Difference between revisions of "Activity-trigonometry problems"
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# Idea about trignometric identities | # Idea about trignometric identities | ||
==Methos== | ==Methos== | ||
− | + | === '''Generalisation By Verification'''=== | |
When A=60° | 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> | + | 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>''' |
− | + | ==='''By Deductive Proof'''=== | |
− | + | LHS=<math>\frac{1-\tan^2 A}{1+\tan^2 A}</math><br>=<math>\frac{1-\frac{\sin^2A}{\cos^2A}}{\sec^2A}</math><br> | |
+ | =<math>\frac{[\cos^2A-\sin^2A]}{cos^2A}\cos^2A</math><br> | ||
+ | =<math>1-\sin^2A-\sin^2A</math><br>=<math>1-2\sin^2A</math>=RHS |
Revision as of 00:50, 1 August 2014
Problem-1
prove that
Interpretation of problems
- It is to prove the problem based on trigonometric identities
- the function of one trigonometric ratio is relates to other
Concept development
Develop the skill of proving problem based trigonometric identity
Skill development
Problem solving
Pre Knowledge require
- Idea about trignometric ratios
- Idea about trignometric identities
Methos
Generalisation By Verification
When A=60°
LHS=Failed to parse (syntax error): {\displaystyle \frac{1-\tan^2 60°}{1+\tan^2 60°}}
=
=
=
=-----(1)
RHS=
=Failed to parse (syntax error): {\displaystyle 1-2\sin^260° }
=
=
=------(2)
from eqn1 & eqn2
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \frac{1-\tan^2 60°}{1+\tan^2 60°}}
=Failed to parse (syntax error): {\displaystyle 1-2\sin^260° }
By Generalisation
By Deductive Proof
LHS=
=
=
=
==RHS