If A = arc sin -3/5 and B = arccos 7/25 what is sin(A+B) & cos(A-B)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have A = arc sin (-3/5) and B = arc cos (7/25)

=> sin A = -3/5 and cos B = 7/25

cos A = sqrt ( 1 - (-3/5)^2)

=> 4/5

sin B = sqrt ( 1 - (7/25)^2)

= 24/25

sin (A + B) = sin A*cos B + cos A*sin B

=> (-3/5)*(7/25) + (4/5)*(24/25)

=> 6/10

cos ( A - B) = cos A * cos B + sin A * sin B

=> (4/5)(7/25) + (-3/5)(24/25)

=> -44/125

The required value of sin(A + B) = 3/5 and cos(A - B) = -44/125

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll apply the formula:

sin(A+B) = sin A*cos B + sinB*cos A

sin A = sin[arcsin (-3)/5] = -3/5

cos B = cos [arccos 7/25] = 7/25

sin B = sqrt(1 - 49/625) = sqrt(576/625) = 24/25

cos A = sqrt(1 - 9/25) = sqrt(16/25) = 4/5

sin(A+B) = (-3/5)*(7/25) + (24/25)*(4/5) = (96-21)/125 = 75/125 = 3/5

We'll apply the formula for the cosine of difference of 2 angles:

cos(A-B) = cos A*cos B + sin A*sin B

We'll just substitute the values found out earlier:

cos(A-B) = 4*7/5*25 + (-3*24)/5*25

cos(A-B) = (28 - 72)/5*25 = -44/125

The values for sin(A+B) and cos(A-B) are: sin(A+B) = = 3/5 , cos(A-B) = -44/125.

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