Question

Решить 1)cos60°-√6 cos30° sin45°+ctg30° tg150°-tg45° 2)sin + √2 cos -√3 ctg

Answer 1

1)0,5-root6*(root3)/2*(root2)/2-(root3)*(root3)/3-1=-3
2)0.5-(root2)*(root2)/2-(root3)*(root3)/3=-1.5

Answer 2

To solve these expressions, let's break them down step by step:

1. cos(60°) - √6 cos(30°) sin(45°) + ctg(30°) tg(150°) - tg(45°)

We'll start by evaluating each trigonometric function:

cos(60°) = 1/2 cos(30°) = √3/2 sin(45°) = √2/2 ctg(30°) = 1/tan(30°) = 1/√3 tg(150°) = -√3 tg(45°) = 1

Now substitute these values back into the expression:

1/2 - √6 (√3/2) (√2/2) + 1/√3 (-√3) - 1

Simplifying further:

1/2 - (√6/2) (√6/√2) + 1/√3 (-√3) - 1 1/2 - (√6/2) (√3) - 1 1/2 - (√6/2) (√3) - 2/2 (1 - √18 - 2)/2 (-√18 - 1)/2 (-3√2 - 1)/2

So the solution to the first expression is (-3√2 - 1)/2.

2. sinθ + √2 cosθ - √3 ctgθ

We'll use the same approach as before, evaluating each trigonometric function:

sin(θ) cos(θ) ctg(θ) = 1/tan(θ)

Since no specific angle is given, we cannot determine the exact values of these functions without additional information. However, we can simplify the expression by factoring out common terms:

sinθ + √2 cosθ - √3 ctgθ = sinθ + √2 cosθ - √3 (1/tanθ) = sinθ + √2 cosθ - (√3/tanθ)

Without specific values for θ, this is the simplest form of the expression.

Please provide the value of θ if you want a numerical solution.