Some Trig Questions I Need Help With (1 Viewer)

kevinsta

Member
Joined
Nov 24, 2013
Messages
62
Gender
Male
HSC
2015
How would i find the exact value of cos(-225) and Sin(315), can you please show me how you got to it, i used to know this but when i was revising i was completely lost.
 

aDimitri

i'm the cook
Joined
Aug 22, 2013
Messages
505
Location
Blue Mountains
Gender
Male
HSC
2014
cos(-225) = cos(360-225) = cos(135) = -cos(45) = -1/sqrt(2)
sin(315) = sin(360-45) = -sin(45) = -1/sqrt(2)
 

bongoli

Active Member
Joined
Mar 7, 2013
Messages
165
Gender
Male
HSC
2014
if it's negative, just do it opposite of how you normally do it. Draw up your quadrants and then draw 150 degrees in the opposite direction. Find out what quadrant it is in... ASTC > (All(1st), Sine(2nd), Tan(3rd), Cosine(4th) then do 360- your angle which would be 210 degrees. This means it is in the 3rd quadrant. Cosine is negative in this quadrant and it follows the rule of 180+theta.

210 = 180 + theta
Therefore it is -cos(30) and this equates to -sqrt(3)/2

follow the same train of thought for these kind of equations, gl
 

kevinsta

Member
Joined
Nov 24, 2013
Messages
62
Gender
Male
HSC
2015
if it's negative, just do it opposite of how you normally do it. Draw up your quadrants and then draw 150 degrees in the opposite direction. Find out what quadrant it is in... ASTC > (All(1st), Sine(2nd), Tan(3rd), Cosine(4th) then do 360- your angle which would be 210 degrees. This means it is in the 3rd quadrant. Cosine is negative in this quadrant and it follows the rule of 180+theta.

210 = 180 + theta
Therefore it is -cos(30) and this equates to -sqrt(3)/2


follow the same train of thought for these kind of equations, gl
Omg man thanks so much i fully understand it now have a nice day
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top