If \(\sqrt{1+\frac{\sqrt{3}}{2}}- \sqrt{1-\frac{\sqrt{3}}{2}}= c\) , then the value of c is:
1
Understanding the Problem: Calculating the Value of c
The question asks us to find the value of the variable \(c\), which is defined by the expression: \(c = \sqrt{1+\frac{\sqrt{3}}{2}}- \sqrt{1-\frac{\sqrt{3}}{2}}\). This involves simplifying a mathematical expression containing nested square roots and subtraction.
Analyzing the Expression with Square Roots
The expression for \(c\) involves two terms, each being the square root of \(1\) plus or minus a fraction involving \(\sqrt{3}\). Specifically, we have \(\sqrt{1+\frac{\sqrt{3}}{2}}\) and \(\sqrt{1-\frac{\sqrt{3}}{2}}\). To simplify such expressions, we often look for ways to express the terms inside the square roots as perfect squares, i.e., in the form \((a \pm b)^2\).
Recall the perfect square formulas:
- \((a+b)^2 = a^2 + 2ab + b^2\)
- \((a-b)^2 = a^2 - 2ab + b^2\)
We need to make the terms \(1 \pm \frac{\sqrt{3}}{2}\) fit the form \(a^2 \pm 2ab + b^2\). The term \(\frac{\sqrt{3}}{2}\) in the original expression looks like it could be related to \(2ab\). We can write \(\frac{\sqrt{3}}{2}\) as \(2 \times \frac{\sqrt{3}}{2} \times \frac{1}{2}\).
Now, let's consider the '1' term. We know that \(\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\) and \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\). Their sum is \(\frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1\). This is perfect! We can write \(1 = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2\).
Step-by-Step Solution to Find the Value of c
Let's substitute these observations into the terms inside the square roots:
- Simplify the first term inside the square root: \[ 1 + \frac{\sqrt{3}}{2} = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \] This exactly matches the form \(a^2 + 2ab + b^2\) with \(a = \frac{\sqrt{3}}{2}\) and \(b = \frac{1}{2}\). \[ 1 + \frac{\sqrt{3}}{2} = \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)^2 \]
- Simplify the second term inside the square root: \[ 1 - \frac{\sqrt{3}}{2} = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \] This matches the form \(a^2 - 2ab + b^2\) with \(a = \frac{\sqrt{3}}{2}\) and \(b = \frac{1}{2}\). \[ 1 - \frac{\sqrt{3}}{2} = \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right)^2 \]
- Substitute these perfect squares back into the expression for \(c\): \[ c = \sqrt{\left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right)^2} - \sqrt{\left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right)^2} \]
- Evaluate the square roots. Remember that \(\sqrt{x^2} = |x|\). \[ c = \left|\frac{\sqrt{3}}{2} + \frac{1}{2}\right| - \left|\frac{\sqrt{3}}{2} - \frac{1}{2}\right| \]
- Determine the signs of the terms inside the absolute values. The first term \(\frac{\sqrt{3}}{2} + \frac{1}{2}\) is the sum of two positive numbers, so it is positive. \(\frac{\sqrt{3}}{2} \approx \frac{1.732}{2} = 0.866\). The second term is \(\frac{\sqrt{3}}{2} - \frac{1}{2} \approx 0.866 - 0.5 = 0.366\). This is also positive.
- Since both terms inside the absolute values are positive, we can remove the absolute value signs: \[ c = \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right) - \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) \]
- Simplify the expression: \[ c = \frac{\sqrt{3}}{2} + \frac{1}{2} - \frac{\sqrt{3}}{2} + \frac{1}{2} \] The terms \(\frac{\sqrt{3}}{2}\) and \(-\frac{\sqrt{3}}{2}\) cancel out. \[ c = \frac{1}{2} + \frac{1}{2} \] \[ c = 1 \]
Thus, the value of \(c\) is 1.
Verification
We found \(c=1\). Let's quickly check if this makes sense. The original expression is a difference of two square roots. The first term \(\sqrt{1+\frac{\sqrt{3}}{2}}\) is clearly greater than the second term \(\sqrt{1-\frac{\sqrt{3}}{2}}\) since \(1+\frac{\sqrt{3}}{2} > 1-\frac{\sqrt{3}}{2}\) (as \(\frac{\sqrt{3}}{2} > 0\)). So, \(c\) must be a positive value. Our result, \(c=1\), is positive, which is consistent.
Revision Table: Key Concepts Used
| Concept | Description | Formula/Example |
|---|---|---|
| Perfect Square Identity | Used to rewrite expressions inside square roots into the form \((a \pm b)^2\). | \((a+b)^2 = a^2+2ab+b^2\) \((a-b)^2 = a^2-2ab+b^2\) |
| Square Root Property | The square root of a squared term is its absolute value. | \(\sqrt{x^2} = |x|\) |
| Absolute Value Definition | Determining the value of \(|x|\) based on the sign of \(x\). | \(|x| = x\) if \(x \ge 0\) \(|x| = -x\) if \(x < 0\) |
| Trigonometric Identity | The identity \(\cos^2 \theta + \sin^2 \theta = 1\) helps recognize '1' as a sum of squares, often used in these types of problems (specifically \(\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = 1\)). | \(\cos^2 \theta + \sin^2 \theta = 1\) \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), \(\sin(30^\circ) = \frac{1}{2}\) |
Additional Information: Simplifying Nested Radicals
Problems like this often involve simplifying expressions of the form \(\sqrt{A \pm \sqrt{B}}\), which are called nested radicals. A common technique, as used here, is to look for a perfect square. Another general formula for simplifying nested radicals of the form \(\sqrt{A \pm \sqrt{B}}\) exists:
If \(A^2 - B\) is a perfect square, say \(K^2\), then:
\[ \sqrt{A + \sqrt{B}} = \sqrt{\frac{A+K}{2}} + \sqrt{\frac{A-K}{2}} \] \[ \sqrt{A - \sqrt{B}} = \sqrt{\frac{A+K}{2}} - \sqrt{\frac{A-K}{2}} \]In our problem, the terms are in the form \(\sqrt{1 \pm \frac{\sqrt{3}}{2}}\), which is not directly \(\sqrt{A \pm \sqrt{B}}\). However, we can rewrite \(\sqrt{1 \pm \frac{\sqrt{3}}{2}}\) as \(\sqrt{\frac{2 \pm \sqrt{3}}{2}} = \frac{\sqrt{2 \pm \sqrt{3}}}{\sqrt{2}}\). Now we have \(\sqrt{2 \pm \sqrt{3}}\) which is in the form \(\sqrt{A \pm \sqrt{B}}\) with \(A=2\) and \(B=3\). Let's apply the formula:
For \(\sqrt{2 + \sqrt{3}}\): \(A=2\), \(B=3\). \(A^2 - B = 2^2 - 3 = 4 - 3 = 1\). \(K = \sqrt{1} = 1\).
\[ \sqrt{2 + \sqrt{3}} = \sqrt{\frac{2+1}{2}} + \sqrt{\frac{2-1}{2}} = \sqrt{\frac{3}{2}} + \sqrt{\frac{1}{2}} = \frac{\sqrt{3}}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{\sqrt{3}+1}{\sqrt{2}} \]For \(\sqrt{2 - \sqrt{3}}\): \(A=2\), \(B=3\). \(A^2 - B = 2^2 - 3 = 1\). \(K = \sqrt{1} = 1\).
\[ \sqrt{2 - \sqrt{3}} = \sqrt{\frac{2+1}{2}} - \sqrt{\frac{2-1}{2}} = \sqrt{\frac{3}{2}} - \sqrt{\frac{1}{2}} = \frac{\sqrt{3}}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{\sqrt{3}-1}{\sqrt{2}} \]Now substitute these back into the expression for \(c = \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2}} - \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2}}\):
\[ c = \frac{1}{\sqrt{2}}\left(\sqrt{2 + \sqrt{3}} - \sqrt{2 - \sqrt{3}}\right) \] \[ c = \frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}+1}{\sqrt{2}} - \frac{\sqrt{3}-1}{\sqrt{2}}\right) \] \[ c = \frac{1}{\sqrt{2}}\left(\frac{(\sqrt{3}+1) - (\sqrt{3}-1)}{\sqrt{2}}\right) \] \[ c = \frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}+1 - \sqrt{3}+1}{\sqrt{2}}\right) \] \[ c = \frac{1}{\sqrt{2}}\left(\frac{2}{\sqrt{2}}\right) \] \[ c = \frac{2}{2} = 1 \]This confirms our earlier result using an alternative method for simplifying nested radicals.
Important Questions from Rational or Irrational Numbers
If \(\frac{\sqrt{38-5\sqrt{3} } }{\sqrt{26+7\sqrt{3} } }= \frac{a+b\sqrt{3} }{23} \) , b > 0, then the value of (b – a) is:
If \( \frac{5}{4{\sqrt 2 }} + \frac{{3 + 2\sqrt 2 }}{{3 - 2\sqrt 2 }} - \frac{{3 - 2\sqrt 2 }}{{3 + 2\sqrt 2 }} = a + b\sqrt 2 \) , then what is the value of (3a + 4b)?
If \(\frac {8 + 2\sqrt 3}{3\sqrt 3 + 5} = a\sqrt 3 - b,\) then the value of a + b is equal to:
If \(\frac{\sqrt{26-7\sqrt{3} } }{\sqrt{14+5\sqrt{3} } } = \frac{b+a\sqrt{3} }{11}\) , b > 0, then what is the value of \(\sqrt{(b-a)} \) ?
If \(\frac{{22\sqrt 2 }}{{4\sqrt 2 - \sqrt {3\, + \,\sqrt 5 }}}\) = \(a + \sqrt 5 b\) , with a, b > 0, then what is the value of (ab) ∶ (a + b)?
