2018年3月6日 · Let #theta = tan ^-1 (sqrt3/3)# #tan theta = sqrt3 / 3# #tan theta = cancel sqrt 3 / ( sqrt 3 * cancel sqrt3# #tan theta = 1 / sqrt 3# From the above table, #tan theta = 1 / sqrt 3 # or #theta = =( pi / 6)^c # or #30^@#

2015年10月24日 · Note #tan^(-1)# is defined as a function with values restricted to the range #[0,pi/2)# and if #theta = tan^(-1)(sqrt(3))# then #tan(theta) = sqrt(3)# A basic reference triangle looks like: #theta= 60^@ = pi/3#

2016年5月23日 · General solution is npi+pi/3, wheren is an integer tantheta=sqrt3=tan(pi/3) i.e. theta=pi/3 A tangent function has a cycle of pi and repeats after every pi Hence General solution is npi+pi/3, wheren is an integer

2016年9月29日 · Consider a triangle with sides #1#, #sqrt(3)/2# and #2#. This is a right angled triangle and one half of an equilateral triangle... Now #tan theta = "opposite"/"adjacent"# So looking at our diagram, #tan (pi/3) = sqrt(3)/1 = sqrt(3)# So one solution of the given equation is #x = pi/3# Note that:

2015年11月14日 · #tan(theta) = sin(theta)/cos(theta)# #sin(theta) = +-sqrt(1 - cos^2(theta))# During the arccosine range, the sine is always positive so

2016年4月26日 · arctan(sqrt(3)/3)= pi/6=30^@ Note that sqrt(3)/3 = 1/sqrt(3) Here is a standard trigonometric triangle with this ratio for the tan: Note that by definition the arctan function has a range of [0,pi) Trigonometry

2015年2月13日 · arctan( sqrt (3) ) = 60^@ or pi/3 radians If theta = arctan ( sqrt (3)) then tan (theta) = sqrt (3) by definition of tan-arctan relationship

2016年11月11日 · tanx=-sqrt3 x=tan^-1(-sqrt3) Radians: x=(2pi)/3, x=(5pi)/3 Degrees: x=120, x=300

2016年10月3日 · -pi/6 tan^-1(-sqrt3/3) tan^-1x means find the ANGLE that has a tangent of x The range of tan^-1 is -pi/2 to pi/2 -sqrt3/3 would fall in the fourth quadrant, so the value of tan^-1 is between -pi/2 and 0 and is a negative angle.

2016年7月18日 · And realizing $3 * tan(30) = \sqrt 3$, I started investigating trig functions... And Taylor series... If this question has no "answer" at this moment in time, it might take "new mathematics" to find an answer, or existing maths may be able to solve it... Either way, I feel that there should be a way to represent these irrational numbers.