Find the Exact Value sin((7pi)/12)cos(pi/4)-cos((7pi)/12)sin(pi/4)

$\sin (\frac{7 π}{12}) \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

Step 1

Simplify each term.

The exact value of $\sin (\frac{7 π}{12})$ is $\frac{\sqrt{2 + \sqrt{3}}}{2}$ .

Rewrite $\frac{7 π}{12}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$\sin (\frac{\frac{7 π}{6}}{2}) \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

Apply the sine half-angle identity.

$(\pm \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{2}}) \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

Change the $\pm$ to $+$ because sine is positive in the second quadrant.

$\sqrt{\frac{1 - \cos (\frac{7 π}{6})}{2}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

Simplify $\sqrt{\frac{1 - \cos (\frac{7 π}{6})}{2}}$ .

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 third quadrant.

$\sqrt{\frac{1 - - \cos (\frac{π}{6})}{2}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

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

$\sqrt{\frac{1 - - \frac{\sqrt{3}}{2}}{2}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

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

Multiply $- 1$ by $- 1$ .

$\sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{2}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

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

$\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

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

$\sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{2}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

Combine the numerators over the common denominator.

$\sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{2}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

Multiply the numerator by the reciprocal of the denominator.

$\sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{1}{2}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

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

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

$\sqrt{\frac{2 + \sqrt{3}}{2 \cdot 2}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

Multiply $2$ by $2$ .

$\sqrt{\frac{2 + \sqrt{3}}{4}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

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

$\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

Simplify the denominator.

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

$\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2^{2}}} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

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

$\frac{\sqrt{2 + \sqrt{3}}}{2} \cos (\frac{π}{4}) - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

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

$\frac{\sqrt{2 + \sqrt{3}}}{2} \cdot \frac{\sqrt{2}}{2} - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

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

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

$\frac{\sqrt{2 + \sqrt{3}} \sqrt{2}}{2 \cdot 2} - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

Combine using the product rule for radicals.

$\frac{\sqrt{(2 + \sqrt{3}) \cdot 2}}{2 \cdot 2} - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

Multiply $2$ by $2$ .

$\frac{\sqrt{(2 + \sqrt{3}) \cdot 2}}{4} - \cos (\frac{7 π}{12}) \sin (\frac{π}{4})$

The exact value of $\cos (\frac{7 π}{12})$ is $- \frac{\sqrt{2 - \sqrt{3}}}{2}$ .

Rewrite $\frac{7 π}{12}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$\frac{\sqrt{(2 + \sqrt{3}) \cdot 2}}{4} - \cos (\frac{\frac{7 π}{6}}{2}) \sin (\frac{π}{4})$

Apply the cosine half-angle identity $\cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + \cos (x)}{2}}$ .

$\frac{\sqrt{(2 + \sqrt{3}) \cdot 2}}{4} - (\pm \sqrt{\frac{1 + \cos (\frac{7 π}{6})}{2}}) \sin (\frac{π}{4})$

Change the $\pm$ to $-$ because cosine is negative in the second quadrant.

$\frac{\sqrt{(2 + \sqrt{3}) \cdot 2}}{4} - - \sqrt{\frac{1 + \cos (\frac{7 π}{6})}{2}} \sin (\frac{π}{4})$

Simplify $- \sqrt{\frac{1 + \cos (\frac{7 π}{6})}{2}}$ .

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 third quadrant.

$\frac{\sqrt{(2 + \sqrt{3}) \cdot 2}}{4} - - \sqrt{\frac{1 - \cos (\frac{π}{6})}{2}} \sin (\frac{π}{4})$

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

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

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

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

Combine the numerators over the common denominator.

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

Multiply the numerator by the reciprocal of the denominator.

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

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

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

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

Multiply $2$ by $2$ .

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

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

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

Simplify the denominator.

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

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

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

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

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

Multiply $- 1$ by $- 1$ .

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

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

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

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

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

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

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

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

Combine using the product rule for radicals.

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

Multiply $2$ by $2$ .

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

Step 2

Combine the numerators over the common denominator.

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

Step 3

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$0.86602540 \dots$

Find the Exact Value sin((7pi)/12)cos(pi/4)-cos((7pi)/12)sin(pi/4)

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