Question

Sin^4(π/12) - cos^4(π/12)

Answer 1

$sin^4\frac{\pi}{12}-cos^4\frac{\pi}{12}=(sin^2\frac{\pi}{12}-cos^2\frac{\pi}{12})(sin^2\frac{\pi}{12}+cos^2\frac{\pi}{12})=\\\\=-(cos^2\frac{\pi}{12}-sin^2\frac{\pi}{12})=-cos\frac{\pi}{6}=-\frac{\sqrt3}{2}$

Answer 2

To calculate the value of $\sin^4(\frac{\pi}{12}) - \cos^4(\frac{\pi}{12})$, you can use trigonometric identities to simplify the expression.

First, let's write $\sin^4(\frac{\pi}{12})$ and $\cos^4(\frac{\pi}{12})$ in terms of a double-angle formula:

Recall the double-angle identities for sine and cosine:

1. $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$
2. $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$

In this case, $\theta = \frac{\pi}{12}$. So, we have:

$\sin(2 \cdot \frac{\pi}{12}) = 2\sin(\frac{\pi}{12})\cos(\frac{\pi}{12})$

Now, let's solve for $\sin(\frac{\pi}{12})\cos(\frac{\pi}{12})$:

$\sin(\frac{\pi}{12})\cos(\frac{\pi}{12}) = \frac{1}{2} \sin(2\cdot \frac{\pi}{12})$

$\sin(\frac{\pi}{12})\cos(\frac{\pi}{12}) = \frac{1}{2} \sin(\frac{\pi}{6})$

Now, we know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$, so:

$\sin(\frac{\pi}{12})\cos(\frac{\pi}{12}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$

Now, we can calculate $\sin^4(\frac{\pi}{12})$ and $\cos^4(\frac{\pi}{12})$ using this result:

$\sin^4(\frac{\pi}{12}) = \left(\frac{1}{4}\right)^4 = \frac{1}{256}$

$\cos^4(\frac{\pi}{12}) = \left(\cos^2(\frac{\pi}{12})\right)^2$

To find $\cos^2(\frac{\pi}{12})$, you can use the double-angle formula for cosine:

$\cos^2(\frac{\pi}{12}) = \frac{1}{2}\left(1 + \cos(\frac{\pi}{6})\right)$

Now, we know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$