Question

sina= I√10 sinß = 1/√5 and
a,ß are acute then show that

a+ß= π\4​

Answer 1

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$\: \:$

$\sf {sin}^{2} \theta \: + \: {cos}^{2} \theta = 1$

$\sf {cos}^{2} \theta = 1 - {sin}^{2} \theta$

$\sf \: cos\theta \: = \sqrt{1 - {sin}^{2}\theta }$

$\: \:$

$\sf \: cos \: a \: = \sqrt{1 - {( \frac{1}{10} )}^{2} } \\ = \sqrt{1 - \frac{1}{10} } \\ = \sqrt{ \frac{9}{10} } \\ = \frac{3}{ \sqrt{10} }$

$\: \:$

$\sf \: cos \: b \: = \sqrt{1 - {( \frac{1}{ \sqrt{5} } )}^{2} } \\ = \sqrt{1 - \frac{1}{5} } \\ = \sqrt{ \frac{4}{5} } \\ = \frac{2}{ \sqrt{5} }$

$\: \: \:$

Finding sin( a + b),

= sin a cos b + cos a sin b

$\sf \: = \frac{1}{ \sqrt{10} } \times \frac{2}{ \sqrt{5} } + \frac{3}{ \sqrt{10} } \times \frac{1}{ \sqrt{5} }$

$\sf = \frac{2}{5 \sqrt{2} } + \frac{3}{5 \sqrt{2} }$

$\sf = \frac{5}{5 \sqrt{2} }$

$\sf = \frac{1}{ \sqrt{2} }$

= sin 45°

$\sf = sin \frac{\pi}{4}$

Now,

$\sf \: sin \: (a + b) = sin \frac{\pi}{4}$

$\sf \: a + b = \frac{\pi}{4}$

------- Proved

$\: \:$

${\huge{\underline{\sf {\pink{More \:\: Information :}}}}}$

$\: \:$

• sin ( a + b ) = sin a cos b + cos a sin b

• sin ( a - b ) = sin a cos b - cos a sin b

• cos ( a + b ) = cos a cos b - sin a sin b

• cos ( a - b ) = cos a cos b + sin a sin b