Please help with math question below:
Express the triple angle formula of tangent (i.e. tan〖(3x)〗) in terms of tanx.
easy to look up, here is one such page
http://www.trans4mind.com/personal_development/mathematics/trigonometry/multipleAnglesTangent.htm
If you have to actually develop it,
use
tan(A+B) = (tanA + tanB)/(1 - tanAtanB)
on tan(3x) = tan(2x+1)
after doing tan2x first.
To express the triple angle formula of tangent (tan(3x)) in terms of tangent (tan(x)), we can use the angle addition formula for tangent, which states that:
tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A) * tan(B))
Here, we can set A = 2x and B = x:
tan(3x) = tan(2x + x)
Using the angle addition formula, we get:
tan(3x) = (tan(2x) + tan(x)) / (1 - tan(2x) * tan(x))
Now, we need to express tan(2x) in terms of tan(x). Again, using the angle addition formula, we set A = x and B = x:
tan(2x) = tan(x + x)
tan(2x) = (tan(x) + tan(x)) / (1 - tan(x) * tan(x))
tan(2x) = 2tan(x) / (1 - tan^2(x))
Substituting this expression back into the formula for tan(3x), we get:
tan(3x) = (2tan(x) / (1 - tan^2(x)) + tan(x)) / (1 - (2tan(x) / (1 - tan^2(x))) * tan(x))
tan(3x) = (2tan(x) + (1 - tan^2(x)) * tan(x)) / (1 - 2tan^2(x) / (1 - tan^2(x)))
Simplifying further:
tan(3x) = (2tan(x) + tan(x) - tan^3(x)) / (1 - 2tan^2(x) / (1 - tan^2(x)))
tan(3x) = (3tan(x) - tan^3(x)) / (1 - 2tan^2(x) / (1 - tan^2(x)))
Therefore, the triple angle formula of tangent (tan(3x)) in terms of tan(x) is:
tan(3x) = (3tan(x) - tan^3(x)) / (1 - 2tan^2(x) / (1 - tan^2(x)))