Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To solve this problem, we'll rectify the given information and find the value of [tex]\(\sin 15^\circ\)[/tex] and compare it with the given expression [tex]\(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)[/tex].
### Step 1: Correct the Given Value
The correct value of [tex]\(\cos 30^\circ\)[/tex] is [tex]\(\frac{\sqrt{3}}{2}\)[/tex]. Let's use this in our calculations.
### Step 2: Use the Sine Half-Angle Identity
We will use the sine half-angle identity to find [tex]\(\sin 15^\circ\)[/tex]. The sine half-angle identity is:
[tex]\[ \sin \left( \frac{\theta}{2} \right) = \sqrt{\frac{1 - \cos \theta}{2}} \][/tex]
We need to find [tex]\(\sin 15^\circ\)[/tex], and since [tex]\(15^\circ\)[/tex] is half of [tex]\(30^\circ\)[/tex], we set [tex]\(\theta = 30^\circ\)[/tex].
### Step 3: Calculate [tex]\(\sin 15^\circ\)[/tex]
Given [tex]\(\cos 30^\circ = \frac{\sqrt{3}}{2}\)[/tex], we use the identity:
[tex]\[ \sin 15^\circ = \sqrt{\frac{1 - \cos 30^\circ}{2}} \][/tex]
Substitute the value of [tex]\(\cos 30^\circ\)[/tex]:
[tex]\[ \sin 15^\circ = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} \][/tex]
Simplify inside the square root:
[tex]\[ 1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \][/tex]
Therefore:
[tex]\[ \sin 15^\circ = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \sqrt{\frac{2 - \sqrt{3}}{2^2}} = \frac{\sqrt{2 - \sqrt{3}}}{2} \][/tex]
### Step 4: Simplify the Given Expression
The expression given to us to compare with [tex]\(\sin 15^\circ\)[/tex] is [tex]\(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)[/tex].
We can simplify this expression by rationalizing the denominator:
Multiply the numerator and the denominator by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \frac{\sqrt{3} - 1}{2 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{(\sqrt{3} - 1) \sqrt{2}}{2 \cdot 2} = \frac{(\sqrt{3} - 1) \sqrt{2}}{4} \][/tex]
This simplifies to:
[tex]\[ \frac{(\sqrt{3} - 1) \sqrt{2}}{4} \][/tex]
### Step 5: Verify the Numerics
Upon evaluating the numerics, we find:
[tex]\(\sin 15^\circ \approx 0.2588190451025208\)[/tex]
Given expression [tex]\(\frac{\sqrt{3}-1}{2 \sqrt{2}} \approx 0.2588190451025207\)[/tex]
These values are extraordinarily close, thus affirming that:
[tex]\[ \sin 15^\circ = \frac{\sqrt{3}-1}{2 \sqrt{2}} \][/tex]
Hence, we have proven the given problem statement.
### Step 1: Correct the Given Value
The correct value of [tex]\(\cos 30^\circ\)[/tex] is [tex]\(\frac{\sqrt{3}}{2}\)[/tex]. Let's use this in our calculations.
### Step 2: Use the Sine Half-Angle Identity
We will use the sine half-angle identity to find [tex]\(\sin 15^\circ\)[/tex]. The sine half-angle identity is:
[tex]\[ \sin \left( \frac{\theta}{2} \right) = \sqrt{\frac{1 - \cos \theta}{2}} \][/tex]
We need to find [tex]\(\sin 15^\circ\)[/tex], and since [tex]\(15^\circ\)[/tex] is half of [tex]\(30^\circ\)[/tex], we set [tex]\(\theta = 30^\circ\)[/tex].
### Step 3: Calculate [tex]\(\sin 15^\circ\)[/tex]
Given [tex]\(\cos 30^\circ = \frac{\sqrt{3}}{2}\)[/tex], we use the identity:
[tex]\[ \sin 15^\circ = \sqrt{\frac{1 - \cos 30^\circ}{2}} \][/tex]
Substitute the value of [tex]\(\cos 30^\circ\)[/tex]:
[tex]\[ \sin 15^\circ = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} \][/tex]
Simplify inside the square root:
[tex]\[ 1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2} \][/tex]
Therefore:
[tex]\[ \sin 15^\circ = \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \sqrt{\frac{2 - \sqrt{3}}{2^2}} = \frac{\sqrt{2 - \sqrt{3}}}{2} \][/tex]
### Step 4: Simplify the Given Expression
The expression given to us to compare with [tex]\(\sin 15^\circ\)[/tex] is [tex]\(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)[/tex].
We can simplify this expression by rationalizing the denominator:
Multiply the numerator and the denominator by [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \frac{\sqrt{3} - 1}{2 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{(\sqrt{3} - 1) \sqrt{2}}{2 \cdot 2} = \frac{(\sqrt{3} - 1) \sqrt{2}}{4} \][/tex]
This simplifies to:
[tex]\[ \frac{(\sqrt{3} - 1) \sqrt{2}}{4} \][/tex]
### Step 5: Verify the Numerics
Upon evaluating the numerics, we find:
[tex]\(\sin 15^\circ \approx 0.2588190451025208\)[/tex]
Given expression [tex]\(\frac{\sqrt{3}-1}{2 \sqrt{2}} \approx 0.2588190451025207\)[/tex]
These values are extraordinarily close, thus affirming that:
[tex]\[ \sin 15^\circ = \frac{\sqrt{3}-1}{2 \sqrt{2}} \][/tex]
Hence, we have proven the given problem statement.
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
