Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (goformit100):

find the value of tan (-330) degrees

OpenStudy (vishweshshrimali5):

First of all apply this: \[\large{\tan(-x) = -\tan(x)}\]

OpenStudy (vishweshshrimali5):

What will you get ?

OpenStudy (kainui):

I would instead suggest that since tangent is a trig function it's periodic every 360 degrees. that means you can add or subtract 360 to any value and it's the same. tan(x)=tan(x+360) That's another way of looking at it.

OpenStudy (primeralph):

tan (x) = tan(x+ 180 deg)

OpenStudy (primeralph):

tan is periodic with period pi.

OpenStudy (anonymous):

oh it will be \[\tan(-330)= -\frac{ 1 }{ \sqrt{3} }\]

OpenStudy (kainui):

True, but there's no reason to over complicate this @primeralph .

OpenStudy (primeralph):

It's not getting over-complicated.

OpenStudy (kainui):

Yeah, I guess you're right. Alright, well @goformit100 are you working on it or still stuck?

OpenStudy (elizabeths):

He's offline

OpenStudy (goformit100):

will the answer have -ve sign or +ve sign ?

OpenStudy (neer2890):

Tan is odd function so, tan(-330)=-tan(330)=- tan(90*3+60) (90*3+60) indicates 4th quadrant. in which tan theta is negative. so answer is negative. also multiple of 90 is odd, so tan changes to cot theta. so - tan(330)=-(-cot 60)=cot 60

OpenStudy (goformit100):

@Koikkara

OpenStudy (goformit100):

tan(-330)=-tan(330) How ?

OpenStudy (neer2890):

because tan theta is odd function. and odd function means f(-x)=-f(x)

OpenStudy (kainui):

I just realized @primeralph 's picture isn't a lava monster or some sort of Tron character. Lol wtf.

OpenStudy (neer2890):

also, \[\tan \Theta=\frac{ \sin \Theta }{ cos \Theta }\] sin is odd function and cos is even. so tan is also odd fxn.

OpenStudy (goformit100):

Will the answer be |dw:1403154923058:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!