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Find the exact value of 6cos(105°)​

User Suther
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1 Answer

16 votes
16 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 Kbosak
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