Question

Prove that a cosB - b sinB = √a^2+b^2-c^2, when a sinB + b sinB = c.

I know it's too hard.

Answer 1

Solution:

It is given that $a \: sinB + B \: sinB = c \:$

$a \: sinB + B \: sinB = c \: (given)$

Squaring both sides,

$So, \: {(a \: sinB + b \: sinB) }^{2} = {c}^{2} .$

$= > {a}^{2} {sin}^{2} B + {b}^{2} {sin}^{2} B + 2ab \: sinB \: cosB = {c}^{2}$

$= > {a}^{2} (1 - {cos}^{2} B) + {b}^{2} (1 - {sin}^{2} B) + 2ab \: sinB \: cosB = {c}^{2}$

$= > {a}^{2} - {a}^{2} {cos}^{2} + {b}^{2}\: - {b}^{2} {sin}^{2} B + 2ab \: sinB \: cosB = {c}^{2}$

$= > {a}^{2} {cos}^{2} B - 2ab \: sinB \: cosB \: + {b}^{2} {sin}^{2} B= {a}^{2} + {b}^{2} - {c}^{2}. \:$

$= > ({a \: cosB - b \: sinB})^{2} = {a}^{2} + {b}^{2} - {c}^{2}$

$= > ({a \: cosB - b \: sinB}) = \sqrt{ {a}^{2} + {b}^{2} - {c}^{2}.}$

Hence, it is proved.

Answer 2

Error :-

→ It will be a sinB + b cosB = c , instead of a sinB + b sinB = c .

Given :-

→ a sinB + b cosB = c .......(1) .

Now,

→ ( a sinB + b cosB )² + ( a cosB - b sinB )² .

= a²sin²B + b²cos²B + 2a sinB b cosB + a²cos²B + b²sin²B - 2a sinB b cosB .

= a²sin²B + a²cos²B + b²cos²B + b²sin²B .

= a²( sin²B + cos²B ) + b²( cos²B + sin²B ) .

= a² + b² . [ ∵ sin²B + cos²B = 1 ] .

Thus, ( a sinB + b cosB )² + ( a cosB - b sinB )² = ( a² + b² ) .

⇒ c² + ( a cosB - b sinB )² = ( a² + b² ) .

⇒ ( a cosB - b sinB )² = ( a² + b² - c² ) .

⇒ ( a cosB - b sinB ) = √( a² + b² - c² ) .

Hence, $\sf \pink { (a \: { \cos }^{2} B - b \: { \sin }^{2} B) = \sqrt{ {a}^{2} + {b}^{2} - {c}^{2} } }.$