Question

Sin 45 cos 45 tg 60 - cos 30 tg 45

Answer 1

$\sin 45^{\circ} \cdot \cos 45^{\circ}\cdot \text {tg } 60^{\circ}- \cos 30^{\circ}\cdot \text {tg } 45^{\circ} = \\ =\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} \cdot \sqrt{3} - \frac{\sqrt{3}}{2} \cdot 1=\frac{2}{4} \cdot \sqrt{3} - \frac{\sqrt{3}}{2}=\\=\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=0$

Ответ: 0

Answer 2

To calculate the expression, let's break it down step by step:

1. sin(45) = cos(45) = √2 / 2 (since sin(x) = cos(90 - x) and cos(45) = √2 / 2)
2. tan(60) = √3 (since tan(60) = sin(60) / cos(60) = √3 / 1 = √3)
3. cos(30) = √3 / 2 (standard value for cos(30))
4. tan(45) = 1 (standard value for tan(45))

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

sin(45) * cos(45) * tan(60) - cos(30) * tan(45)

= (√2 / 2) * (√2 / 2) * √3 - (√3 / 2) * 1

= (√2 * √2 * √3) / 4 - (√3 / 2)

= (√6 / 4) - (√3 / 2)

To simplify further, find a common denominator:

= (√6 / 4) - (2√3 / 4)

Now, combine like terms:

= (√6 - 2√3) / 4

So, the final value of the expression is (√6 - 2√3) / 4.