Question

Prove that sin³A + sin³(60° + A) + sinº(240° + A) = -3/4sin3A ​

Answer 1

See below for proof.

Explanation:

`\boxed{\textsf{Prove that}\;\;\sin^3A + \sin^3(60^(\circ) + A) + \sin^3(240^(\circ)+ A) = \sin^3A}`

Step 1

Rewrite 240° as (180° + 60°):

`\sin^3A + \sin^3(60^(\circ) + A) + \sin^3(180^(\circ)+60^(\circ)+ A)`

Step 2

As sin(180° + x) = -sin(x), we can rewrite sin³(180° + 60° + A) as:

`\sin^3(180^(\circ)+60^(\circ)+ A)=-\sin^3(60^(\circ)+ A)`

Step 3

Substitute this into the expression:

`\sin^3A + \sin^3(60^(\circ) + A) -\sin^3(60^(\circ)+ A)`

Step 4

As the last two terms cancel each other, we have:

`\sin^3A`

Hence proving that:

`\sin^3A + \sin^3(60^(\circ) + A) + \sin^3(240^(\circ)+ A) = \sin^3A`

As one calculation:

`\sin^3A + \sin^3(60^(\circ) + A) + \sin^3(240^(\circ)+ A)`

`=\sin^3A + \sin^3(60^(\circ) + A) + \sin^3(180^(\circ)+60^(\circ)+ A)`

`=\sin^3A + \sin^3(60^(\circ) + A) -\sin^3(60^(\circ)+ A)`

`=\sin^3A`

`\hrulefill`

`\boxed{\textsf{Prove that}\;\;\sin^3A + \sin^3(120^(\circ) + A) + \sin^3(240^(\circ)+ A) = -(3)/(4)\sin 3A}`

Step 1

Use the sine and cos double angle identities to rewrite sin(3x) in terms of sin(x):

`\begin{aligned}\sin(3x)&=\sin(2x+x)\\&=\sin2 (x)\cos (x)+\sin (x)\cos2 (x)\\&=(2\sin (x)\cos (x))\cos (x)+\sin (x)(1-2\sin^2 (x))\\&=2\sin (x)\cos^2 (x)+\sin (x)-2\sin^3 (x)\\&=2\sin (x)(1-\sin^2 (x))+\sin (x)-2\sin^3 (x)\\&=2\sin (x)-2\sin^3 (x)+\sin (x)-2\sin^3 (x)\\&=3\sin (x)-4\sin^3 (x)\end{aligned}`

Rearrange to isolate sin³x:

`\begin{aligned}\sin(3x)&=3\sin (x)-4\sin^3 (x)\\\\4\sin^3 (x)&=3\sin (x)-\sin (3x)\\\\\sin^3 (x)&=(3\sin (x)-\sin (3x))/(4)\end{aligned}`

Step 2

Use this expression to rewrite the terms in sin³A on the left side of the equation:

`\sin^3A + \sin^3(120^(\circ) + A) + \sin^3(240^(\circ)+ A)`

`=(3\sin A-\sin3A)/(4)+ (3\sin (120^(\circ) + A)-\sin (3(120^(\circ) + A)))/(4)+(3\sin (240^(\circ)+ A)-\sin (3(240^(\circ)+ A)))/(4)`

`=(3\sin A-\sin3A+3\sin (120^(\circ) + A)-\sin (360^(\circ) + 3A)+3\sin (240^(\circ)+ A)-\sin (720^(\circ)+ 3A))/(4)`

Step 3

As sin(x ± 360°n) = sin(x), we can simplify:

`\sin(360^(\circ)+3A) = \sin (3A)`

`\sin(720^(\circ)+3A) = \sin (3A)`

Therefore:

`=(3\sin A-\sin3A+3\sin (120^(\circ) + A)-\sin (3A)+3\sin (240^(\circ)+ A)-\sin (3A))/(4)`

`=(3\sin A-3\sin3A+3\sin (120^(\circ) + A)+3\sin (240^(\circ)+ A))/(4)`

Factor out the 3 in the numerator:

`=(3\left(\sin A-\sin3A+\sin (120^(\circ) + A)+\sin (240^(\circ)+ A)\right))/(4)`

Step 4

Rewrite 240° = 180° + 60°:

`\sin(240^(\circ) + A) = \sin(180^(\circ) + 60^(\circ) + A)`

As sin(180° + x) = -sin(x), we can rewrite sin(180° + 60° + A) as:

`- \sin(60^(\circ) + A)`

Therefore:

`=(3\left(\sin A-\sin3A+\sin (120^(\circ) + A)-\sin (60^(\circ)+ A)\right))/(4)`

Step 5

As sin(120° + x) = sin(60° - x) then:

`=(3\left(\sin A-\sin3A+\sin (60^(\circ) -A)-\sin (60^(\circ)+ A)\right))/(4)`

Step 6

As sin(60° - x) - sin(60° + x) = -sin(x), then:

`=(3\left(\sin A-\sin3A-\sin A\right))/(4)`

Step 7

Simplify:

`=(-3\sin3A)/(4)`

`=-(3)/(4)\sin3A`

Hence proving that:

`\sin^3A + \sin^3(120^(\circ) + A) + \sin^3(240^(\circ)+ A) = -(3)/(4)\sin 3A`