Given that Sin(A+B)=2Cos(A-B) and tan A = 1/3 ,
Find the exact value of tan B.
if tan A = 1/3, then by Pythagoras,
sin A = 1/√10
cos A = 3/√10
if sin(A+B) = 2cos(A-B)
sinAcosB + cosAsinB = 2(cosAcosB + sinAsinB)
(1/√10)cosB + (3/√10)sinB = (6/√10)cosB + (2/√10)sinB
multiply by √10 and simplify
cosB + 3sinB = 6cosB + 2sinB
sinB = 5cosB
sinB/cosB = 5/1
tanB = 5/1
then cosB = 1/√26
To find the exact value of tan B, we need to utilize the given equation and the given value of tan A.
Let's start by rearranging the given equation: Sin(A+B) = 2Cos(A-B)
Using the trigonometric identity: Sin(A+B) = Sin A * Cos B + Cos A * Sin B, we can rewrite the equation as:
Sin A * Cos B + Cos A * Sin B = 2Cos(A-B)
Since we know that tan A = 1/3, we can use the definition of tan to express Sin A and Cos A in terms of tan A:
tan A = Sin A / Cos A
1/3 = Sin A / Cos A
By cross-multiplying, we get:
Sin A = (1/3) * Cos A
Now let's substitute the expression for Sin A in our rearranged equation:
(1/3) * Cos A * Cos B + Cos A * Sin B = 2Cos(A-B)
Next, let's divide both sides of the equation by Cos A to simplify it:
[ (1/3) * Cos A * Cos B + Cos A * Sin B ] / Cos A = 2Cos(A-B) / Cos A
After simplifying further:
(1/3) * Cos B + Sin B = 2Cos(A-B) / Cos A
Now, let's substitute the value of tan A = 1/3 into the equation, remembering that tan A = Sin A / Cos A:
(1/3) * Cos B + Sin B = 2Cos(A-B) / [ (1/3) * Cos A ]
Multiplying both sides of the equation by (1/3) to get rid of the fraction on the left side:
Cos B + 3Sin B = 6Cos(A-B) / Cos A
Now, let's bring all terms involving B to one side of the equation:
Cos B + 3Sin B - 6Cos(A-B)/Cos A = 0
Finally, we have an equation in terms of B that we can solve to find the exact value of tan B. However, this equation might not have a simple solution, so it could require further algebraic manipulation or the use of numerical methods to find an approximate value for tan B.