Simplification (sin & cos) 15pts~

b. 1-sin^2(3x)
=cos^2(3x)

since there is a identity
sin^2 x + cos^2 x =1
係依度個x係指任何degree都ok

c. cos(pi/2 + 2x )
=-sin(2x)

係cos入面見到pi/2, 後面無論+定- ,都要轉左sin先
然後在係個sin 出面睇下駛唔駛加番個negative sign
係依個case, pi/2 + 2x 個第二個quad,
cos 係第二個quad係-的, 所以我就係個sin出面加番個負號

d. 開方[(1+cos 70)/2]
cos 2x = cos^2 x - sin^2 x
cos 2x = 2cos^2 x -1
so cos^2 x =(1+cos2x)/2

so 開方[(1+cos 70)/2]
=開方(cos^2 35)
=cos 35

e. 1-2sin^2 x
=cos2x

用番上面條式
cos 2x = cos^2 x - sin^2 x
cos 2x = 1- 2sin^2 x

唔知夠唔夠詳盡, 如果都唔明可以同我講

b. 1-cos²3x=sin²3x

c. cos(pi/2+2x)=-sin(2x) [cos converted to sin because it is pi/2+something, -is needed because cos is negetive in second quadrant]

d. sqrt[(1+cos70°)/2]=cos35 [cos2x=2cos²x-1 implies that cosx=sqrt(1+cos2x)/2]

e. 1-2sin²x=sin²x+cos²x-2sin²x=cos²x-sin²x=cos2x