Trigonometry
Tan20.Tan40.Tan60.Tan80=?
This seems like a textbook question, but i want able to find it in the 6 books that I own.
This seems like a textbook question, but i want able to find it in the 6 books that I own.
Labels: mathschallenge, SharedPuzzle
posted by Vago at
2:21 AM
7 Comments:
The answer is 3.
I'll let the others find out how to do it.
just remember : tan=sin/cos
and
sin(a).sin(b)=(-1/2).(cos(a+b)-cos(a-b)).
-El Tistou
i got 3 or -0.0894746808. 3 i got by tan(20)x tan(40)x tan(60)x tan(80)=3, and tan(20 x tan(40 x tan(60 x tan(80)))) = -0.0894746808
thank you 9 year old ti83+
i think the question was to do it using just your pain, a sheet of paper and your brain.
Now using the calculator is the easy way, but try it the hard way!
much more fun...
El Tistou
Well, El Tistou, plz tell me how you get it.
Also your answer is wrong, the correct answer is rt of 3 (aka tan 60)
In other words u gotta prove that tan20.tan40.tan80=1
I do not know how though so can any1 shed any light?
Also i am in 10th and we havent done any cidentities like that sinA.sinB thing, and it should be possible to solve it using SIMPLE trig. like the ratios, the 90-A, or the sin^2+cos^2=1 and such other identities.
None of those A+B or anything.
Correction: answer is 3 but i still want a simple explanation
First note tan(a+b)=(tan(a) + tan(b))/(1-tan(a) tan(b))
and tan(a-b)=(tan(a) - tan(b))/(1+tan(a) tan(b))
Let t=tan(20), then tan(40) = tan(20+20) = 2t/(1-t^2)
and tan(60)=tan(40+20)=t((3-t^2)/(1-3t^2)
Also tan(60) = sqrt(3)
But tan(40)=tan(60-20)=(sqrt(3)-t)/(1+t*sqrt(3))
and tan(80)=tan(60+20)=(sqrt(3)+t)/(1-t*sqrt(3))
So tan(40)*tan(80) = (3-t^2)/(1-3t^2)
(The last used the identity (a-b)(a+b)=a^2-b^2 (twice))
Compare with tan(40+20) above => t*tan(40)*tan(80) = tan(60)
and as t = tan(20), tan(20)*tan(40)*tan(80)=tan(60)
so tan(20)*tan(40)*tan(60)*tan(80)
= tan(60)*tan(60) = sqrt(3)*sqrt(3) = 3
Post a Comment
Links to this post:
Create a Link
<< Home