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Given that sin a = 3/5, cos b = 24/25. a is in (0,90) and b is in (180, 270 )....

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corrcorina | Student, College Freshman | eNoter

Posted October 2, 2010 at 8:39 PM via web

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Given that sin a = 3/5, cos b = 24/25. a is in (0,90) and b is in (180, 270 ). Calculate sin ( a + b ) and sin ( a - b ).

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giorgiana1976 | College Teacher | Valedictorian

Posted October 2, 2010 at 8:43 PM (Answer #1)

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We notice that the angle a belongs to the first quadrant, where the values of the functions sine and cosine are both positive. We also notice that the angle b is in the third quadrant, where the values of the functions sine and cosine are both negative.

We'll write the formula for sin (a+b).

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

We know the values for sin a and cos b but we'll have to calculate the values for sin b and cos a.

We'll apply the fundamental formula of trigonometry:

(sin a)^2 + (cos a)^2 = 1

(cos a)^2 = 1 - (sin a)^2

Since a is in the first quadrant, when we'll calculate the sqrt of 1 - (sin a)^2, we'll keep only the positive value for cos a.

cos a = sqrt (1 - 9/25)

cos a = sqrt 16/25

cos a = 4/5

(sin b)^2 + (cos b)^2 = 1

(sin b)^2 = 1 - (cos b)^2

Since b is in the third quadrant, when we'll calculate the sqrt of 1 - (cos b)^2, we'll keep only the negative value for sin b.

sin b = -sqrt(1 - 24^2/25^2)

sin b = - sqrt[(25-24)(25+24)/25^2]

sin b = - sqrt(1*49/25^2)

sin b = -7/5

Now, we can calculate sin (a+b)

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

sin (a+b) = (3/5)*(24/25) + (-7/5)*(4/5)

sin (a+b) = 72/125 - 28/25

sin (a+b) = (72-140)/25

sin (a+b) = -68/25

We'll write the formula for sin (a-b).

sin (a-b) = sin a*cos b - sin b*cos a

sin (a-b) = 72/125 + 28/25

sin (a-b) = (72+140)/25

sin (a-b) = 212/25

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neela | High School Teacher | Valedictorian

Posted October 2, 2010 at 8:51 PM (Answer #2)

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siana = 3/5 and cosb = 24/25.

a is in (0 ,90) and b in (180 , 270).

Therefore cosa = sqrt (1-sin^2a) = sqrt(1-3^2/5^2) = sqrt(16/25) = 4/5.

cos b = -24/25 . b in (180 to 270 . sinb = -sqrt(1 -24^2/25^1) = - 7/25.

to find sin(a+b) = sin(a+b) = sinacosb +cosasinb

sin(a+b) = (3/5)(-24/25) + (4/5)(-7/25) = (-72- 28)/125 =  =100/125 = -4/5.

sin(a-b) = sinacosb -cosasinb =  (3/5)(-24/25) - (4/5)(-7/25) =  (-72+28)/125 = -44/125.

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william1941 | College Teacher | Valedictorian

Posted October 2, 2010 at 9:05 PM (Answer #3)

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Here we need to find sin (a+b) and sin (a-b). We are given sin a = 3/5 and cos b = 24/25

Now we know that:

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

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

Also as a is in the first quadrant sin and cos are positive. In the third quadrant sin and cos are negative

Now as we are given sin a  and cob b we need to calculate cos a and sin b.

cos a = sqrt [ 1- (sin a)^2] = sqrt (1- 9 /25) = 4 / 5

sin b = sqrt [ 1- (cos a)^2] = sqrt [1 - (24/25)^2] = - 7 / 25

So sin (a+b) = sin a*cos b + sin b*cos a

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

=72/ 125 - 28/ 25

=-68/125

sin (a-b) = (3/5)(24/25) -(-7/5)(4/5)

= 72/ 125 + 28/ 25

= 212/125

Therefore sin (a+b)= -68/125 and sin (a-b) = 212/125

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