807 𝑥^2𝑛−2𝑥^𝑛𝑦^𝑛cos𝑛𝜃+𝑦^2𝑛=𝑥²−2𝑥𝑦cos𝜃+𝑦²𝑥²−2𝑥𝑦cos𝜃+2𝜋𝑛+𝑦²⋯ to 𝑛 factors, adding 2𝜋𝑛 to the angle successively.
Proof: By solving the quadratic on the left, we get 𝑥=𝑦(cos𝑛𝜃+𝑖sin𝑛𝜃)^1𝑛. The 𝑛 values of 𝑥 are found by (757) and (626), and thence tha factors. For the factors 𝑥^𝑛±𝑦^𝑛 see (480). 808 sin𝑛𝜙=2^𝑛−1sin𝜙sin𝜙+𝜋𝑛sin𝜙+2𝜋𝑛⋯ as far as 𝑛 factors of sines.
Proof: By putting 𝑥=𝑦=1 and 𝜃=2𝜙 in the last. 809 If 𝑛 be even, sin𝑛𝜙=2^𝑛−1sin𝜙cos𝜙sin²𝜋𝑛−sin²𝜙sin²2𝜋𝑛−sin²𝜙⋯ 810 If 𝑛 be odd, omit cos𝜙 and make up 𝑛 factors, reckoning two factors for each pair of terms in brackets.
Proof: From (808), by collecting equidistant factors in pairs, and applying (659). 811 cos𝑛𝜙=2^𝑛−1sin𝜙+𝜋2𝑛sin𝜙+3𝜋2𝑛⋯ to 𝑛 factors. Proof: Put 𝜙+𝜋2𝑛 for 𝜙 in (808). 812 Also, if 𝑛 be odd, cos𝑛𝜙=2^𝑛−1cos𝜙sin²𝜋2𝑛−sin²𝜙sin²3𝜋2𝑛−sin²𝜙⋯ 813 If 𝑛 be even, omit cos𝜙,
Proof: as in (809) 814 𝑛=2^𝑛−1sin𝜋𝑛sin2𝜋𝑛sin3𝜋𝑛⋯sin(𝑛−1)𝜋𝑛 Proof: divide (809) by sin𝜙, and make 𝜙 vanish; then apply (754). 815 sin𝜃=𝜃1−𝜃𝜋²1−𝜃2𝜋²1−𝜃3𝜋²⋯ 816 cos𝜃=1−2𝜃𝜋²1−2𝜃3𝜋²1−2𝜃5𝜋²⋯ Proof: Put 𝜙=𝜃𝑛 in (809) and (842); divide by (814) and make 𝑛 infinite. 817 𝑒^𝑥−2cos𝜃+𝑒^−𝑥=4sin²𝜃21+𝑥²𝜃²1+𝑥²(2𝜋±𝜃)²1+𝑥²(4𝜋±𝜃)²⋯ Proved by substituting 𝑥=1+𝑧2𝑛, 𝑦=1−𝑧2𝑛, and 𝜃𝑛 for 𝜃 in (807) Making 𝑛 infinite, and reducing one series of factors to 4sin²𝜃2 by putting 𝑧=0.

De Moivre's Property of the Circle

Circle: Take 𝑃 any point, and 𝑃𝑂𝐵=𝜃 any angle, 𝐵𝑂𝐶=𝐶𝑂𝐷=⋯=2𝜋𝑛; 𝑂𝑃=𝑥; 𝑂𝐵=𝑟 819 𝑥^2𝑛−2𝑥^𝑛𝑟^𝑛cos𝑛𝜃+𝑟^2𝑛=𝑃𝐵²𝑃𝐶²𝑃𝐷²⋯ to 𝑛 factors By (807) and (702), since 𝑃𝐵²=𝑥²−2𝑥𝑟cos𝜃+𝑟², ⋯ 820 If 𝑥=𝑟, 2𝑟^𝑛sin𝑛𝜃2=𝑃𝐵⋅𝑃𝐶⋅𝑃𝐷⋯ 821

Cotes's Properties

If 𝜃=2𝜋2, 𝑥^𝑛∼𝑟^𝑛=𝑃𝐵⋅𝑃𝐶⋅𝑃𝐷⋯ 822 𝑥^𝑛+𝑟^𝑛=𝑃𝑎⋅𝑃𝑏⋅𝑃𝑐⋯

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive