Question

Please help! I'm stuck on this Precal/Trig problem :O Click attached to view!

Answer 1

it can be written as
sin(180+75)=-sin(75)
and sin(75)=sin(30+45)=sin(30)cos(45)+sin(45)cos(30)
calculate the above and multiply with -

Answer 2

we both were at the same time:) @xavier123

Answer 3

sin(255)=sin(180+75)=-sin(75)=-sin(45+30)=-(sin45*cos30+cos45*sin30)={-sqrt(2)-sqrt(6)}/4

FORMULAE USED:
sin(A+B)=sinAcosB+cosAsinB.
sin(180+x)=-sinx

Answer 4

Sin (255)= Sin (210 +45) =Sin 210 * Cos 45 + Cos 210 * Sin 45 = 1/ (\[\sqrt{2}\] (Sin 210 + Cos 210)

Now, Sin (210) = Sin (180 + 30) = - Sin 30 = -1/2
and, Cos (210)= Cos (180+30)= -Cos 30 = -sqrt (3)/2

Now, Sin 255

= \[1/\sqrt{2} (-1/2 -\sqrt{3}/2)\]
=\[-1/2\sqrt{2}(1+\sqrt{3})\]

Answer 5

now for tan(255)
tan(180+75)=tan(75)
tan(30+45)
\[\Large \tan(30+45)=\frac{\tan(30)+\tan(45)}{1-\tan(30)\tan(45)}\]

Answer 6

answer's everywhere.. you gotta take it