Step-by-step explanation:
this is a 45 45 90. I have drawn a picture for you to see how you find lengths of a 45 45 90 triangle.
so, your base and height are the same, but let us denote the triangle as o, a, and h for opposite, adjacent, and hypotenuse. Because we have two 45 degree angles we can choose either one. I have drawn an image showing these labels and underlined the angle I am using in yellow. Now,
This is the base (in the picture)
[tex]o= \frac{ \sqrt[]{2} }{2} [/tex]
and this is the height (in the drawing)
[tex]a= \frac{ \sqrt{2} }{2} [/tex]
But we need the hypotenuse to answer the questions.
To find the hypotenuse, we can use the first picture I drew and see the hypotenuse is
[tex]x \sqrt{2} [/tex]
and we already have x, which is one of the other 2 sides or
[tex] \frac{ \sqrt{2} }{2} [/tex]
so,
[tex] \frac{ \sqrt[]{2} }{2} \times \sqrt{2 } = \frac{ \sqrt{2} \times \sqrt{2} }{2} = \frac{2}{2} = 1[/tex]
because the radical cancels when it is multiplied by itself (these is because we would multiply 2x2 under the radical to get 4, and the square root of 4 is 2).
Now, we use SOH CAH TOA to determine Sin and Cos of 45 degrees.
SOH means
[tex] \sin(x) = \frac{opposite}{hypotenuse} [/tex]
CAH means
[tex] \cos(x) = \frac{opposite}{hypotenuse} [/tex]
and we aren't worried about tangent, so TOA is irrelevant here. Now, for sin(45°) we just plug in the information from our labeled triangle.
[tex] \sin(45) = \frac{ \frac{ \sqrt{2} }{2} }{1} = \frac{ \sqrt{2} }{2} [/tex]
and Cos will be the same
[tex]\cos(45) = \frac{ \frac{ \sqrt{2} }{2} }{1} = \frac{ \sqrt{2} }{2} [/tex]
finally (although we could have done this first), the coordinates are
([tex]\frac{ \sqrt{2} }{2}[/tex], [tex]\frac{ \sqrt{2} }{2}[/tex])
Hope this helps!
