Question

sin^3x + sin^3(120°+x) + sin^3(240°+x)​

Answer 1

Answer:

f(x)=sin

3

x+sin

3

(120+x)+sin

3

(240+x)

=

4

3sinx−sin3x+3sin(120+x)−sin(2π+3x)+3sin(240+x)−sin(4π+x)

[∵sin3x=3sinx−4sin

3

x]

=

4

3[sinx+sin(120+x)+sin(240+x)]−3sin3x

=

4

3

[[sinx+2sin(180+x)cos60

o

]−sin3x]

=

4

3

[[sinx−sinx]−sin3x]

=

4

3

[−sin3x]

f(x)=−

4

3

sin3x

∴ range of f(x)ϵ[−

4

3

,

4

3

]

So, P=3