Given cos A = -3/5, π<A<3π/2, and tan B = 7/24, B in Quadrant I, find: (a) sin (A+B), (b) cos (A+B), and (c) tan (A+B)
To do these type of problems, you need to know the following:
the expansion of
1. sin(A+B) = sinAcosB + cosAsinB
2. cos(A+B) = cosAcosB - sinAsinB
3. and tan(A+B) = (tanA + tanB)/( 1 - tanAtanB)
4. the CAST rule
5. the definitions of the basic trig ratios based on the sides
of a right-angled triangle
given: cos A = -3/5 , cosA = adjacent/hypotenuse and A is in III
so sinA = -4/5 , tanA = 4/3
tan B = 7/24, B is in I, so sinB = 7/25 and cosB = 24/25
sin(A+B) = (-4/5)(24/25) + (-3/5)((7/25) = -117/125
do the others in the same way.
You can always check your answers with your calculator,
my sin(A+B) is correct after checking.
Thank you
To find the values of trigonometric functions for the sum of angles A and B, we need to use the trigonometric sum formulas:
(a) sin(A + B) = sin A * cos B + cos A * sin B
(b) cos(A + B) = cos A * cos B - sin A * sin B
(c) tan(A + B) = (tan A + tan B) / (1 - tan A * tan B)
Let's start by calculating the values.
Given:
cos A = -3/5 and π < A < 3π/2
tan B = 7/24 and B is in Quadrant I
To find sin A, we can use the Pythagorean identity sin^2(A) + cos^2(A) = 1. Since cos A = -3/5, we can calculate sin A.
sin A = √(1 - cos^2(A))
sin A = √(1 - (-3/5)^2)
sin A = √(1 - 9/25)
sin A = √(16/25)
sin A = 4/5
Now, we can substitute the values of sin A, cos A, and tan B into the formulas.
(a) sin(A + B) = sin A * cos B + cos A * sin B
sin(A + B) = (4/5) * cos B + (-3/5) * sin B
(b) cos(A + B) = cos A * cos B - sin A * sin B
cos(A + B) = (-3/5) * cos B - (4/5) * sin B
(c) tan(A + B) = (tan A + tan B) / (1 - tan A * tan B)
tan(A + B) = ((-3/5) + (7/24)) / (1 - (-3/5) * (7/24))
Now, we can simplify the expressions to the final values.