well, that is certainly one way. Since 13π/12 = 3π/4 + π/3
sin(3π/4 + π/3) = (1/√2)(1/2) + (-1/√2)(√3/2)
cos(3π/4 + π/3) = (-1/√2)(1/√2) - (-1/2)(√3/2)
Now simplify, divide and simplify the result. You should ned up with 2-√3
How could you evaluate tan (13pi/12) if you did not know the sum and difference formula for tangent? Would you use the sin and cos sum and difference formulas, and if so, can someone walk me through it? Thank you!!!
5 answers
OR
13 π / 12 = 12 π / 12 + π / 12 = π + π / 12
tan ( A + B ) = ( tan A + tan B ) / ( 1 - tan A ∙ tan B )
In this case:
tan ( 13 π / 12 ) = tan ( π + π / 12 ) = ( tan π + tan π / 12 ) / ( 1 - tan π ∙ tan π / 12 )
Since tan π = 0
tan ( 13 π / 12 ) = ( 0 + tan π / 12 ) / ( 1 - 0 ∙ tan π / 12 )
tan ( 13 π / 12 ) = ( tan π / 12 ) / 1
tan ( 13 π / 12 ) = tan π / 12
tan ( θ / 2 ) = ( 1 - cos θ ) / sin θ
In this case π / 12 = ( π / 6 ) / 2
tan ( π / 12 ) = tan ( π / 6 ) / 2 = [ 1 - cos ( π / 6 ) ] / sin ( π / 6 ) =
( 1 - √3 / 2 ) / ( 1 / 2 ) =
( 2 / 2 - √3 / 2 ) ∙ 2 =
( 2 - √3 ) / 2 ∙ 2 = 2 - √3
_____________________
Remark:
1 / ( 1 / 2 ) = 2
so
( 1 - √3 / 2 ) / ( 1 / 2 ) =
( 1 - √3 / 2 ) ∙ 1 / ( 1 / 2 ) =
( 2 / 2 - √3 / 2 ) ∙ 2
_____________________
Solution:
tan 13 π / 12 = 2 - √3
13 π / 12 = 12 π / 12 + π / 12 = π + π / 12
tan ( A + B ) = ( tan A + tan B ) / ( 1 - tan A ∙ tan B )
In this case:
tan ( 13 π / 12 ) = tan ( π + π / 12 ) = ( tan π + tan π / 12 ) / ( 1 - tan π ∙ tan π / 12 )
Since tan π = 0
tan ( 13 π / 12 ) = ( 0 + tan π / 12 ) / ( 1 - 0 ∙ tan π / 12 )
tan ( 13 π / 12 ) = ( tan π / 12 ) / 1
tan ( 13 π / 12 ) = tan π / 12
tan ( θ / 2 ) = ( 1 - cos θ ) / sin θ
In this case π / 12 = ( π / 6 ) / 2
tan ( π / 12 ) = tan ( π / 6 ) / 2 = [ 1 - cos ( π / 6 ) ] / sin ( π / 6 ) =
( 1 - √3 / 2 ) / ( 1 / 2 ) =
( 2 / 2 - √3 / 2 ) ∙ 2 =
( 2 - √3 ) / 2 ∙ 2 = 2 - √3
_____________________
Remark:
1 / ( 1 / 2 ) = 2
so
( 1 - √3 / 2 ) / ( 1 / 2 ) =
( 1 - √3 / 2 ) ∙ 1 / ( 1 / 2 ) =
( 2 / 2 - √3 / 2 ) ∙ 2
_____________________
Solution:
tan 13 π / 12 = 2 - √3
Thank you both sooo much! It was the last problem on my hw and I've been stuck for hours.
OR
so we can't use tan expansions, ok
tan(13pi/12) = sin(13pi/12) / cos(13pi/12)
When I was learning this stuff many eons ago, I personally found that
I could work it out better in degrees than all that radian stuff with π's
(13pi/12) = 195°
several ways to get 195 with basic angles, oobleck did 60 + 135
I am going to try 180+15
we need sin15 and cos15.
I know cos30 = √3/2 and cos30 = 1 - 2sin^2 15
2sin^2 15 = 1 - √3/2 = (2 - √3)/2
sin15= √((2-√3)/4)
similarly, cos15 = √((2+√3)/4)
so tan15 = √((2-√3)/4) / √((2+√3)/4) = √(2-√3) / √(2+√3)
but tan 195° = - tan15° = -√(2-√3) / √(2+√3)
which does reduce to 2 - √3, but that's another story.
Don't claim this is easier, just another way. What else do I have to do
after 4 weeks of quarantine lock-down ??
so we can't use tan expansions, ok
tan(13pi/12) = sin(13pi/12) / cos(13pi/12)
When I was learning this stuff many eons ago, I personally found that
I could work it out better in degrees than all that radian stuff with π's
(13pi/12) = 195°
several ways to get 195 with basic angles, oobleck did 60 + 135
I am going to try 180+15
we need sin15 and cos15.
I know cos30 = √3/2 and cos30 = 1 - 2sin^2 15
2sin^2 15 = 1 - √3/2 = (2 - √3)/2
sin15= √((2-√3)/4)
similarly, cos15 = √((2+√3)/4)
so tan15 = √((2-√3)/4) / √((2+√3)/4) = √(2-√3) / √(2+√3)
but tan 195° = - tan15° = -√(2-√3) / √(2+√3)
which does reduce to 2 - √3, but that's another story.
Don't claim this is easier, just another way. What else do I have to do
after 4 weeks of quarantine lock-down ??
Haha you're right Reiny. Thank you very much and I hope you continue to entertain yourself during quarantine.
1 answer