If sin Theta sq rt of 6 4 and it is in quadrant I what doe

If sin \\Theta = (sq rt of 6) / (4), and it is in quadrant I, what does... sin 2\\Theta= cos 2\\Theta= tan 2\\Theta= If the explanation could please be handwritten, that would be great. It\'s a bit easier to understand than when it\'s typed. However, if it\'s more suitable for you to type it, then that\'s alright too.

Solution

given sin =(6)/4 , in first quadrant =>0<=<=90^o

=>0<=2<=180^o

sin =(6)/4 =oppositeside/hypotenuse side

=>oppositeside²+adjacentside²=hypotenuse side²

=>(6)²+adjacentside²=4²

=>6+adjacentside²=16

=>adjacentside²=10

=>adjacentside=10

cos =adjacentside/hypotenuse =(10)/4

sin(2) =2sincos

sin(2) =2((6)/4)((10)/4)

sin(2) =(60)/8

sin(2) =(15)/4

cos(2) =cos² -sin²

cos(2) =((10)/4)²-((6)/4)²

cos(2) =(10/16)-(6/16)

cos(2) =4/16

cos(2) =1/4

tan(2) =sin(2)/cos(2)

tan(2) =((15)/4)/(1/4)

tan(2) =15