Question

finding exact values of trig functions.

285 degrees
FIND EXACT VALUE OF TANGENT

Answer 1

Your answer is -3.4874
Use calculator here:
http://www.1728.org/trigcalc.htm

Answer 2

It can't be in decimal form

Answer 3

On my teacher' s answer key, this is what she put:
-sqr6 - sqr2 divided by sqr6 + sqr2 = -2-sqr3

Answer 4

But I don't get how she got that

Answer 5

you can't mean 286 then

Answer 6

maybe you mean 285

Answer 7

285=45+240

Answer 8

use sum rule for tangent

Answer 9

My mistake, you're right, I meant 285

Answer 10

If you don't have any questions, then I assume you have this.

Answer 11

How do you convert -sqr6 - sqr2 divided by sqr6 + sqr2 into -2-sqr3??
I know the tangent sum and difference identity but I'm just on this part...

Answer 12

\[\frac{-\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}=-\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}}=-1 \neq -2 -\sqrt{3}\]

Answer 13

\[\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\]
we have
\[\tan(45+240)\]
so we have a=45 and b=240
what is tan(a) or tan(45)
what is tan(b) or tan(240)

Answer 14

Yes, I got the same answer as you, -1, but my teachers answer key, says -2 -sqr3.
Is it possible that my teacher got it wrong?

Answer 15

it shouldn't be 1

Answer 16

or -1

Answer 17

do you know what tan(45) and tan(240) equals

Answer 18

Tan(45) = 1 and tan(240)= sqr3
Right?

Answer 19

yeah

Answer 20

\[\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)} \\ a=45 \\ b=240 \\ \tan(a)=1 \\ \tan(b)=\sqrt{3} \\ \tan(45+240)=\frac{1+\sqrt{3}}{1-1 \sqrt{3}}=\frac{1+\sqrt{3}}{1-\sqrt{3}}\]

Answer 21

now multiply top and bottom by the conjugate of the bottom

Answer 22

\[\frac{1+ \sqrt{3}}{1-\sqrt{3}} \cdot \frac{1+\sqrt{3}}{1+\sqrt{3}}=?\]

Answer 23

this part rationalizes the bottom (the denominator)

Answer 24

do you know how to multiply the top and bottom?

Answer 25

I'm not sure... Is it
1+3/ 1-3

Answer 26

I'm sorry I'm not that good at math

Answer 27

\[\frac{(1+\sqrt{3})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}=\frac{1(1+\sqrt{3})+\sqrt{3}(1+\sqrt{3})}{1(1+\sqrt{3})-\sqrt{3}(1+\sqrt{3})}\]
use the distribute property twice for top and bottom
also there is a short cut for doing (A-B)(A+B)=A^2-B^2
or even (A-B)(A-B)=A^2-2AB+B^2

Answer 28

Please try this identities:
\[\begin{gathered}
\tan (285) = \tan (180 + 105) = \frac{{\tan (180) + \tan (105)}}{{1 - \tan (180)\tan (105)}} = \tan (105) \hfill \\
\tan \left( {105} \right) = \tan \left( {180 - 75} \right) = \frac{{\tan \left( {180} \right) - \tan \left( {75} \right)}}{{1 + \tan \left( {180} \right)\tan \left( {75} \right)}} = - \tan \left( {75} \right) \hfill \\
\end{gathered} \]
so:
\[\tan \left( {285} \right) = - \tan \left( {75} \right)\]

Answer 29

Ohhhhh I get it now!! Then all of that equal.s -2-sqr3!

Answer 30

It is...did you actually finish the multiplying and addition/subtract then a little division

Answer 31

and finally:
\[\tan (75) = \tan (45 + 30) = \frac{{\tan (45) + \tan (30)}}{{1 - \tan (45)\tan (30)}} = \frac{{1 + \left( {1/\sqrt 3 } \right)}}{{1 - \left( {1/\sqrt 3 } \right)}}\]

Answer 32

Yes, it'll be
4+2sqr3 / -2 = -2-sqr3