(a) Convert #cos2x-sqrt3sin2x# as a cosine ratio only; (b) solve the equation #y=f(x)=cos2x-sqrt3sin2x-1# and (c) draw the graph of the function to check result?

1 Answer
Jul 12, 2017

Please see below.

Explanation:

(a) #cos2x-sqrt3sin2x=2(cos2x xx1/2-sin2x xxsqrt3/2)#

= #2(cos2xcos(pi/3)-sin2xsin(pi/3))#

= #2cos(2x+pi/3)#

(b) #y=f(x)=cos2x-sqrt3sin2x-1#, will cut #x#-axis where #y=0# i.e. where #cos2x-sqrt3sin2x=1# or #2cos(2x+pi/3)=1# i.e. when

#cos(2x+pi/3)=1/2=cos(2npi+-pi/3)# i.e.

#x+pi/6=npi+-pi/6#, where #n# is an integer or #{-pi,-pi/3,0,(2pi)/3,pi}# in #[-pi,pi]# (choose #n=-1,0,1# for this).

It will cut #y#-axis where #x=0# i.e. #y=0# - note that it means that #f(x)# passes through origin.

(c) The graph of #y=f(x)=cos2x-sqrt3sin2x-1# appears as follows:

graph{cos(2x+pi/3)-1/2 [-5.05, 4.95, -3.09, 1.91]}