If a cos theta +b sin …
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Sia ? 4 years, 10 months ago
Given, acos{tex} \theta{/tex} - b sin{tex} \theta{/tex} = c
Squaring on both sides
(a cos{tex} \theta{/tex} - b sin{tex} \theta{/tex})2 = c2
By Adding (a sin{tex} \theta{/tex} + b cos{tex} \theta{/tex})2 on both sides, we get
(a cos{tex} \theta{/tex} - b sin{tex} \theta{/tex})2 + (a sin{tex} \theta{/tex} + b cos{tex} \theta{/tex})2 =c2 + (a sin{tex} \theta{/tex} + b cos{tex} \theta{/tex})2
(a2cos2{tex} \theta{/tex} + b2sin2{tex} \theta{/tex} - 2ab sin{tex} \theta{/tex} cos{tex} \theta{/tex}) + (a2sin2{tex} \theta{/tex} + b2cos2{tex} \theta{/tex} + 2ab sin{tex} \theta{/tex} cos{tex} \theta{/tex}) =c2 + (a sin{tex} \theta{/tex} + b cos{tex} \theta{/tex})2
a2(cos2{tex} \theta{/tex} + sin2{tex} \theta{/tex}) + b2(sin2{tex} \theta{/tex} + cos2{tex} \theta{/tex})=c2 + (a sin{tex} \theta{/tex} + b cos{tex} \theta{/tex})2
a2 + b2 = c2 + (a sin{tex} \theta{/tex} + b cos{tex} \theta{/tex})2
{tex} \Rightarrow{/tex} (a sin{tex} \theta{/tex} + b cos{tex} \theta{/tex})2 = a2 + b2 - c2
{tex} \Rightarrow{/tex} a sin{tex} \theta{/tex} + b cos{tex} \theta{/tex} = {tex} \pm \sqrt { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } }{/tex}
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