Plane Trigonometry

Additional Formula

823 cot𝐴+tan𝐴=2cosec2𝐴=sec𝐴cosec𝐴 824 cosec2𝐴+cot2𝐴=cot𝐴⋅sec𝐴=1+tan𝐴tan𝐴2 826 cos𝐴=cos⁴𝐴2−sin⁴𝐴2 827 tan𝐴+sec𝐴=tan45°+𝐴2 828 tan𝐴+tan𝐵cot𝐴+cot𝐵=tanAtan𝐵 829 sec²𝐴cosec²𝐴=sec²𝐴+cosec²𝐴 830 If 𝐴+𝐵+𝐶=𝜋2 tan𝐵tan𝐶+tan𝐶tan𝐴+tan𝐴tan𝐵=1 831 If 𝐴+𝐵+𝐶=𝜋, cot𝐵cot𝐶+cot𝐶cot𝐴+cot𝐴cot𝐵=1 832 sin⁻¹35+sin⁻¹45=𝜋2, tan⁻¹12+tan⁻¹13=𝜋4 833 In a right-angled triangle 𝐴𝐵𝐶, 𝐶 being the right angle, cos2𝐵=𝑎²−𝑏²𝑎²+𝑏², tan2𝐵=2𝑏𝑏𝑎²−𝑏² 834 tan12𝐴=𝑐−𝑏𝑐+𝑏. 𝑅+𝑟=12(𝑎+𝑏). 835 In any triangle, sin12(𝐴−𝐵)=𝑎−𝑏𝑐cos12𝐶 cos12(𝐴−𝐵)=𝑎+𝑏𝑐sin12𝐶 836 sin𝐴−𝐵sin𝐴+𝐵=𝑎²−𝑏²𝑐² tan12𝐴+tan12𝐵tan12𝐴−tan12𝐵=𝑐𝑎−𝑏 837 12(𝑎²+𝑏²+𝑐²)=𝑏𝑐cos𝐴+𝑐𝑎cos𝐵+𝑎𝑏cos𝐶 838 Area of triangle 𝐴𝐵𝐶=12𝑏𝑐sin𝐴=12𝑎²sin𝐵sin𝐶sin𝐴=12(𝑎²−𝑏²)sin𝐴sin𝐵sin(𝐴−𝐵) 839 Area of triangle 𝐴𝐵𝐶=2𝑎𝑏𝑐𝑎+𝑏+𝑐cos12𝐴cos12𝐵cos12𝐶 840 Area of triangle 𝐴𝐵𝐶=14(𝑎+𝑏+𝑐)²tan12𝐴tan12𝐵tan12𝐶 841 𝑟=12(𝑎+𝑏+𝑐)tan12𝐴tan12𝐵tan𝐶 842 2𝑅𝑟=𝑎𝑏𝑐𝑎+𝑏+𝑐, △=𝑟𝑟_𝑎𝑟_𝑏𝑟_𝑐 843 𝑎cos𝐴+𝑏cos𝐵+𝑐cos𝐶=4𝑅sin𝐴sin𝐵sin𝐶 844 𝑅+𝑟=12(𝑎cot𝐴+𝑏cot𝐵+𝑐cot𝐶) =sum of perpendiculars on the sides from centre of circumscribing circle.
This may also be shown by applying Enc. VI. D. to the circle described on 𝑅 as diameter and the quadrilateral so formed. 845 𝑟_𝑎𝑟_𝑏𝑟_𝑐=𝑎𝑏𝑐cos12𝐴cos12𝐵cos12𝐶 846 𝑟=(𝑟_𝑏𝑟_𝑐)+(𝑟_𝑐𝑟_𝑎)+(𝑟_𝑎𝑟_𝑏) 847 1𝑟=1𝑟_𝑎+1𝑟_𝑏+1𝑟_𝑐. tan12𝐴=𝑟𝑟_𝑎𝑟_𝑏𝑟_𝑐 849 If 𝑂 be the centre of inscribed circle, 𝑂𝐴=2𝑏𝑐𝑎+𝑏+𝑐cos12𝐴 850 𝑎(𝑏cos𝐶−𝑐cos𝐵)=𝑏²−𝑐² 851 𝑏cos𝐵+𝑐cos𝐶=𝑐cos(𝐵−𝐶) 852 𝑎cos𝐴+𝑏cos𝐵+𝑐cos𝐶=2𝑎sin𝐵sin𝐶 853 cos𝐴+cos𝐵+cos𝐶=1+2𝑎sin𝐵sin𝐶𝑎+𝑏+𝑐 854 If 𝑠=12(𝑎+𝑏+𝑐), 1−cos²𝑎−cos²𝑏−cos²𝑐+2cos𝑎cos𝑏cos𝑐=4sin𝑠sin(𝑠−𝑎)sin(𝑠−𝑏)sin(𝑠−𝑐) 855 −1+cos²𝑎+cos²𝑏+cos²𝑐+2cos𝑎cos𝑏cos𝑐=4cos𝑠cos(𝑠−𝑎)cos(𝑠−𝑏)cos(𝑠−𝑐) 856 4cos𝑎2cos𝑏2cos𝑐2=cos𝑠+cos(𝑠−𝑎)+cos(𝑠−𝑏)+cos(𝑠−𝑐) 857 4sin𝑎2sin𝑏2sin𝑐2=−sin𝑠+sin(𝑠−𝑎)+sin(𝑠−𝑏)+sin(𝑠−𝑐) 858 𝜋²=61+12²+13²+⋯=81+12³+15²+⋯ Proof: Equate coefficients of 𝜃² in the expansion of sin𝜃𝜃 by (764) and (815) or of cos𝜃 by (765) and (816).

Examples of the solutions of Triangles

Example 1

Case II. (724): Two sides of a triangle 𝑏, 𝑐, being 900 and 700 feet, and the included angle 47°55′, to find the remaining angles. tan𝐵−𝐶2=𝑏−𝑐𝑏+𝑐cot𝐴2=18cot23°42ʹ30ʺ; therefore logtan12(𝐵−𝐶)=logcot𝐴2−log8 therefore 𝐿tan12(𝐵−𝐶)=𝐿cot23°42ʹ30ʺ−3log2 10 being added to each side of the equation. { ∴ 12(𝐵−𝐶)=15°53ʹ19.55ʺ* and 12(𝐵+𝐶)=66°17ʹ30.00ʺ∴ 𝐵=82°10ʹ49.55ʺ And, by subtraction, 𝐶=50°24ʹ10.45ʺ

Example 2

Case III. (732). Given the sides 𝑎, 𝑏, 𝑐= 7, 8, 9 respectively, to find the angles. tan𝐴2=(𝑠−𝑏)(𝑠−𝑐)𝑠(𝑠−𝑎)=4⋅312⋅5=210 therefore 𝐿tan𝐴2=10+12(log2−1)=9.650515 therefore 12𝐴=24°5ʹ41.43ʺ* 12𝐵 is found in a similar manner, and 𝐶=180°−𝐴−𝐵.

Example 3

In a right-angled triangle, given the hypotenuse 𝑐=6953 and a side 𝑏=3, to find the remaoning angles.
Here cos𝐴=36953. But, since 𝐴 is nearly a right angle, it cannot be determined accurately from logcos𝐴. Therefore take sin𝐴2=1−cos𝐴2=34756953 therefore 𝐿sin𝐴2=10+12(log3475−log6953)=9.8493913 therefore 𝐴2=44°59ʹ15.52ʺ* therefore 𝐴=89°58ʹ31.04ʺ and 𝐵=0°1ʹ28.96ʺ

Remark

* See Chambers's Mathematieal Tables for a concise explanation of the method of obtaining thes figures.

Sources and References

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