first expand it
cosxcosπ/4 - sinxsinπ/4 - (cosxcoxπ/4 + sinxsinπ/4) = 1
-2sinxsinπ/4 = 1
-2sinx(√2/2) = 1 , since we know sinπ/4 = 1/√2 or √2/2
sinx = -1/√2
the sine is negative in III and IV
so
x = π + π/4 = 5π/4</b? or
x = 2π - π/4 = 7π/4
solve the equation on the interval [0,2pi)
cos(x+pi/4)-cos(x-pi/4)=1
3 answers
x = π + π/4 = 5π/4 or
x = 2π - π/4 = 7π/4
x = 2π - π/4 = 7π/4
Use the identity formulas for cos(a+b) and cos(a-b):
cosx*cos(pi/4) - sinx*sin(pi/4)-cosx*cos(pi/4) - sinx*sin(pi/4) = 1
-2 sinx*sin(pi/4)= 1
sinx (1/sqrt2)= -1/2
sinx = -(sqrt2)/2
x = 5 pi/4 or 7 pi/4
cosx*cos(pi/4) - sinx*sin(pi/4)-cosx*cos(pi/4) - sinx*sin(pi/4) = 1
-2 sinx*sin(pi/4)= 1
sinx (1/sqrt2)= -1/2
sinx = -(sqrt2)/2
x = 5 pi/4 or 7 pi/4
1 answer