Question

Oues : If a sinΦ + b cos Φ = c, then prove that a cos Φ - b sin Φ = ± √a² + b² -c². ​
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Answer 1

To Prove :

$\tt a \cos \phi - b \sin \phi = \pm \sqrt{ {a}^{2} + {b}^{2} - {c}^{2} }$

Proof :

$\tt a \sin \phi + b \cos \phi = c.$

Squaring both sides, we get.

$\tt {a \sin \phi + b \cos \phi }^{2} = {c }^{2} .$

$\tt\implies {a}^{2} { \sin}^{2} \phi + {b}^{2} { \cos }^{2} \phi + 2ab \sin \phi. \cos \phi = {c}^{2}$

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

$\tt\implies {a}^{2} - {a}^{2} { \cos}^{2} \phi+ {b}^{2} - { {b}^{2} \sin }^{2} \phi + 2ab \sin \phi. \cos \phi = {c}^{2} .$

$\tt\implies {a}^{2} + {b}^{2} - {c}^{2} = {a}^{2} \cos \phi + {b}^{2} \sin \phi - 2ab \sin \phi \cos \phi.$

$\tt\implies a \cos \phi - b \sin \phi = \pm \sqrt{ {a}^{2} + {b}^{2} - {c}^{2} }$

$\tt{ \red{H}ance \: \red{P}roved.}$

Some information :

$\blacksquare \tt { \sin }^{2} \theta + { \cos}^{2} \theta = 1.$

$\blacksquare \tt{(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab.$

$\blacksquare \tt {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab.$

$\blacksquare \tt{ \sin }^{2} \theta = 1 - { \cos}^{2} \theta.$