Math, asked by kaaraksh, 9 months ago

cos(A+B)=0, sin(A-B)=?

Answers

Answered by Anonymous

25

AnswEr :

Given that,

$\sf \: cos(A + B) = 0$

We have to find the value of sin(A - B)

Here,

$\sf \: cos(A + B) = 0 \\ \\ \longrightarrow \: \sf cos(A + B) = cos90 \\ \\ \longrightarrow \: \sf \ \: A + B = 90^{\circ} \\ \\ \longrightarrow \boxed{\sf B = 90 - A }$

Now,

$\sf \: sin(A - B) \\ \\ \dashrightarrow \: \sf \: sin \bigg(A - (90 - A )\bigg) \\ \\ \dashrightarrow \sf \: sin(2A - 90) \\ \\ \dashrightarrow \: \sf \: sin2Acos90 - sin90cos2A \\ \\ \dashrightarrow \: \sf \: - cos2A \:$

Thus,sin(A - B) = - cos2A

NoTE

sin(A + B) = sinAcosB + cosAsinB

sin(A - B) = sinAcosB - cosAsinB

cos(A + B) = cosAcosB - sinAsinB

cos(A - B) = cosAcosB + sinAsinB

Answered by MarshmellowGirl

28

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