Answer:
[tex]-\frac{3(\sqrt{6}-\sqrt{2})}{2}\text{ or } \frac{-3\sqrt{6}+3\sqrt{2}}{2}}\text{ or }\frac{3(\sqrt{2}-\sqrt{6})}{2}[/tex]
Step-by-step explanation:
There are multiple ways to achieve and even express the exact answer to this problem. Because the exact value of [tex]6\cos(105^{\circ}})[/tex] 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:
[tex]\cos(\alpha +\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta[/tex]
Notice that [tex]45+60=105[/tex] and therefore we can easily solve this problem if we know values of [tex]\cos(45^{\circ})[/tex], [tex]\cos(60^{\circ})[/tex], [tex]\sin (45^{\circ})[/tex], and [tex]\sin(60^{\circ})[/tex], which is plausible as they are all key angles on the unit circle.
Recall from either memory or the unit circle that:
- [tex]\cos(45^{\circ})=\sin(45^{\circ})=\frac{\sqrt{2}}{2}[/tex]
- [tex]\cos(60^{\circ})=\frac{1}{2}[/tex]
- [tex]\sin(60^{\circ})=\frac{\sqrt{3}}{2}[/tex]
Therefore, we have:
[tex]\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}})=\frac{\sqrt{2}}{2}\cdot \frac{1}{2}-\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2},\\\cos(105^{\circ})=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4},\\\cos(105^{\circ})={\frac{-\sqrt{6}+\sqrt{2}}{4}}[/tex]
Since we want the value of [tex]6\cos 105^{\circ}[/tex], simply multiply this by 6 to get your final answer:
[tex]6\cdot {\frac{-\sqrt{6}+\sqrt{2}}{4}}=\frac{-3\sqrt{6}+3\sqrt{2}}{2}}=\boxed{\frac{3(\sqrt{2}-\sqrt{6})}{2}}[/tex]
Solution 2 (Combination of trig. identities):
Although less plausible, you may have the following memorized:
[tex]\sin 15^{\circ}=\cos75^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4},\\\sin 75^{\circ}=\cos15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]
If so, we can use the following trig. identity:
[tex]\cos(\theta)=\sin(90^{\circ}-\theta)[/tex] (the cosine of angle theta is equal to the sine of the supplement of angle theta - the converse is also true)
Therefore,
[tex]\cos (105^{\circ})=\sin (90^{\circ}-105^{\circ})=\sin(-15^{\circ})[/tex]
Recall another trig. identity:
[tex]\sin(-\theta)=-\sin (\theta)[/tex] and therefore:
[tex]\sin (-15^{\circ})=-\sin (15^{\circ})[/tex]
Multiply by 6 to get:
[tex]6\cos (105^{\circ})=-6\sin (15^{\circ})=-6\cdot \frac{\sqrt{6}-\sqrt{2}}{4}=\boxed{-\frac{3(\sqrt{6}-\sqrt{2})}{2}}[/tex] (alternative final answer).
