2016年9月30日 · Use the known trigonometric identity . #cos(a+b)=cosa*cosb-sina*sinb# we have that. #cos(x+pi/2)=cosx*cos(pi/2)-sinx*sin(pi/2)=cosx*0-sinx*1=-sinx#

2016年5月1日 · This is a well used trig. relation along with #sin(pi/2-theta)#. that is : #cos(pi/2-theta)=sintheta" and " sin(pi/2-theta)=costheta#

Evaluate the parenthetical expression first, then the cosine follows. cos (pi/4) = 0.707 The cosine function just applies to whatever is defined to it. It may be a function itself. In this case it is just and expression. (pi/2)/2 = pi/4 cos (pi/4) = 0.707 (assuming radian measures)

2016年3月18日 · How do you apply the fundamental identities to values of #theta# and show that they are true

2018年4月5日 · cos(pi/2-x)*csc(-x)=-1 for x not a multiple of pi. We know that the angles of a right triangle must sum to pi radians. Since the right angle is pi/2 radians, the other two angles must sum to pi/2 radians. Therefore pi/2 minus the adjacent angle must be the opposite angle.

2018年2月23日 · \cos (-\frac{\pi }{2})=0 Remember that cos is an even function so that: cos (-x)=cos(x) So by applying this property we have: =\cos (\frac{\pi }{2}) =0 That's it!

2015年9月5日 · cos((3pi)/2)=0 You can calculate it using reduction formula. cos((3pi)/2)=cos(2pi-pi/2)=cos(pi/2)=0

2016年1月27日 · x=pi/2 sin A=cos (pi/2-A) Therefore, from the given equation sin (x-pi/2)=cos x If A=x-pi/2, then sin (x-pi/2)=cos (pi/2-(x-pi/2)) it follows that cos x=cos (pi/2-(x ...

2015年9月15日 · If P(x,y) submits alpha at the centre of unit circle, x = cos alpha , y = sinalpha, tanalpha = y/x. etc. For alpha = pi/2 , x = 0, y = 1. => cospi/2 = 0, sinpi/2 = 1 and tanpi/2= 1/0 (undefined) cscpi/2 = 1,secpi/2 = 1/0(undefined) and cotpi/2 = 0.

2015年4月23日 · for the "true" proof you need to use matrice, but this is acceptable : sin(a+b) = sin(a)cos(b)+cos(a)sin(b) sin(pi/2+x) = sin(pi/2)*cos(x)+cos(pi/2)*sin(x) sin(pi/2) = 1 cos(pi/2) = 0 So we have : sin(pi/2+x) = cos(x) Since this answer is very usefull for student here the full demonstration to obtain sin(a+b) = sin(a)cos(b)+cos(a)sin(b) (do not read this if you are not …