Question

Please someone help me to prove this. ​

Answer 1

Step-by-step explanation:

8 cos 80° cos 140° cos 160°

4 (2 cos 80° cos 140°) cos 160°

4 cos 160° (2 cos 80° cos 140°)

Use product to sum formula.

4 cos 160° (cos 60° + cos 220°)

4 cos 160° (½ + cos 220°)

2 cos 160° + 4 cos 220° cos 160°

2 cos 160° + 2 (2 cos 220° cos 160°)

Use product to sum formula again.

2 cos 160° + 2 (cos 60° + cos 380°)

2 cos 160° + 2 (½ + cos 380°)

2 cos 160° + 1 + 2 cos 380°

2 cos 160° + 1 + 2 cos 20°

Use shift identity cos(180 − θ) = -cos θ.

-2 cos 20° + 1 + 2 cos 20°

1

Answer 2

Our question that we have at hand here is 8(cos 80°)(cos 140°)(cos 160°). We need to convert that into 'decimal form' to prove that it equals 1. However decimal form doesn't generally give us a decimal, in this case.

Let's start by using the identity cos(s)cos(t) = cos(s + t) + cos(s - t) / 2. In this case we are looking particularly at the expression (cos 80°)(cos 140°), where s = 80, and t = 140.

$\cos \left(80^{\circ \:}\right)\cos \left(140^{\circ \:}\right)=\frac{\cos \left(80^{\circ \:}+140^{\circ \:}\right)+\cos \left(80^{\circ \:}-140^{\circ \:}\right)}{2}$

$=8\cdot \frac{\cos \left(80^{\circ \:}+140^{\circ \:}\right)+\cos \left(80^{\circ \:}-140^{\circ \:}\right)}{2}\cos \left(160^{\circ \:}\right)$

$= 4\cos \left(160^{\circ \:}\right)\left(\cos \left(220^{\circ \:}\right)+\cos \left(-60^{\circ \:}\right)\right)$

To further simplify this expression, we know that cos(- 60°) = cos(60°). This may not seem like a very effective step, but we can apply a few trivial identities here. Remember that cos(60°) = 1 / 2,

$4\cos \left(160^{\circ \:}\right)\left(\cos \left(220^{\circ \:}\right)+\frac{1}{2}\right)$

$= 4(0.25)\\= 1$

And hence we proved that 8(cos 80°)(cos 140°)(cos 160°) in decimal form, = 1.