Again, 3u Cambo Book... challenging (for me lol). (1 Viewer)

[currysauce]

1.

In triangle ABC, b/c = 4/3.

(a) If B=2C show that cosC = 2/3
(b) If B=3C show that sinC = root(15)/6

2.

By writing sin^4x as (sin²x)², show that sin^4x = 3/8 - 1/2 cos2x + 1/8 cos4x

 

[rama_v]

(a) sin2@/4 = sin@/3
6sin@cos@=4sin@
divide by sin@
6cos@=4
therefore cos@=4/6=2/3
where alpha is angle C

(b) sin3@/4=sin@/3
Remember that sin3@ = 3sin@-4sin^3@
so 9sin@-4sin^3@=4sin@
sin@(9-4sin^2@)=4sin@
divide by sin@
so 9-4sin^2@=4
sin^2@=5/12
so sin@=sqrt5/sqrt12
rationalise the denominator
sin@=(sqrt15)/6

 

[FinalFantasy]

2.

By writing sin^4x as (sin²x)², show that sin^4x = 3/8 - 1/2 cos2x + 1/8 cos4x

(sin x)^4= (sin²x)²=(1-cos²x)²=1-2cos²x+cos^4 x

sin^4x = 3/8 - 1/2 cos2x + 1/8 cos4x
=3\8-1\2(2cos²x-1)+1\8(2cos²2x-1)
=3\8-cos²x+1\2+1\4 cos²2x-1\8
=3\4+1\4cos²2x-cos²x
=3\4+1\4(2cos²x-1)²-cos²x
=3\4+1\4(4cos^4 x-4cos²x+1)-cos²x
=3\4+cos^4x-cos²x+1\4-cos²x
=1-2cos²x+cos^4 x=(sin x)^4= (sin²x)²

 

[currysauce]

thanks guys

 

[currysauce]

New problems,

Use compound-angle formulae to solve for 0<=x<=2pi

sin(x+pi/6) = 2sin(x-pi/6)

 

[shafqat]

New problems,

Use compound-angle formulae to solve for 0<=x<=2pi

sin(x+pi/6) = 2sin(x-pi/6)

sinxcospi/x + cosxsinpi/6 = 2sinxcospi/6 - 2cosxsinpi/6
sqrt3/2.sinx + cosx/2 = sqrt3.sinx - cosx
sqrt3/2.sinx = 3/2 .cosx
sinx/cosx = sqrt3
tanx = sqrt3
x = pi/3, 4pi/3

 

[currysauce]

thanks man