Saved Bookmarks
| 1. |
To prove sin3A = 3sinA - 4sin^3A |
| Answer» Given that x = 30o\xa0we have to verify that sin 3x = 3sin x - 4sin3x.{tex}\\Rightarrow 3 x = 90 ^ { \\circ }{/tex}{tex}\\therefore\\sin 3 x{/tex}{tex}= \\sin 90 ^ { \\circ } = 1{/tex}and, 3sin x - 4sin3x{tex}= 3 \\sin 30 ^ { \\circ } - 4 \\sin ^ { 3 } 30 ^ { \\circ }{/tex}{tex}= 3 \\times \\frac { 1 } { 2 } - 4 \\left( \\frac { 1 } { 2 } \\right) ^ { 3 }{/tex}{tex}= \\frac { 3 } { 2 } - \\frac { 1 } { 2 } = 1{/tex}. | |