Question

Find the exact value of 6cos(105°)​

Answer 1

`-(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).