Question

1. Find cos 255

2. Find sin 195



with the following answer in the picture for both

Answer 1

Your answer is the second one.

Both cos 255 and sin 195 are -√6-√2/4

Answer 2

`\sf \cos(255^\circ) = (√(2) -√(6))/(4)`

`\sf \sin(125^\circ) = (√(2) -√(6))/(4)`

Explanation:

To find the cosine and sine values of the given angles, let's use the properties of trigonometric functions.

Cosine of 255 degrees:

The angle 255 degrees is in the third quadrant where the cosine function is negative. We can find the cosine value using the reference angle in the first quadrant.

`\sf \cos(255^\circ) = -\cos(255^\circ - 180^\circ)`

The reference angle is
`\sf 255^\circ - 180^\circ = 75^\circ`, and in the first quadrant,
`\sf \cos(75^\circ) = \cos(30^\circ + 45^\circ)`.

Using the cosine addition formula
`\sf \cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta`:

`\sf \cos(75^\circ) = \cos(30^\circ)\cos(45^\circ) - \sin(30^\circ)\sin(45^\circ)`

Now, let's substitute the values:
`\sf \cos(30^\circ) = (√(3))/(2)`,
`\sf \cos(45^\circ) = (√(2))/(2)`,
`\sf \sin(30^\circ) = (1)/(2)`,
`\sf \sin(45^\circ) = (√(2))/(2)`.

`\sf \cos(75^\circ) = (√(3))/(2) \cdot (√(2))/(2) - (1)/(2) \cdot (√(2))/(2)`

Simplifying the expression, we get:

`\sf \cos(75^\circ) = (√(6) - √(2))/(4)`

Cos(225°) is in the third quadrant. So, it must be negative.

`\sf \cos(225^\circ) = -\left((√(6) - √(2))/(4) \right)`

Distribute negative sign:

`\sf \cos(225^\circ) = (-√(6) +√(2))/(4)`

`\sf \cos(225^\circ) = (√(2) -√(6))/(4)`

`\hrulefill`

Sine of 195 degrees:

The angle 195 degrees is in the third quadrant where the sine function is negative. We can find the sine value using the reference angle in the first quadrant.

`\sf \sin(195^\circ) = -\sin(195^\circ - 180^\circ)`

The reference angle is
`\sf 195^\circ - 180^\circ = 15^\circ`, and in the first quadrant,
`\sf \sin(15^\circ) = \sin(45^\circ - 30^\circ)`.

Using the sine subtraction formula
`\sf \sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta`:

`\sf \sin(15^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ)`

Now, let's substitute the values:
`\sf \sin(45^\circ) = (√(2))/(2)`,
`\sf \cos(30^\circ) = (√(3))/(2)`,
`\sf \cos(45^\circ) = (√(2))/(2)`,
`\sf \sin(30^\circ) = (1)/(2)`.

`\sf \sin(15^\circ) = (√(2))/(2) \cdot (√(3))/(2) - (√(2))/(2) \cdot (1)/(2)`

Simplifying the expression, we get:

`\sf \sin(15^\circ) = (√(6) - √(2))/(4)`

sin(195°) is in the third quadrant. So, it must be negative.

So,

`\sf \sin(195^\circ) = -\left((√(6) - √(2))/(4) \right)`

Distribute negative sign:

`\sf \sin(125^\circ) = (-√(6) +√(2))/(4)`

`\sf \sin(125^\circ) = (√(2) -√(6))/(4)`

`\hrulefill`

So, the answers are:

`\sf \cos(255^\circ) = (√(2) -√(6))/(4)`

`\sf \sin(125^\circ) = (√(2) -√(6))/(4)`