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# If cos x +cos^2x =1 then sin ^12 x +3sin^10 x +3sin^8 x +sin^6 x+1=

Ajay
209 Points
5 years ago
Here is solution........................................................................................................
$cosx\quad +\quad { cos }^{ 2 }x\quad =\quad 1\quad \quad implies\\ cosx\quad =\quad { sin }^{ 2 }x\quad and\quad { cos }^{ 2 }x\quad =\quad 1-cosx\quad \\ { sin }^{ 12 }x\quad =\quad cos^{ 6 }x,\quad { sin }^{ 10 }x\quad =\quad cos^{ 5 }x,\quad \quad { sin }^{ 8 }x\quad =\quad cos^{ 4 }x,\quad and\quad { sin }^{ 6 }x\quad =\quad cos^{ 3 }x,\\ substituting\quad above\quad in\quad orignal\quad expression\\ F(x)\quad =\quad cos^{ 6 }x\quad +\quad 3cos^{ 5 }x\quad +\quad \quad 3cos^{ 4 }x\quad +\quad cos^{ 3 }x\quad +1\\ \quad \quad \quad \quad \quad \quad \quad put\quad { cos }^{ 2 }x\quad =\quad 1-cosx$
To be continued
Ajay
209 Points
5 years ago
Remaining solution..............................................................................................................
$=\quad { (1-cosx) }^{ 3 }\quad +\quad 3cosx{ (1-cosx) }^{ 2 }\quad +3{ (1-cosx) }^{ 2 }\quad +\quad cosx({ (1-cosx) }\quad +\quad 1\\ \quad \quad \quad \quad \quad =\quad 2{ cos }^{ 3 }x\quad -{ cos }^{ 2 }x\quad -5cosx\quad +\quad 5\\ \quad \quad \quad \quad \quad \quad =\quad 2cosx(1-cosx)\quad -\quad (1-cosx)\quad -5cosx\quad +\quad 5\\ \quad \quad \quad \quad \quad \quad =\quad -2{ cos }^{ 2 }x\quad -2cosx\quad +\quad 4\quad \\ \quad \quad \quad \quad \quad =\quad -2({ cos }^{ 2 }x\quad +\quad cosx)\quad +\quad 4\\ \quad \quad \quad \quad \quad =\quad -2\quad +\quad 4\quad =\quad 2$