tan((60+45)/3)how do i find the exact value of tan35, cant form the double angles atm
And how do you do that?tan((60+45)/3)
There's probably a shorter way than using tan(105/3) but find the expansion of tan3@ and then let @=x/3. You'll get x^3-3Kx^2-3x+K=0 where K=tan(105). And then just use the cubic formula lol. And sub in a hell lot of values.And how do you do that?
I doubt a hsc level student knows cubic formula. Is it even in the syllabus?There's probably a shorter way than using tan(105/3) but find the expansion of tan3@ and then let @=x/3. You'll get x^3-3Kx^2-3x+K=0 where K=tan(105). And then just use the cubic formula lol. And sub in a hell lot of values.
Nah, it isn't lol. I don't think they'll ask such crap questions. They're more likely to ask tan(15)=tan(45-30)I doubt a hsc level student knows cubic formula. Is it even in the syllabus?
Yeah, only way there'd be a nice solution is if x E Q. And there's isn't one for this cubic.hmmm..... yes I agree. They'd probably just make you apply a single double-angle formula rather than such multi-step questions, espcially if the cubic is as inelegant as the above.
Where is this question from? I don't think it is easy at all.how do i find the exact value of tan35, cant form the double angles atm
If there's an exact value for tan(35), i'd like to hear it lol.i still cant do it with the double angle result, i use the expansion of tan angles over 3 and i dont get the answer still..
lol dw making it over 3 is wrong method
how would you do it? solution please
But you state: "find the exact value..."LOLOL the question wasnt find exact values, dw i got it
was a show question xP
exploitable where do you get these epic animations, lags typing as well
tan(pi/8)=sqrt((2-sqrt(2))/(2+sqrt(2)))i got a exact value question now
find exact value of tan (pi / 8)
using 1-cos2x / 1+cos2x = tan^2 x
i got the answer but i dont know how to get it into a simpler form, help ^^
root2 - 1 is the answer
i got root (3-2root 2)
yes i know its just a subbing in question, just dont know what to sub