Question

${\circ \: \text {If} \: \: a \: \text{cos} \theta - b \text{sin} \theta = c, \text{then show that}} \\ {a \: \: \text{sin} \: \theta + b \: \text{cos} \: \theta = \pm \: \sqrt{{a² + b² + c²}}} \:$​

Answer 1

Step-by-step explanation:

We have,

$a \cos( \theta) - \sin( \theta) = c$

$\implies \: (a \cos( \theta) - b \sin( \theta) )^{2} = {c}^{2}$

$\implies \: a ^{2} \cos^{2} ( \theta) + b^{2} \sin^{2} ( \theta) - 2ab \sin( \theta) \cos( \theta) = {c}^{2}$

$\implies \: a ^{2} (1 - \sin^{2} ( \theta) ) + b^{2} (1 - \cos^{2} ( \theta) )- 2ab \sin( \theta) \cos( \theta) = {c}^{2}$

$\implies \: a ^{2} - {a}^{2} \sin^{2} ( \theta) + b^{2} - {b}^{2} \cos^{2} ( \theta) - 2ab \sin( \theta) \cos( \theta) = {c}^{2}$

$\implies \: - {a}^{2} \sin^{2} ( \theta) - b^{2} \cos^{2} ( \theta) - 2ab \sin( \theta) \cos( \theta) = {c}^{2} - {a}^{2} - {b}^{2}$

$\implies \: - ( {a}^{2} \sin^{2} ( \theta) + b^{2} \cos^{2} ( \theta) + 2ab \sin( \theta) \cos( \theta) ) = - ( {a}^{2} + {b}^{2} - {c}^{2} )$

$\implies \: ( {a}^{2} \sin^{2} ( \theta) + b^{2} \cos^{2} ( \theta) + 2ab \sin( \theta) \cos( \theta) ) = ( {a}^{2} + {b}^{2} - {c}^{2} )$

$\implies \: ( a \sin( \theta) + b \cos ( \theta) )^{2} = ( {a}^{2} + {b}^{2} - {c}^{2} )$

$\implies \: a \sin( \theta) + b \cos ( \theta) = \pm \sqrt{ {a}^{2} + {b}^{2} - {c}^{2} }$