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Line 15:
Line 15: − When A=60°+
Activity-trigonometry problems (view source)
Revision as of 19:11, 16 August 2015
, 19:11, 16 August 2015→Generalisation By Verification
==Methos Of Solutions==
==Methos Of Solutions==
=== '''Generalisation By Verification'''===
=== '''Generalisation By Verification'''===
When A=60°
LHS=<math>\frac{1-\tan^2 60°}{1+\tan^2 60°}</math> <br>=<math>\frac{1-{(\sqrt{3})}^2 }{1+{(\sqrt{3})}^2 }</math><br>=<math>\frac{1-3}{1+3}</math><br>=<math>\frac{-2}{4}</math><br>=<math>\frac{-1}{2}</math>-----(1)<br>RHS=<math>1-2\sin^2 A</math><br>=<math>1-2\sin^260° </math><br><math>1-2{(\frac{\sqrt{3}}{2})}^2</math><br>=<math>1-2(\frac{3}{4})</math><br>=<math>\frac{4-2(3)}{4}</math><br>=<math>\frac{-1}{2}</math>------(2)<br>from eqn1 & eqn2<br><math>\frac{1-\tan^2 60°}{1+\tan^2 60°}</math>=<math>1-2\sin^260° </math><br>'''By Generalisation'''<br> '''<math>\frac{1-\tan^2 A}{1+\tan^2 A}=1-2\sin^2 A</math>'''
LHS=<math>\frac{1-\tan^2 60°}{1+\tan^2 60°}</math> <br>=<math>\frac{1-{(\sqrt{3})}^2 }{1+{(\sqrt{3})}^2 }</math><br>=<math>\frac{1-3}{1+3}</math><br>=<math>\frac{-2}{4}</math><br>=<math>\frac{-1}{2}</math>-----(1)<br>RHS=<math>1-2\sin^2 A</math><br>=<math>1-2\sin^260° </math><br><math>1-2{(\frac{\sqrt{3}}{2})}^2</math><br>=<math>1-2(\frac{3}{4})</math><br>=<math>\frac{4-2(3)}{4}</math><br>=<math>\frac{-1}{2}</math>------(2)<br>from eqn1 & eqn2<br><math>\frac{1-\tan^2 60°}{1+\tan^2 60°}</math>=<math>1-2\sin^260° </math><br>'''By Generalisation'''<br> '''<math>\frac{1-\tan^2 A}{1+\tan^2 A}=1-2\sin^2 A</math>'''