1/2 cos(cos^-1(1/2)) . let cos^-1(1/2)=A :.cosA=1/2=cos(pi/3) :.A=pi/3 or cos^-1(1/2)=pi/3 So , cos(cos^-1(1/2)) = cos (pi/3) =1/2[Ans]

The angle is 30º or pi/6+2pin or -30º or-pi/6+2pin Let theta=cos^(-1)(sqrt3/2) So costheta=sqrt3/2 cos is >=0 in the 1st and 4th quadrant So theta=30º (pi/6 rad) and theta =-30º (-pi/6 rad) Trigonometry

Refer to Explanation cos^-1(1/2) = 60^@ Trigonometric Ratios of Special Angles:

Mar 20, 2017 · For the #color(magenta)(cos^-1(-4/5)# the triangle is in quadrant 2 since #color(magenta)(cos^-1 x)# is restricted to quadrants 1 and 2 and since the argument is negative the triangle has to be in quadrant 2.

May 26, 2016 · Let b = cos^(-1)(4/5). Then cos b = 4/5 > 0. b is in the 1st quadrant or in the 4th. So, sin b =+-3/5. ...

May 1, 2018 · Let #theta = cos^(-1)(1/2)#. In other words, #cos(theta)=1/2# But we know that #theta=pi/3# is a solution to this equality; #cos(pi/3)=1/2# Now, since #cos# is a periodic function with period #2npi#, for integer #n#, we can rewrite this as: #cos(pi/3+2npi)=1/2# Or #pi/3+2npi=cos^(-1)(1/2)#, #forall n in ZZ#.

Oct 13, 2016 · #cos(tan^-1(4/3)-sin^-1(12/13))# The restriction for the range of #tan^-1x # is #(-pi/2,pi/2)# and the restriction for #sin^-1x# is # [-pi/2,pi/2]#.

Feb 1, 2017 · 0.60 Method 1: Use calculator cos ^-1 (6/7) --> arccos (6/7) Calculator gives: cos x = 6/7 --> arc x = 31^@ tan 31^@ = 0.60 Method 2: Use right triangle OMA with horizontal leg OA = 6 hypotenuse OM = 7 Vertical leg AM = sqrt(49 - 36) = sqrt13 Call t the angle < AOM: tan t = AM/OA = sqrt13/6 = 0.60

Apr 20, 2016 · ( 2 k + 1 )pi, k =0, +-1, +-2, +-3... If a = cos^(-1)(-1), cos a = -1 If the range of a is specified as , say, [0, 2 pi ], a = pi.

Aug 1, 2016 · Expand and simplify LHS - Apply identity sin^2x+cos^2x = 1 To verify: (cotx - cscx)(cosx +1) =-sinx LHS=(cosx/sinx - 1/sinx)(cosx+1) =1/sinx(cosx-1)(cosx+1) =1/sinx ...