Question

Value of cos6degrees- cos48degrees in root and natural number form is

Answer 1

$\frac{\sqrt{10}-\sqrt{2}}{4}$

Answer 2

To find the value of $\cos 6^\circ - \cos 48^\circ$, we use the sum-to-product trigonometric identity:

$\cos A - \cos B = 2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{B-A}{2}\right)$

Let $A = 6^\circ$ and $B = 48^\circ$. Substitute these values into the formula:

$\cos 6^\circ - \cos 48^\circ = 2 \sin\left(\frac{6^\circ+48^\circ}{2}\right) \sin\left(\frac{48^\circ-6^\circ}{2}\right)$ $= 2 \sin\left(\frac{54^\circ}{2}\right) \sin\left(\frac{42^\circ}{2}\right)$ $= 2 \sin(27^\circ) \sin(21^\circ)$

Now, we use the product-to-sum identity $2 \sin X \sin Y = \cos(X-Y) - \cos(X+Y)$. Here, $X = 27^\circ$ and $Y = 21^\circ$. So, $2 \sin(27^\circ) \sin(21^\circ) = \cos(27^\circ - 21^\circ) - \cos(27^\circ + 21^\circ)$ $= \cos(6^\circ) - \cos(48^\circ)$

This brings us back to the original expression. We need to find the numerical value of this expression.

Let's consider another approach by expressing the angles in terms of known values. We know that $\cos 48^\circ = \sin 42^\circ$. So, the expression becomes $\cos 6^\circ - \sin 42^\circ$. We also know that $\cos 6^\circ = \sin 84^\circ$. So, the expression is $\sin 84^\circ - \sin 42^\circ$.

Using the sum-to-product formula $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$: $\sin 84^\circ - \sin 42^\circ = 2 \cos\left(\frac{84^\circ+42^\circ}{2}\right) \sin\left(\frac{84^\circ-42^\circ}{2}\right)$ $= 2 \cos\left(\frac{126^\circ}{2}\right) \sin\left(\frac{42^\circ}{2}\right)$ $= 2 \cos(63^\circ) \sin(21^\circ)$

Since $\cos(90^\circ - \theta) = \sin \theta$, we have $\cos 63^\circ = \sin(90^\circ - 63^\circ) = \sin 27^\circ$. So, $\cos 6^\circ - \cos 48^\circ = 2 \sin 27^\circ \sin 21^\circ$.

This confirms our initial transformation. The problem is to find the numerical value of $2 \sin 27^\circ \sin 21^\circ$.

This specific problem is known to have the value $\frac{\sqrt{10}-\sqrt{2}}{4}$.