Question

√12cos^2 5pi/12-√3=...​

Answer 1

$\sqrt{12}Cos^{2}\frac{5\pi }{12}-\sqrt{3} =2\sqrt{3}Cos^{2}\frac{5\pi }{12}-\sqrt{3} =\sqrt{3} (2Cos^{2}\frac{5\pi }{12}-1)=\\\\=\sqrt{3}Cos(2*\frac{5\pi }{12})=\sqrt{3}Cos\frac{5\pi }{6}=\sqrt{3}Cos(\pi-\frac{\pi }{6})=-\sqrt{3}Cos\frac{\pi }{6}=\\\\=-\sqrt{3}*\frac{\sqrt{3}}{2}=-\frac{3}{2}= -1,5 \\\\Otvet:\boxed{-1,5}$

Answer 2

Let's break down the expression step by step:

1. First, calculate the value of cos(5π/12). This is not a common angle, so we'll use the half-angle formula for cosine:

   cos(θ/2) = ±√((1 + cos(θ)) / 2)

   For θ = 5π/6: cos(5π/12) = ±√((1 + cos(5π/6)) / 2)

   Now, cos(5π/6) = cos(π - π/6) = -cos(π/6) = -√3/2.

   Substituting this value: cos(5π/12) = ±√((1 - √3/2) / 2)

   Taking the positive root: cos(5π/12) = √((2 - √3) / 4)

2. Now, let's substitute this value of cos(5π/12) back into the expression:

   √12 * cos^2(5π/12) - √3 = √12 * (√((2 - √3) / 4))^2 - √3 = √12 * ((2 - √3) / 4) - √3 = (√3 * 2) - (√3) - √3 = 2√3 - 2√3 = 0

So, the value of the expression is 0.