Use a sum or difference formula to find the exact value of the trig function; tan(105*) Please show detailed work, thank you...
tan105=(tan135-tan30)/(1+tan135tan30)
I've gotten to the point above, but I do not know how to simplify.
(-1-sqrt3/3)/(1+(-1)(sqrt3/3))
105 comes from what angles?
135-30
and 135 is also known as?
180-135 = 45
(3pi)/4
45 is simpler to evaluate with
i see you went straight for the tan(a+b) instead of wasting time with the sin(a+b)/cos(a+b)
i can never really recall which tan is the right one so I just go by the sin cos sin(a+b) = sin(a)cos(b) + sin(b)cos(a) ------------------------------- cos(a+b) = cos(a)cos(b) -sin(a)sin(b) sin(30)cos(45) + sin(45)cos(30) ---------------------------- cos(30)cos(45) -sin(30)sin(45)
1/2 sqrt(2)/2 + sqrt(2)/2 sqrt(3)/2 ------------------------------ sqrt(3)/2 sqrt(2)/2 - 1/2 sqrt(2)/2 sqrt(2)/4 + sqrt(6)/4 ------------------ sqrt(6)/4 - sqrt(2)/4 sqrt(2) + sqrt(6) ------------- sqrt(6) - sqrt(2) seems to fit
do a conjugate of the bottom to see if we can simplify it more
(-sqrt2+sqrt6)-(sqrt6-sqrt2)
I do not know what a conjugate is
sqrt(2) + sqrt(6) sqrt(6) + sqrt(2) ------------- -------------- sqrt(6) - sqrt(2) sqrt(6) + sqrt(2) 8 + 4 sqrt(3) ----------- 6 - 2 8 + 4 sqrt(3) ----------- 4 2 + sqrt(3) with any luck that checks out with the wolf
http://www.wolframalpha.com/input/?i=tan%28105%29 yep, just gotta place it in the correct quadrant for sign
a conjugate is a name for a pair of ..... well (a+b) (a-b) is a conjugate pair
OK, thank you for your time and effort
youre welcome
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