Precalculus Examples

$\sin (\frac{17 π}{12}) \cos (\frac{π}{4})$

Step 1

The exact value of $\sin (\frac{17 π}{12})$ is $- \frac{\sqrt{2 + \sqrt{3}}}{2}$ .

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Step 1.1

Rewrite $\frac{17 π}{12}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$\sin (\frac{\frac{17 π}{6}}{2}) \cos (\frac{π}{4})$

Step 1.2

Apply the sine half-angle identity.

$(\pm \sqrt{\frac{1 - \cos (\frac{17 π}{6})}{2}}) \cos (\frac{π}{4})$

Step 1.3

Change the $\pm$ to $-$ because sine is negative in the third quadrant.

$- \sqrt{\frac{1 - \cos (\frac{17 π}{6})}{2}} \cos (\frac{π}{4})$

Step 1.4

Simplify $- \sqrt{\frac{1 - \cos (\frac{17 π}{6})}{2}}$ .

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Step 1.4.1

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$- \sqrt{\frac{1 - \cos (\frac{5 π}{6})}{2}} \cos (\frac{π}{4})$

Step 1.4.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$- \sqrt{\frac{1 - - \cos (\frac{π}{6})}{2}} \cos (\frac{π}{4})$

Step 1.4.3

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$- \sqrt{\frac{1 - - \frac{\sqrt{3}}{2}}{2}} \cos (\frac{π}{4})$

Step 1.4.4

Multiply $- - \frac{\sqrt{3}}{2}$ .

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Step 1.4.4.1

Multiply $- 1$ by $- 1$ .

$- \sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{2}} \cos (\frac{π}{4})$

Step 1.4.4.2

Multiply $\frac{\sqrt{3}}{2}$ by $1$ .

$- \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} \cos (\frac{π}{4})$

Step 1.4.5

Write $1$ as a fraction with a common denominator.

$- \sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{2}} \cos (\frac{π}{4})$

Step 1.4.6

Combine the numerators over the common denominator.

$- \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{2}} \cos (\frac{π}{4})$

Step 1.4.7

Multiply the numerator by the reciprocal of the denominator.

$- \sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{1}{2}} \cos (\frac{π}{4})$

Step 1.4.8

Multiply $\frac{2 + \sqrt{3}}{2} \cdot \frac{1}{2}$ .

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Step 1.4.8.1

Multiply $\frac{2 + \sqrt{3}}{2}$ by $\frac{1}{2}$ .

$- \sqrt{\frac{2 + \sqrt{3}}{2 \cdot 2}} \cos (\frac{π}{4})$

Step 1.4.8.2

Multiply $2$ by $2$ .

$- \sqrt{\frac{2 + \sqrt{3}}{4}} \cos (\frac{π}{4})$

Step 1.4.9

Rewrite $\sqrt{\frac{2 + \sqrt{3}}{4}}$ as $\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}}$ .

$- \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}} \cos (\frac{π}{4})$

Step 1.4.10

Simplify the denominator.

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Step 1.4.10.1

Rewrite $4$ as $2^{2}$ .

$- \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2^{2}}} \cos (\frac{π}{4})$

Step 1.4.10.2

Pull terms out from under the radical, assuming positive real numbers.

$- \frac{\sqrt{2 + \sqrt{3}}}{2} \cos (\frac{π}{4})$

Step 2

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$- \frac{\sqrt{2 + \sqrt{3}}}{2} \cdot \frac{\sqrt{2}}{2}$

Step 3

Multiply $- \frac{\sqrt{2 + \sqrt{3}}}{2} \cdot \frac{\sqrt{2}}{2}$ .

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Step 3.1

Multiply $\frac{\sqrt{2}}{2}$ by $\frac{\sqrt{2 + \sqrt{3}}}{2}$ .

$- \frac{\sqrt{2} \sqrt{2 + \sqrt{3}}}{2 \cdot 2}$

Step 3.2

Combine using the product rule for radicals.

$- \frac{\sqrt{2 (2 + \sqrt{3})}}{2 \cdot 2}$

Step 3.3

Multiply $2$ by $2$ .

$- \frac{\sqrt{2 (2 + \sqrt{3})}}{4}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$- \frac{\sqrt{2 (2 + \sqrt{3})}}{4}$

Decimal Form:

$- 0.68301270 \dots$