193k views
3 votes
Find the exact value of 6cos(105°)​

User Eulerfx
by
8.1k points

1 Answer

2 votes

Answer:


-(3(√(6)-√(2)))/(2)\text{ or } (-3√(6)+3√(2))/(2)}\text{ or }(3(√(2)-√(6)))/(2)

Explanation:

There are multiple ways to achieve and even express the exact answer to this problem. Because the exact value of
6\cos(105^(\circ)}) is a non-terminating (never-ending) decimal, it does not have a finite number of digits. Therefore, you cannot express it as an exact value as a decimal, as you'd either have to round or truncate.

Solution 1 (Cosine Addition Identity):

Nonetheless, to find the exact value we must use trigonometry identities.

Identity used:


\cos(\alpha +\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta

Notice that
45+60=105 and therefore we can easily solve this problem if we know values of
\cos(45^(\circ)),
\cos(60^(\circ)),
\sin (45^(\circ)), and
\sin(60^(\circ)), which is plausible as they are all key angles on the unit circle.

Recall from either memory or the unit circle that:


  • \cos(45^(\circ))=\sin(45^(\circ))=(√(2))/(2)

  • \cos(60^(\circ))=(1)/(2)

  • \sin(60^(\circ))=(√(3))/(2)

Therefore, we have:


\cos(105^(\circ))=\cos(45^(\circ)+60^(\circ)}),\\\cos(45^(\circ)+60^(\circ)})=\cos 45^(\circ)\cos 60^(\circ)-\sin 45^(\circ)\sin 60^(\circ),\\\cos(45^(\circ)+60^(\circ)})=(√(2))/(2)\cdot (1)/(2)-(√(2))/(2)\cdot (√(3))/(2),\\\cos(105^(\circ))=(√(2))/(4)-(√(6))/(4),\\\cos(105^(\circ))={(-√(6)+√(2))/(4)}

Since we want the value of
6\cos 105^(\circ), simply multiply this by 6 to get your final answer:


6\cdot {(-√(6)+√(2))/(4)}=(-3√(6)+3√(2))/(2)}=\boxed{(3(√(2)-√(6)))/(2)}

Solution 2 (Combination of trig. identities):

Although less plausible, you may have the following memorized:


\sin 15^(\circ)=\cos75^(\circ)=(√(6)-√(2))/(4),\\\sin 75^(\circ)=\cos15^(\circ)=(√(6)+√(2))/(4)

If so, we can use the following trig. identity:


\cos(\theta)=\sin(90^(\circ)-\theta) (the cosine of angle theta is equal to the sine of the supplement of angle theta - the converse is also true)

Therefore,


\cos (105^(\circ))=\sin (90^(\circ)-105^(\circ))=\sin(-15^(\circ))

Recall another trig. identity:


\sin(-\theta)=-\sin (\theta) and therefore:


\sin (-15^(\circ))=-\sin (15^(\circ))

Multiply by 6 to get:


6\cos (105^(\circ))=-6\sin (15^(\circ))=-6\cdot (√(6)-√(2))/(4)=\boxed{-(3(√(6)-√(2)))/(2)} (alternative final answer).

User Vikash Rathee
by
7.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories