How do you prove that ArcTan(1) + ArcTan(2) + ArcTan(3) = π?
2015年9月24日 · Prove that (arctan (1) + arctan (2) + arctan (3) = pi) Call artan (1) = x; arctan (2) = y; and arctan (3) = z Apply the trig identity: tan (a + b) = (tan a + tan b ...
How do you find the exact value of #arctan(2)#? - Socratic
2018年3月15日 · This is not a rational number of degrees, nor a rational multiple of pi radians. We can write: arctan 2 = pi/2 - sum_(k=0)^oo (-1)^k 1/(2^(2k+1)(2k+1)) arctan(2) is an angle in a …
How do you evaluate # cos(arctan(2) +arctan(3))#? - Socratic
2016年4月16日 · Consider the following two right angled triangles: Then: #{ (tan(alpha) = 2), (tan(beta) = 3) :}# Notice that: ...
How do you find the exact value of #arctan(2) + arctan(3)
2015年5月25日 · In this way: #arctan2+arctan3=alpha#. Let's take now the tangent of both members: #tan(arctan2+arctan3)=tanalpharArr#
How do you evaluate #sec(arctan 2)#? - Socratic
2016年5月15日 · Let a = arc tan 2. Then, tan a = 2 > 0. a is in either the 1st quadrant or the 3rd. So, the given expression #sec a =+-sqrt 5#,
How do you evaluate #arctan(2/5)#? - Socratic
2016年11月27日 · #arctan(2/5) ~=2/5-8/375+32/15625~=0.381# rounded to the third decimal figure Explanation: The mac Laurin series for #arctan(x)# is
How do you find the exact value of sin(arctan(2))? - Socratic
2015年9月7日 · Let alpha = arctan(2), then, we require the value of sinalpha . now tan alpha = 2 => sinalpha = 2/sqrt5. tanalpha = 2 => y= 2a, x = 2a & r = sqrt (5a^2 ) = sqrt5 a sinalpha = 2a/ …
What is the derivative of #arctan(2x)#? - Socratic
2015年6月6日 · The general rule for derivating this type of function is y=arctan(u) y'=(u')/(1+u^2) Thus, considering that here u=2x y'=2/(1+(2x)^2)=2/(1+4x^2) Done!
How do you evaluate #arctan(1/2)+arctan(1/3)#? - Socratic
2016年7月21日 · By using the formula for the tangent of a sum, we find that the given sum is just \pi/4. Let tan\theta_1=1/2, tan\theta_2=1/3.
How do you evaluate #cos(Arctan 2)#? - Socratic
2016年6月10日 · Set x=arctan2 hence tanx=2 square both sides to get tan^2x=4=>tan^2x+1=5=>1/cos^2x=5 Hence cosx=1/sqrt5 or ...