Question

Sin60°- 8tg45°- cos30°-8tg135°;

Answer 1

$Sin60^{o}-8tg45^{o}-Cos30^{o}-8tg135^{o}=\frac{\sqrt{3} }{2}-8*1-\frac{\sqrt{3}}{2}-8tg(180^{o}-45^{o})=-8+8tg45^{o}=-8+8*1=-8+8=0\\\\Otvet:\boxed{0}$

Answer 2

To solve the expression Sin60° - 8tg45° - cos30° - 8tg135°, let's break it down step by step:

1. Evaluate Sin60°: Sin60° = √3/2

2. Evaluate 8tg45°: tg45° = 1 8tg45° = 8

3. Evaluate cos30°: cos30° = √3/2

4. Evaluate 8tg135°: tg135° = -1 8tg135° = -8

Now, let's substitute the values back into the expression:

Sin60° - 8tg45° - cos30° - 8tg135° = (√3/2) - 8 - (√3/2) - (-8) = √3/2 - 8 + √3/2 + 8 = 2√3/2

Simplifying further, we get:

2√3/2 = √3

Therefore, Sin60° - 8tg45° - cos30° - 8tg135° is equal to √3.