stuck

[rasta]

hey people
i m a newbie here and i m stuck with this
tan(arctan3+pi/4).arctan(tan(-31pi/11) = ?
using the tan(pi + b) = tan b i got :
tan(arctan3+pi/4).arctan(tan(-31pi/11) = tan(arctan3+pi/4). 2pi/11 .
also i dont know how to prove this one :
arctan -1/2 + arctan -1/3 = - pi/4
thx in adavance
peace

[hotvette]

also i dont know how to prove this one: arctan -1/2 + arctan -1/3 = - pi/4

I assume you are allowed to use trig identities. If so, using the trig identity of tan(a + b) = xxxxxx should get you there.

[Fermat]

...
also i dont know how to prove this one :
arctan -1/2 + arctan -1/3 = - pi/4
thx in adavance
peace
Take the tan of both sides.

[rasta]

Take the tan of both sides.
i did but i got 5/6= 1 wich s not true

[hotvette]

Suggestion by Fermat is much easier than mine, although both will yield the same result. You made a math mistake in taking the tangent of the left hand side.

[rasta]

how shall i do it?

[hotvette]

Show us how you got 5/6.

[rasta]

i did this
tan(arctan-1/3) + tan(arctan -1/2) = tan-pi/4
-1/3 + -1/2 = -1
5/6 = 1

[hotvette]

There's the math mistake: tan(a+b) is NOT tan(a) + tan(b). The solution is much simpler if you can use the actual values of arctan(-1/3) and arctan(-1/2). If not, then you need a trig identity like I said originally.

[Fermat]

You have A + B = C

use,

tan(A+B) = (tanA + tanB)/(1 - tanA.tanB)

and equate to tanC

[rasta]

There's the math mistake: tan(a+b) is NOT tan(a) + tan(b). The solution is much simpler if you can use the actual values of arctan(-1/3) and arctan(-1/2). If not, then you need a trig identity like I said originally.
never mind i got it :smile:
thx

[mohlam12]

that s good