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# Content

`Plane Trigonometry Additional Formula Examples of the solutions of Triangles  Example 1  Example 2 Example 3    Remark Sources and References`

# Plane Trigonometry

823 cot𝐴+tan𝐴=2cosec2𝐴=sec𝐴cosec𝐴 824 cosec2𝐴+cot2𝐴=cot𝐴⋅sec𝐴=1+tan𝐴tan𝐴2 826 cos𝐴=cos4𝐴2sin4𝐴2 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, △=𝑟𝑟𝑎𝑟𝑏𝑟𝑐 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 𝑟𝑎𝑟𝑏𝑟𝑐=𝑎𝑏𝑐cos12𝐴cos12𝐵cos12𝐶 846 𝑟=(𝑟𝑏𝑟𝑐)+(𝑟𝑐𝑟𝑎)+(𝑟𝑎𝑟𝑏) 847 1𝑟=1𝑟𝑎+1𝑟𝑏+1𝑟𝑐. tan12𝐴=𝑟𝑟𝑎𝑟𝑏𝑟𝑐 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)=logcot𝐴2log8 therefore 𝐿tan12(𝐵−𝐶)=𝐿cot23°42ʹ30ʺ−3log2 10 being added to each side of the equation.
∴ 𝐿cot23°42ʹ30ʺ=10.3573942* 3log2=0.9030900 ∴ 𝐿tan12(𝐵−𝐶)=9.4543042
{ 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

ID: 210900013 Last Updated: 9/13/2021 Revision: 0 Ref: References

1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science &amp; Engineering
2. Ayres, F. JR, Moyer, R.E., 1999, Schaum's Outlines: Trigonometry
3. Hopkings, W., 1833, Elements of Trigonometry  Home 5

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