2016年11月9日 · 1 Unit circle --> cos (5pi) = cos (pi + 4pi) = cos pi = - 1

2015年7月23日 · #{5pi}/12 \times {360^circ}/{2 pi} = 75^circ = 150^circ/2# How? First I sigh and point out that this is yet another trig problem that assumes the student knows only the two cliche triangles, 30/60/90 and 45/45/90.

2016年4月30日 · The angle #(5pi)/3" is located in the 4th quadrant "# where the cos ratio has a positive value. The 'related' acute angle is #(2pi - (5pi)/3) = pi/3 # and so # cos((5pi)/3) = cos(pi/3) # Using the #color(blue)" Exact value triangle for this angle" # #rArr cos((5pi)/3)=cos(pi/3)=1/2#

2015年12月31日 · sin(5pi) = 0 cos(5pi) = -1 tan(5pi) = 0 sec(5pi) = -1 csc(5pi) is "undefined" cot(5pi) is "undefined" First, let's note that sine and cosine are periodic with period 2pi, meaning for any integer k, sin(x) = sin(x+k2pi) and cos(x) = cos(x+k2pi) With this, we get the first two desired values: sin(5pi) = sin(pi + 2(2pi)) = sin(pi) = 0 and cos(5pi) = cos(pi+2(2pi)) = cos(pi) = -1 The remaining ...

2016年6月1日 · cos((5pi)/4)= -1/sqrt(2) or -sqrt(2)/2 (5pi)/4 is an angle in Quadrant III and as such (based on CAST) its cos is negative.

2016年1月8日 · cos ((5pi)/8) = - (sqrt(2 - sqrt2))/2 cos ((5pi)/8) = cos ((-3pi)/8 + pi) = -cos ((3pi)/8) Find cos ((3pi)/8). Call cos x = cos ((3pi)/8) cos 2x = cos ((3pi)/4 ...

2015年7月29日 · #arccos(cos((5pi)/4))# is an angle between #0# and #pi# whose cosine is the same as the cosine of #(5pi)/4#. The angle we want is #(3pi)/4# We know that #cos((5pi)/4) = -sqrt2/2# and the Quadrant II angle with cosine equal to #-sqrt2/2# is #(3pi)/4#

2018年5月17日 · Use trig table and unit circle --> #sin ((5pi)/3) = sin (-pi/3 + (6pi)/3) = sin (-pi/3) = - sin pi/3 = - sqrt3/2#

2015年7月17日 · Evaluate cos ((-5pi)/2) Answer: zero cos ((-5pi)/2) = cos (-pi/2 - 2pi) = cos (-pi/2) = 0

2015年9月21日 · cos^-1[cos((-5pi)/3)]=pi/3 First get the value of cos((-5pi)/3) For this we need to find the acute angle ...