Question

sin alpha=3/5 cos beta =9/41 find sin(alpha-beta)

Answer 1

hey mate here is ur answer

Answer 2

$Given \: sin \alpha = \frac{3}{5} \: --(1)$

$i )cos^{2} \alpha = 1- sin^{2} \alpha \\= 1 - \big(\frac{3}{5}\big)^{2} \\= 1 - \frac{9}{25} \\= \frac{25-9}{25} \\= \frac{16}{25}$

$cos \alpha = \sqrt{\frac{16}{25} }\\= \frac{4}{5} \: --(2)$

$and \: cos \beta = \frac{9}{41} \: --(3)$

$ii )sin^{2} \beta = 1- cos^{2} \alpha \\= 1 - \big(\frac{9}{41}\big)^{2} \\= 1 - \frac{81}{1681} \\= \frac{1681-81}{1681} \\= \frac{1600}{1681}$

$sin \beta = \sqrt{\frac{1600}{1681}} \\= \frac{40}{41} \: --(4)$

$Now, \red { Value \:of \: sin (\alpha - \beta )} \\= sin \alpha cos \beta - cos \alpha sin \beta \\= \frac{3}{5} \times \frac{9}{41} - \frac{4}{5} \times \frac{40}{41}\\= \frac{27}{205} - \frac{160}{205} \\= \frac{27-160}{205} \\= \frac{-133}{205}$

Therefore.,

$\red { Value \:of \: sin (\alpha - \beta ) } \green {= \frac{-133}{205}}$

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