Question

Cos105º-sin195º+sin(-135º)​

Answer 1

Ответ:   $-\frac{\sqrt2}{2}$ .

Объяснение:

$cos105^\circ -sin195^\circ +sin(-135^\circ )=\\\\cos(90^\circ +15^\circ )-sin(180^\circ +15^\circ )-sin135^\circ =\\\\=-sin15^\circ -(-sin15^\circ )-sin(90^\circ +45^\circ )=\\\\=-sin15^\circ +sin15^\circ -cos45^\circ =-cos45^\circ =-\frac{\sqrt2}{2}$

Answer 2

ответ:

$- \frac{ \sqrt{2} }{2}$

Answer 3

To evaluate the expression cos(105º) - sin(195º) + sin(-135º), we can use the trigonometric identities and properties to simplify each term and then perform the calculations.

1. cos(105º): Using the cosine of the difference formula, we have: cos(105º) = cos(45º + 60º) = cos(45º)cos(60º) - sin(45º)sin(60º) = (√2/2)(1/2) - (√2/2)(√3/2) = √2/4 - √6/4 = (√2 - √6)/4

2. sin(195º): Using the sine of the difference formula, we have: sin(195º) = sin(45º + 150º) = sin(45º)cos(150º) + cos(45º)sin(150º) = (√2/2)(-√3/2) + (√2/2)(1/2) = -√6/4 + √2/4 = (-√6 + √2)/4

3. sin(-135º): The sine function is periodic, so sin(-135º) is the same as sin(225º): sin(-135º) = sin(225º) = sin(180º + 45º) = -sin(45º) = -√2/2

Now we can substitute the values back into the original expression:

cos(105º) - sin(195º) + sin(-135º) = (√2 - √6)/4 - (-√6 + √2)/4 + (-√2/2) = (√2 - √6 + √6 - √2 - √2)/4 - √2/2 = 0/4 - √2/2 = -√2/2

Therefore, cos(105º) - sin(195º) + sin(-135º) equals -√2/2.