if COS A=3/5 and Sin B=7/25, when A us acute and B is obtuse, find without using tables. the value of COS(A+B)
Wrong solution
To find the value of cos(A+B), we can use the cosine addition formula:
cos(A+B) = cos(A)cos(B) - sin(A)sin(B)
Given that cos(A) = 3/5 and sin(B) = 7/25, we need to find cos(B) and sin(A).
Since A is acute, cos(A) = 3/5 implies sin(A) = sqrt(1 - cos^2(A)) = sqrt(1 - (3/5)^2) = 4/5 (using the Pythagorean identity).
Since B is obtuse, sin(B) = 7/25 implies cos(B) = sqrt(1 - sin^2(B)) = sqrt(1 - (7/25)^2) = 24/25.
Now, substituting the values into the formula:
cos(A+B) = (3/5)(24/25) - (4/5)(7/25)
= 72/125 - 28/125
= 44/125
Therefore, cos(A+B) = 44/125.
To find the value of COS(A+B) without using tables, we can make use of the trigonometric identities and the given information.
1. Start by applying the sum identity for cosine: COS(A+B) = COS(A) * COS(B) - SIN(A) * SIN(B).
2. We already know COS(A) = 3/5 and SIN(B) = 7/25. We need to find the values of COS(B) and SIN(A) to proceed.
3. Since A is acute, we know that SIN(A) = √(1 - COS^2(A)). Calculate SIN(A) using this formula.
SIN(A) = √(1 - (3/5)^2) = √(1 - 9/25) = √(16/25) = 4/5
4. Now, we need to find COS(B). As B is obtuse, we can use the fact that COS(B) = -√(1 - SIN^2(B)). Calculate COS(B) using this formula.
COS(B) = -√(1 - (7/25)^2) = -√(1 - 49/625) = -√(576/625) = -24/25
5. Substitute the values of COS(A), SIN(A), COS(B), and SIN(B) into the sum identity for cosine:
COS(A+B) = (3/5 * -24/25) - (4/5 * 7/25)
Simplify the expression:
COS(A+B) = -72/125 - 28/125
COS(A+B) = -100/125
Now, simplify the fraction:
COS(A+B) = -4/5
Therefore, the value of COS(A+B) is -4/5.
DDDDDDRAW IT !!!!!!!
first triangle 3,4, 5 in quad 1
second triangle 24, 7 , 25 in quad II
now look at it
with triangle similar to A in quadrant III but 8 tiems lengths
x = - 24, same as triangle A
y = -32
hypotenuse = 40
so
I have a triangle in Quadrant III with hypotenuse 40 and leg 24 along -x axis
what is angle below -y axis ?
cos T = -24/40 = -3/5