Question

please someone help me to prove this...
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Answer 1

Step-by-step explanation:

cos 20° cos 40° cos 60° cos 80°

Simplify cos 60° to ½.

½ cos 20° cos 40° cos 80°

We can use double angle formula, sin(2θ) = 2 sin θ cos θ.  Multiply and divide by 2 sin 20°.

½ (2 sin 20° cos 20° cos 40° cos 80°) / (2 sin 20°)

½ (sin 40° cos 40° cos 80°) / (2 sin 20°)

Multiply and divide by 2 and use double angle formula again.

½ (2 sin 40° cos 40° cos 80°) / (4 sin 20°)

½ (sin 80° cos 80°) / (4 sin 20°)

Multiply and divide by 2 and use double angle formula again.

½ (2 sin 80° cos 80°) / (8 sin 20°)

½ (sin 160°) / (8 sin 20°)

Use phase shift identity sin θ = sin(180−θ).

½ (sin 20°) / (8 sin 20°)

1/16

Answer 2

In this case we have the equation cos(20°)(cos(40°)(cos(60°)(cos(80°) = 1 / 16, and are asked to prove that this equation is true. Let's start by using the property 'cos(s)cos(t) = cos(s + t) + cos(s - t) / 2.' Taking the bit 'cos(20°)(cos(40°)' we know that s should = 20°, and t should = 40°.

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

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

$\mathrm{Substituting\:the\:value\:back}:\frac{\cos \left(20^{\circ \:}+40^{\circ \:}\right)+\cos \left(20^{\circ \:}-40^{\circ \:}\right)}{2}\cos \left(60^{\circ \:}\right)\cos \left(80^{\circ \:}\right)$

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

This expression is indeed not simplified, but remember that we can use the identities 'cos(- 20°) = cos(20°)' and 'cos(60°) = 1 / 2.' Let's substitute one step and a time.

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

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

$\frac{\frac{1}{2}\cos \left(80^{\circ \:}\right)\left(\frac{1}{2}+\cos \left(20^{\circ \:}\right)\right)}{2} = \frac{\cos \left(80^{\circ \:}\right)\left(1+2\cos \left(20^{\circ \:}\right)\right)}{8}$

Now that we have this 'simplified expression,' if we take the numerator as a whole in decimal form, it will = 0.5. Respectively 0.5 / 8 = 1 / 2 / 8 = 1 / 16. Hence our equation is true.

$\cos \left(80^{\circ \:\:}\right)\left(1+2\cos \left(20^{\circ \:\:}\right)\right) = 0.5,\\0.5 / 8 = (1 / 2) / 8 = 1 / 16$