root3/1-cos30
leave your answer in surd form
well, I happen to remember that sin 30 = 1/2
so draw a right triangle with small side 1 and
hypotenuse 2
then the big side is sqrt(4-1) = sqrt3
so
cos 30 = sqrt3 / 2
so
sqrt3 / (1 - sqrt3 / 2) = 2 sqrt3 /(2-sqrt3)
=2 sqrt3 /(2-sqrt3) * (2+sqrt3)/(2+sqrt3)
=(4 sqrt3 + 3) / (4-3)
= 4 sqrt3 + 3
=2 sqrt3/(2-sqrt3)*(2+sqrt3)/(2+sqrt3)
is going to be equal to (4 sqrt3+6)/(4-3)
=4 sqrt3 + 6
thank you though. I didn't know how it was supposed to be done.🙏
To simplify the expression \(\frac{\sqrt{3}}{1-\cos 30}\), we need to remember the values of the trigonometric function cosine (\(\cos\)) at special angles.
The special angles for cosine are commonly known as the "unit circle" values. For example, \(\cos 30\) is a special angle because it lies on the unit circle, which makes it easier to calculate.
So, let's calculate \(\cos 30\) first. Using the unit circle or reference triangles, we know that at \(\theta = 30^\circ\), \(\cos \theta = \frac{\sqrt{3}}{2}\).
Now, substitute \(\cos 30\) into the expression \(\frac{\sqrt{3}}{1-\cos 30}\):
\[
\frac{\sqrt{3}}{1-\frac{\sqrt{3}}{2}}
\]
To simplify this expression further, we need to rationalize the denominator. Multiplying the numerator and denominator by the conjugate of the denominator, which is \(1+\frac{\sqrt{3}}{2}\), we get:
\[
\frac{\sqrt{3} \cdot (1+\frac{\sqrt{3}}{2})}{(1-\frac{\sqrt{3}}{2}) \cdot (1+\frac{\sqrt{3}}{2})}
\]
Now, simplify by applying the distributive property:
\[
\frac{\sqrt{3} + \frac{3}{2}}{(1-\frac{3}{4})}
\]
Combining like terms in the numerator:
\[
\frac{\sqrt{3} + \frac{3}{2}}{\frac{1}{4}}
\]
To divide by a fraction, take the reciprocal of the denominator and multiply:
\[
(\sqrt{3} + \frac{3}{2}) \cdot 4
\]
Applying the distributive property:
\[
4\sqrt{3} + 6
\]
Hence, the simplified form of \(\frac{\sqrt{3}}{1-\cos 30}\) is \(4\sqrt{3} + 6\).