1. Find all solutions to the equation in the interval [0, 2π).
cos 2x - cos x = 0
using a graphical tool
x1=0
x2=2π/3
x3=4π/3
the answer is the letter A) 0, two pi divided by three. , four pi divided by three.
2. Rewrite with only sin x and cos x.
sin 3x
sin(3x)=sin(2x+x)
sin(A+B)=sinAcosB+cosAsinB
sin(x+2x)=sinxcos2x+cosxsin2x
sin2x =
2sinxcosx
cos2x = (cosx)^2 -
(sinx)^2
sin(x+2x)=sinx((cosx)^2 - (sinx)^2)+cosx(2sinxcosx)
we have (sinx)^2 =1- (cosx)^2
sin(x+2x)=sinx((cosx)^2 - (1- (cosx)^2)+cosx(2sinxcosx)
sin(x+2x)=sinx((2cosx)^2 - 1)+2sinx(cosx)^2
sin(x+2x)=2sinx(cosx)^2 - sinx+2sinx(cosx)^2
(cosx)^2 =1- (sinx)^2
sin(x+2x)=2sinx(cosx)^2 - sinx+2sinx(1- (sinx)^2)
sin(x+2x)=2sinx(cosx)^2 - sinx+2sinx- 2(sinx)^3)
sin(x+2x)=2sinx(cosx)^2 +sinx- 2(sinx)^3)
the answer is the letter D) 2 cos2x sin x + sin x - 2 sin3x
3. Find the exact value by using a half-angle identity.
cosine of five pi divided by twelve
cos(x/2)=±(√1+cos(x))/2
We know that cos(π/6)=√3/2.
So
cos(π/12)=(√2+√3)/2 See attached file problem 13
the answer is one half times the square root of quantity
two plus square root of three
