give that sin( A + B )= 2cos(A - B ) and tan A = 1/3,Find the exact value of tanB
if tanA = 1/3, then by the Pythagorean triangle
sinA = 1/√10 , and cosA = 3/√10
sin(A+B) = 2cos(A-B)
sinAcosB + cosAsinB = 2cosAcosB + 2sinAsinB
(1/√10)cosB + (3/√10)sinB = 2(3√10)cosB + 2(1/√10)sinB
times √10 ....
cosB + 3sinB = 6cosB + 2sinB
-sinB = 3cosB
-sinB/cosB = 3
tanB = -3
Given:
sin( A + B )= 2cos(A - B )
Expand by sum and difference formulae:
sinAcosB+cosAsinB=2cosAcosB+2sinAsinB
Divide each side by cosAcosB and simplify:
2tanAtanB-tanA-tanB+2=0
Substitute tanA=1/3
(2/3)tanB-tanB=5/3
Solve for tanB
tanB=5
found a silly arithmetic error in 3rd last line, should have been
sinB = 5cosb
sinB/cosB = 5
tanB = 5
To find the exact value of tan(B), we need to use the given information and apply trigonometric identities.
Given:
sin(A + B) = 2cos(A - B) (Eq. 1)
tan(A) = 1/3
We can start by manipulating Eq. 1 to express sin(A + B) and cos(A - B) in terms of sine and cosine of A and B. We'll use the angle addition and subtraction formulas:
sin(A + B) = sin(A)cos(B) + cos(A)sin(B) (Eq. 2)
cos(A - B) = cos(A)cos(B) + sin(A)sin(B) (Eq. 3)
Let's substitute these equations into Eq. 1:
sin(A)cos(B) + cos(A)sin(B) = 2(cos(A)cos(B) + sin(A)sin(B))
Now, let's rearrange the terms to isolate sin(B):
sin(A)cos(B) + cos(A)sin(B) = 2cos(A)cos(B) + 2sin(A)sin(B)
sin(A)cos(B) - 2sin(A)sin(B) = 2cos(A)cos(B) - cos(A)sin(B)
Factor out sin(B) and cos(B):
sin(A)[cos(B) - 2sin(B)] = cos(A)[2cos(B) - sin(B)]
Divide both sides by cos(A)[2cos(B) - sin(B)]:
sin(A) / cos(A) = cos(A) / [2cos(B) - sin(B)]
Using the identity tan(A) = sin(A) / cos(A), we can substitute the value tan(A) = 1/3:
1/3 = cos(A) / [2cos(B) - sin(B)]
Cross-multiply:
3cos(A) = 2cos(B) - sin(B)
Now, we have an equation involving cos(A) and cos(B). We'll use the Pythagorean identity to relate these trigonometric values:
cos^2(A) + sin^2(A) = 1
Since tan(A) = 1/3, we have sin(A) = 1 and cos(A) = 3/sqrt(10).
Substituting these values into the equation:
3(3/sqrt(10)) = 2cos(B) - sin(B)
9/sqrt(10) = 2cos(B) - sin(B) (Eq. 4)
Now, let's solve Eq. 4 for cos(B) - sin(B):
2cos(B) - sin(B) = 9/sqrt(10)
Move the sin(B) term to the left side:
2cos(B) = sin(B) + 9/sqrt(10)
Now, we'll square both sides of the equation to eliminate the square root:
(2cos(B))^2 = (sin(B) + 9/sqrt(10))^2
4cos^2(B) = sin^2(B) + (9/sqrt(10))^2 + 2sin(B)(9/sqrt(10))
Using the Pythagorean identity again, we substitute cos^2(B) = 1 - sin^2(B):
4(1 - sin^2(B)) = sin^2(B) + (9/sqrt(10))^2 + 2sin(B)(9/sqrt(10))
Distribute and rearrange terms:
4 - 4sin^2(B) = sin^2(B) + 81/10 + (18/10)(sin(B))
Multiply through by 10 to remove the fraction:
40 - 40sin^2(B) = 10sin^2(B) + 81 + 18sin(B)
Combine like terms:
50sin^2(B) + 18sin(B) - 40 = 0
This equation is quadratic in form. We can solve it using factoring, the quadratic formula, or graphing techniques. I will use factoring for simplicity.
Factor the quadratic:
(10sin(B) + 20)(5sin(B) - 2) = 0
Setting each factor equal to zero:
10sin(B) + 20 = 0 or 5sin(B) - 2 = 0
First equation:
10sin(B) = -20
sin(B) = -2
Since the sine function is only defined in the range [-1, 1], there are no solutions for sin(B) = -2.
Second equation:
5sin(B) = 2
sin(B) = 2/5
Now, we can use the identity tan(B) = sin(B) / cos(B) to find the exact value of tan(B):
tan(B) = (2/5) / cos(B)
However, with the given information, we don't have enough data to determine the exact value of cos(B) or find a numerical value for tan(B). More information or additional trigonometric relationships would be required to obtain an exact answer.