[tex]$\sin 45^\circ=\frac{-1}{\sqrt{2} }, \ \cos45^\circ=\frac{1}{\sqrt{2} }, \ \ \tan 45^\circ=-1[/tex]
Solution:
In the given right triangle,
θ = 45°
Opposite side to θ = –1
Adjacent side to θ = 1
Hypotenuse = [tex]\sqrt{2}[/tex]
Using basic trigonometric ratio formulas,
[tex]$\sin \theta=\frac{\text { Opposite }}{\text { Hypotenuse }}[/tex]
[tex]$\sin 45^\circ=\frac{-1}{\sqrt{2} }[/tex]
[tex]$\cos \theta=\frac{\text { Adjacent }}{\text { Hypotenuse }}[/tex]
[tex]$\cos45^\circ=\frac{1}{\sqrt{2} }[/tex]
[tex]$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}[/tex]
[tex]$\tan 45^\circ=\frac{-1}{1}=-1[/tex]
Therefore,
[tex]$\sin 45^\circ=\frac{-1}{\sqrt{2} }, \ \cos45^\circ=\frac{1}{\sqrt{2} }, \ \ \tan 45^\circ=-1[/tex]
