Question

If cosA = $\frac{a}{\sqrt{a^{2}+b^{2} } }$ , prove that: asinA = bcosA.

Answer 1

Step-by-step explanation:

We have,

$\rm \cos(A) = \frac{a}{ \sqrt{ {a}^{2} + {b}^{2} } }$

So,

$\rm \sin(A) = \sqrt{1 - \cos ^{2} (A) } = \sqrt{1 - \bigg(\frac{a}{ \sqrt{ {a}^{2} + {b}^{2} } } \bigg)^{2} }$

$\rm \implies\sin(A) = \sqrt{1 - \frac{a^{2} }{ {a}^{2} + {b}^{2} } }$

$\rm \implies\sin(A) = \sqrt{\frac{ {a}^{2} + {b}^{2} - a^{2} }{ {a}^{2} + {b}^{2} } }$

$\rm \implies\sin(A) = \sqrt{\frac{ {b}^{2} }{ {a}^{2} + {b}^{2} } }$

$\rm \implies\sin(A) = \frac{ b }{ \sqrt{ {a}^{2} + {b}^{2} }}$

Now,

$\rm \implies \: a\sin(A) = \frac{ ab }{ \sqrt{ {a}^{2} + {b}^{2} }}$

$\rm \implies \: a\sin(A) =b \bigg( \frac{ a }{ \sqrt{ {a}^{2} + {b}^{2} }} \bigg)$

$\rm \implies \: a\sin(A) =b \bigg( \cos(A) \bigg)$

$\rm \implies \: a\sin(A) =b \cos(A)$