Solved on Nov 02, 2023

Find the value of $\cos(-\pi/12)$ .

STEP 1

Assumptions1. We know the value of $\cos \left(\frac{\pi}{3}\right)$ , $\cos \left(\frac{\pi}{4}\right)$ and $\cos \left(\frac{\pi}{6}\right)$ . . We will use the formula for $\cos(a+b)$ which is $\cos(a+b) = \cos a \cos b - \sin a \sin b$ .
3. We are aware that $\cos(-x) = \cos(x)$ .

STEP 2

We can express $\frac{\pi}{12}$ as $\frac{\pi}{} - \frac{\pi}{4}$ .

STEP 3

Now, we can use the formula for $\cos(a-b)$ which is $\cos(a-b) = \cos a \cos b + \sin a \sin b$ .
$\cos \left(\frac{\pi}{3} - \frac{\pi}{}\right) = \cos \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{}\right) + \sin \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{}\right)$

STEP 4

We know that $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$ , $\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ , $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ , and $\sin \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ .
Substitute these values into the equation $\cos \left(\frac{\pi}{12}\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}$

STEP 5

implify the equation $\cos \left(\frac{\pi}{12}\right) = \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}$

STEP 6

We can simplify further by multiplying the numerator and denominator by $\sqrt{2}$ $\cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}$

STEP 7

We can factor out $\frac{1}{4}$ $\cos \left(\frac{\pi}{12}\right) = \frac{1}{4}(\sqrt{2} + \sqrt{6})$

STEP 8

Since $\cos(-x) = \cos(x)$ , we can say that $\cos \left(-\frac{\pi}{12}\right) = \cos \left(\frac{\pi}{12}\right)$

STEP 9

So, $\cos \left(-\frac{\pi}{12}\right) = \frac{}{4}(\sqrt{2} + \sqrt{6})$ So, $\cos \left(-\frac{\pi}{12}\right) = \frac{}{4}(\sqrt{2} + \sqrt{6})$ .

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Find the value of $\cos(-\pi/12)$ .

STEP 1

Assumptions1. We know the value of $\cos \left(\frac{\pi}{3}\right)$ , $\cos \left(\frac{\pi}{4}\right)$ and $\cos \left(\frac{\pi}{6}\right)$ . . We will use the formula for $\cos(a+b)$ which is $\cos(a+b) = \cos a \cos b - \sin a \sin b$ .
3. We are aware that $\cos(-x) = \cos(x)$ .

STEP 2

We can express $\frac{\pi}{12}$ as $\frac{\pi}{} - \frac{\pi}{4}$ .

STEP 3

Now, we can use the formula for $\cos(a-b)$ which is $\cos(a-b) = \cos a \cos b + \sin a \sin b$ .
$\cos \left(\frac{\pi}{3} - \frac{\pi}{}\right) = \cos \left(\frac{\pi}{3}\right) \cos \left(\frac{\pi}{}\right) + \sin \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{}\right)$

STEP 4

STEP 5

implify the equation $\cos \left(\frac{\pi}{12}\right) = \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}$

STEP 6

We can simplify further by multiplying the numerator and denominator by $\sqrt{2}$ $\cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}$

STEP 7

We can factor out $\frac{1}{4}$ $\cos \left(\frac{\pi}{12}\right) = \frac{1}{4}(\sqrt{2} + \sqrt{6})$

STEP 8

Since $\cos(-x) = \cos(x)$ , we can say that $\cos \left(-\frac{\pi}{12}\right) = \cos \left(\frac{\pi}{12}\right)$

STEP 9

So, $\cos \left(-\frac{\pi}{12}\right) = \frac{}{4}(\sqrt{2} + \sqrt{6})$ So, $\cos \left(-\frac{\pi}{12}\right) = \frac{}{4}(\sqrt{2} + \sqrt{6})$ .

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Find the value of cos⁡(−π/12)\cos(-\pi/12)cos(−π/12).

STEP 1

STEP 2

We can express π12\frac{\pi}{12}12π​ as π−π4\frac{\pi}{} - \frac{\pi}{4}π​−4π​.

STEP 3

STEP 4

STEP 5

implify the equationcos⁡(π12)=122+322\cos \left(\frac{\pi}{12}\right) = \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}cos(12π​)=22​1​+22​3​​

STEP 6

We can simplify further by multiplying the numerator and denominator by 2\sqrt{2}2​cos⁡(π12)=24+64\cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}cos(12π​)=42​​+46​​

STEP 7

We can factor out 14\frac{1}{4}41​cos⁡(π12)=14(2+6)\cos \left(\frac{\pi}{12}\right) = \frac{1}{4}(\sqrt{2} + \sqrt{6})cos(12π​)=41​(2​+6​)

STEP 8

Since cos⁡(−x)=cos⁡(x)\cos(-x) = \cos(x)cos(−x)=cos(x), we can say thatcos⁡(−π12)=cos⁡(π12)\cos \left(-\frac{\pi}{12}\right) = \cos \left(\frac{\pi}{12}\right)cos(−12π​)=cos(12π​)

STEP 9

So,cos⁡(−π12)=4(2+6)\cos \left(-\frac{\pi}{12}\right) = \frac{}{4}(\sqrt{2} + \sqrt{6})cos(−12π​)=4​(2​+6​)So, cos⁡(−π12)=4(2+6)\cos \left(-\frac{\pi}{12}\right) = \frac{}{4}(\sqrt{2} + \sqrt{6})cos(−12π​)=4​(2​+6​).

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Find the value of $\cos(-\pi/12)$ .

We can express $\frac{\pi}{12}$ as $\frac{\pi}{} - \frac{\pi}{4}$ .

implify the equation $\cos \left(\frac{\pi}{12}\right) = \frac{1}{2\sqrt{2}} + \frac{\sqrt{3}}{2\sqrt{2}}$

We can simplify further by multiplying the numerator and denominator by $\sqrt{2}$ $\cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}$

We can factor out $\frac{1}{4}$ $\cos \left(\frac{\pi}{12}\right) = \frac{1}{4}(\sqrt{2} + \sqrt{6})$

Since $\cos(-x) = \cos(x)$ , we can say that $\cos \left(-\frac{\pi}{12}\right) = \cos \left(\frac{\pi}{12}\right)$

So, $\cos \left(-\frac{\pi}{12}\right) = \frac{}{4}(\sqrt{2} + \sqrt{6})$ So, $\cos \left(-\frac{\pi}{12}\right) = \frac{}{4}(\sqrt{2} + \sqrt{6})$ .