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 …

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 …

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 …

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 …

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 …

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 …

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 …