Show that 12 cos 30° -2 tan 60° can be written in the form square root k where k is an integer.

FIrst of all we need to understand how cos 30° and tan 60° are found. Let's think of an equilateral triangle, all three angles of the triangle are equal and 60° and each side has length 2. Let's half the triangle in the centre, now the triangle has a bottom lenth of 1, hypotenuse  length 2 and side length of square root 3. This is found by Phthagoras' Theorem,
22 = 12+(square root 3) 2. Now the angle between the hypotenuse and the bottom is still 60°, the angle between the side and the hypotenuse is 30° and other angle between the bottom and side is a right angle, 90°. By using the trigonometric functions for the halved triangle, cos 30° is equivalent to (square root 3) / 2 and tan 60° is equivalent to square root 3 by using SOHCAHTOA respectively. Therefore 12 cos 30° -2 tan 60°= 12(square root 3)/2 - 2(square root 3) = 6 square root 3 - 2 square root 3 = 4 square root 3 = square root (4x4x3) = square root 48 where k=48.

AC

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