Question

please someone help me....​

Answer 1

Answer:  see proof below

Step-by-step explanation:

Use the following Product to Sum Identities:

2 sin A sin B = cos (A - B) - cos (A + B)

2 sin A cos B = sin (A + B) + sin (A - B)

Use the Unit Circle to evaluate: cos 120 = -1/2    &      sin 60 = √3/2

Proof LHS → RHS

LHS:                                            sin 20 · sin 40 · sin 80

Regroup:                              (1/2) sin 20 · 2 sin 40 · sin 80

Product to Sum Identity:     (1/2) sin 20 [cos(80-40) - cos (80+40)]

Simplify:                                (1/2) sin 20 [cos 40 - cos 120]

Unit Circle:                            (1/2) sin 20 [cos 40 + (1/2)]

Distribute:                             (1/2) sin 20 cos 40 + (1/4) sin 20

Product to Sum Identity:       (1/4)[sin(20 + 40) + sin (20 - 40)] + (1/4) sin 20

Simplify:                                 (1/4)[sin 60 + sin (-20)] + (1/4) sin 20

                                           = (1/4)[sin 60 - sin 20] + (1/4) sin 20

Unit Circle:                             (1/4)[(√3/2) - sin 20] + (1/4) sin 20

Distribute:                              (√3/8) -  (1/4) sin 20 + (1/4) sin 20

Simplify:                                  √3/8

LHS = RHS:   √3/8 = √3/8  $\checkmark$

Answer 2

We are given the equation cos(20°)(cos(40°)(cos(60°)(cos(80°) = √3 / 8. Let's once again start by applying the identity 'sin(s)sin(t) = - cos(s + t) + cos(s - t) / 2. In this case if we focus on the expression 'cos(20°)(cos(40°),' s would be = 20°, and t = 40°.

$\mathrm{Use\:the\:following\:identity}:\quad \sin \left(s\right)\sin \left(t\right)=\frac{-\cos \left(s+t\right)+\cos \left(s-t\right)}{2}$

$\sin \left(20^{\circ \:}\right)\sin \left(40^{\circ \:}\right)=\frac{-\cos \left(20^{\circ \:}+40^{\circ \:}\right)+\cos \left(20^{\circ \:}-40^{\circ \:}\right)}{2}$

$\mathrm{Substitute}:\frac{-\cos \left(20^{\circ \:}+40^{\circ \:}\right)+\cos \left(20^{\circ \:}-40^{\circ \:}\right)}{2}\sin \left(80^{\circ \:}\right)$

$\mathrm{Multiply\:fractions}:\frac{\sin \left(80^{\circ \:}\right)\left(-\cos \left(60^{\circ \:}\right)+\cos \left(-20^{\circ \:}\right)\right)}{2}$

Remember that cos(- x) = cos(x). Respectively cos(- 20°) = cos(20°). Let's substitute and afterwards apply the identity 'cos(60°) = 1 / 2.'

$\frac{\sin \left(80^{\circ \:}\right)\left(-\cos \left(60^{\circ \:}\right)+\cos \left(20^{\circ \:}\right)\right)}{2} = \frac{\sin \left(80^{\circ \:}\right)\left(-\frac{1}{2}+\cos \left(20^{\circ \:}\right)\right)}{2}$

And if we further simplify the expression, we should receive the following...

$\frac{\sin \left(80^{\circ \:}\right)\left(-1+2\cos \left(20^{\circ \:}\right)\right)}{4}$

Now we want to prove that this expression = √3 / 8. The denominator here is 4 so we can multiply the whole thing by 2 to have a denominator of 8. 2((sin(80°)(- 1 + 2cos(20°)) when simplified = √3. Therefore the expression is true.