how do i find the value of this equation (trig)?

Question

tan beta = 3, how do i find the value of
tan beta + tan(beta + pi) + tan(beta + 2pi)

thanks for the help.

Answer 1

It's 9.
tan(beta + pi) = tan(beta + 2pi) = tan(beta) = 3

So tan beta + tan(beta + pi) + tan(beta + 2pi) = 3+3+3 = 9.

You can see why tan(beta) = tan(beta + pi) = tan(beta + 2pi) by looking at this graph of y = tan x
http://www.intmath.com/TrigGrph/4_TCSC.php
You can see that the graph repeats itself in intervals of pi. So if you look at where y = 3, then move pi spaces to the right on the x-axis, you will see that y=3 there as well.

Answer 2

Use the trig identity

tan(b) = tan(b + n*pi), where n = 0, 1, 2, ...

if tan(b) = 3,
tan (b + pi) = 3,
tan (b + 2 pi) = 3

tan(b) + tan(b+pi) + tan(b + 2pi) = 3 + 3 + 3 = 9

Hope this helps,

-Guru

Answer 3

sin(beta + pi) is -sin(beta). cos(beta + pi) is -cos(beta). Therefore sin/cos(beta + pi), or tan(beta + pi), is the same as tan(beta). As for adding 2pi, anytime you add 2pi to an angle, it's the same angle as before, so tan(beta + 2pi) is tan(beta). Therefore, the equation simplifies to tan(beta) + tan(beta) + tan(beta), or 9.

Answer 4

tan(B) + (tan(b+pi) + tan(b+2pi)

tan(b+pi) = tan(b) + tan(pi) /(1-tan(b).tan(pi) )

= 3 +0/(1-3.0) = 3

tan(b+2pi) = tan(b) + tan(2pi)/(1- tan(b).tan(2pi)
= 3+0/(1-3.0) = 3

ans = 3+3+3 = 9.


tan (pi)= tan(3.14) = 0
tan (2pi)= tan(6.28) =0

Answer 5

You should know than tan(x+π) = -tan(x), and tan(x+2π) = tan(x).
So if tan(beta) = 3, tan(beta+π) = -3, and tan(beta+2π) = 3.