Respuesta :
Given that
[tex]\sin60^o= \frac{ \sqrt{3} }{2} [/tex]
[tex]\cos 30^o =\sin(90-30)^o=\sin60^o=\frac{ \sqrt{3} }{2}[/tex],
[tex]\cos 120^o =-\sin(180-120)^o=-\sin60^o=-\frac{ \sqrt{3} }{2}[/tex],
[tex]\cos30^o\neq0[/tex],
[tex]\cos120^o\neq0[/tex]
Therefore, the true statement is
[tex]\cos 30^o =\sin(90-30)^o=\sin60^o=\frac{ \sqrt{3} }{2}[/tex],
because the cosine and sine are complements.
[tex]\sin60^o= \frac{ \sqrt{3} }{2} [/tex]
[tex]\cos 30^o =\sin(90-30)^o=\sin60^o=\frac{ \sqrt{3} }{2}[/tex],
[tex]\cos 120^o =-\sin(180-120)^o=-\sin60^o=-\frac{ \sqrt{3} }{2}[/tex],
[tex]\cos30^o\neq0[/tex],
[tex]\cos120^o\neq0[/tex]
Therefore, the true statement is
[tex]\cos 30^o =\sin(90-30)^o=\sin60^o=\frac{ \sqrt{3} }{2}[/tex],
because the cosine and sine are complements.
Answer:
cos 30° = square root of three over two, because the cosine and sine are complements
Step-by-step explanation: