Please help with this one ... thanks so much
if cosA = 4/5; sin <0; cosB = 12/13; 0<B<90, determine sin(A-B)
please note that in the heading tere is an error with the B, it says ; 0<B>90 when it should read 0<B<90, like in the above shown way
Please help with this one ... thanks so much
if cosA = 4/5; sin <0; cosB = 12/13; 0<B<90, determine sin(A-B)
please note that in the heading tere is an error with the B, it says ; 0<B>90 when it should read 0<B<90, like in the above shown way
if cosA = 4/5; sin < 0; cosB = 12/13; 0<B<90, determine sin(A-B)
This is only barely a calculation problem. You should be sufficiently familiar with the definitions to pull this off.
You need the Pythagorean Throrem.
cos(A) = 4/5; sin < 0 ==> sin(A) = -3/5 Note that this is Quadrant IV
cos(B) = 12/13; 0<B<90 ==> sin(B) = 5/13
You need the identity:
sin(A-B) = sin(A)cos(B)-cos(A)sin(B)
Let's see what you get.
Hello, Jono Mathematician
Identity: .If , . determine:
We are given: .
Cosine is positive in Quadrants 1 and 4; sine is negative in Quadrants 3 and 4.
. . Hence, is in Quadrant 4.
Since , we have: . (use Pythagorus!)
Since is in Quadrant 4,
. . Hence: .
We are given: . is in Quadrant 1.
Since , we have: .
Since is in Quadrant 1,
. . Hence: .
Substitute into the identity [1]:
. .