Question

If sinA = 3/5 and coB =9/41 then the sin (A-B)​

Answer 1

Answer:

Correct Question:-

If sinA = 3/5 and cosB =9/41, A and B lies in Ist quadrant, then the value of sin (A-B)

Answer:-

Given :-

• sinA = 3/5 and cosB =9/41, A and B lies in Ist quadrant.

To find :-

• The value of sin(A - B)

Identity used:-

$\bf \:★ \: sin(A - B) = sinA cosB - sinB cosA$

$\bf \:★ \: sin²x + cos²x = 1$

$\bf\ ★\:sinx = \sqrt{1 - {cos}^{2}x }$

$\bf \:★ \: cosx = \sqrt{1 - {sin}^{2} x}$

Solution :-

$\bf \: As \: sinA = \dfrac{3}{5}$

$\bf \:We \: know, \: sin²A + cos²A = 1$

$\bf\implies \: {(\dfrac{3}{5} )}^{2} + {cos}^{2} A = 1$

$\bf\implies \: {cos}^{2} A = 1 - \dfrac{9}{25}$

$\bf\implies \: {cos}^{2} A = \dfrac{16}{25}$

$\bf \:A \: lies \: in \: first \: quadrant$

$\bf\implies \:cosA = \dfrac{4}{5}$

$\bf \:As \: cosB = \dfrac{9}{41}$

$\bf \:We \: know, sin²B + cos²B = 1$

$\bf \bf\implies \: {sin}^{2} B + {(\dfrac{9}{41}) }^{2} = 1$

$\bf\implies \: {sin}^{2} B = 1 - \dfrac{81}{1681}$

$\bf\implies \: {sin}^{2} B = \dfrac{1600}{1681}$

$\bf \:B \: lies \: in \: first \: quadrant$

$\bf\implies \:sinB \: = \dfrac{40}{41}$

Now,

$\bf \:sin(A - B) = sinA cosB - sinB cosA$

On substituting the values of sinA, cosA, sinB, cosB, we get

$\bf\implies \:sin(A - B) = \dfrac{3}{5} \times \dfrac{9}{41} - \dfrac{40}{41} \times \dfrac{4}{5}$

$\bf\implies \:sin(A - B) = \dfrac{27 - 160}{205}$

$\bf\implies \:sin(A - B) = - \dfrac{133}{205}$

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