Math Problem Statement

What is the exact value of sin(-11π/12)?

Solution

To solve sin(11π12)\sin \left(-\frac{11\pi}{12} \right), we will use trigonometric identities and simplifications. Here's the step-by-step process:


Step 1: Simplify the angle

We know: sin(x)=sin(x)\sin(-x) = -\sin(x) So, sin(11π12)=sin(11π12)\sin \left(-\frac{11\pi}{12} \right) = -\sin \left(\frac{11\pi}{12} \right)


Step 2: Rewrite 11π12\frac{11\pi}{12}

Express 11π12\frac{11\pi}{12} as: 11π12=ππ12\frac{11\pi}{12} = \pi - \frac{\pi}{12} This means: sin(11π12)=sin(ππ12)\sin \left( \frac{11\pi}{12} \right) = \sin \left( \pi - \frac{\pi}{12} \right) Using the identity sin(πx)=sin(x)\sin(\pi - x) = \sin(x), we have: sin(11π12)=sin(π12)\sin \left( \frac{11\pi}{12} \right) = \sin \left( \frac{\pi}{12} \right) Thus: sin(11π12)=sin(π12)\sin \left(-\frac{11\pi}{12} \right) = -\sin \left( \frac{\pi}{12} \right)


Step 3: Exact value of sin(π12)\sin \left( \frac{\pi}{12} \right)

The angle π12\frac{\pi}{12} can be expressed as: π12=π3π4\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4} Now, use the sine subtraction formula: sin(ab)=sin(a)cos(b)cos(a)sin(b)\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b) Substituting a=π3a = \frac{\pi}{3} and b=π4b = \frac{\pi}{4}, we get: sin(π12)=sin(π3)cos(π4)cos(π3)sin(π4)\sin \left( \frac{\pi}{12} \right) = \sin \left( \frac{\pi}{3} \right) \cos \left( \frac{\pi}{4} \right) - \cos \left( \frac{\pi}{3} \right) \sin \left( \frac{\pi}{4} \right)


Step 4: Substitute exact values

From known trigonometric values:

\sin \left( \frac{\pi}{4} \right) = \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}$$ Substitute these: $$\sin \left( \frac{\pi}{12} \right) = \left( \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \right) - \left( \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \right)$$ Simplify: $$\sin \left( \frac{\pi}{12} \right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$$ $$\sin \left( \frac{\pi}{12} \right) = \frac{\sqrt{6} - \sqrt{2}}{4}$$ --- ### Step 5: Final result $$\sin \left( -\frac{11\pi}{12} \right) = -\sin \left( \frac{\pi}{12} \right) = -\frac{\sqrt{6} - \sqrt{2}}{4}$$ Thus, the correct answer is **(B) $$-\frac{\sqrt{6} - \sqrt{2}}{4}$$**. --- ### Do you want me to expand on any step or provide further details? Let me know! --- ### Related Questions 1. What is the exact value of $$\cos\left(-\frac{11\pi}{12}\right)$$? 2. How does the identity $$\sin(-x) = -\sin(x)$$ affect calculations involving negative angles? 3. How can you use the subtraction formula for other trigonometric functions like cosine and tangent? 4. Why does $$\sin(\pi - x) = \sin(x)$$ hold for all $$x$$? 5. Can you explain the relationship between radians and degrees in trigonometric calculations? --- ### Tip Always check the quadrant of your angle after simplifications to determine the correct sign of the trigonometric value!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sine Function
Angle Simplification
Trigonometric Identities

Formulas

sin(-x) = -sin(x)
sin(π - x) = sin(x)
sin(a - b) = sin(a)cos(b) - cos(a)sin(b)

Theorems

Trigonometric Identity

Suitable Grade Level

Grades 10-12