Question

the value of sin^3 10° + sin^3 50° - sin^3 70° is equal to

Answer 1

The answer is -3/8.

We use identity (sin(x))^3 = (3sin(x) -sin(3x))/4.

All angles below are in degrees.

Then the expression becomes

0.25(3sin10 - sin30 +3sin50 - sin150 -3sin70 +sin210) =

0.25(3sin10 -0.5 + 3sin50 -0.5 -3sin 70 -0.5) =

0.25(3sin10 + 3sin50 -3sin 70 -1.5) = 0.75(sin10 + sin50 -sin 70 -0.5) .

We now apply another identity sinx+siny = 2 sin((x+y)/2)cos((x-y)/2) to

sum (sin10 + sin50).

sin10 + sin50 = 2sin30*cos20 = cos20 = sin70

The expression then becomes

0.75(sin70 -sin 70 -0.5) = -0.75*0.5 = -0/375 = -3/8.



I hope this is helpful to you :)

Answer 2

Answer:

sin^3 10° + sin^3 50° - sin^3 70°= -3/8

Step-by-step explanation:

Given, $sin^{3}10^{\circ}+sin^{3}50^{\circ} -sin^{3}70^{\circ}$

Find the value of the expression.

We know identity, $a^3+ b^3 + c^3 - 3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$ $sin^{3}10^{\circ}+sin^{3}50^{\circ} -sin^{3}70^{\circ}$

Here, $a=sin10^{\circ} , b=sin50^{\circ}, c=-sin70^{\circ}$

$a+b+c=sin10^{\circ}+sin50^{\circ} -sin70^{\circ}$

              =0

$= > a^3+ b^3 + c^3 - 3abc=0$

So, $sin^{3}10^{\circ}+sin^{3}50^{\circ} sin^{3}70^{\circ}=-3sin10^{\circ}sin50^{\circ}sin70^{\circ}$

                                              $=-\frac{3}{4}sin30^{\circ}\\\\=-\frac{3}{8}$

#SPJ2