Trigonometry Flashcards
Find angles between 0 and 360 degrees which are the same as -Π/2, 1180oand 9Π/4
- The first is negative so we add 2Π.
- -Π/2 + 2Π = 3Π/2
- 1180o = 11800 - 3600 = 8200 = 4600 = 100o
- 9Π/4 radians = 9 Π/4 - 2Π radians= Π/4 radians
How do we find the CosΘ and sinΘ are for arbitrary Θ ?
- Determine the quadrant.
- Draw right triangle by drawing an altitude to the the x axis.
- Get the Cartesian co-ordinates (x,y).
- Points to the left of the y axis have a negative * x* and those below have a negative y.
- we then set sin Θ = x
- and cos Θ = y
Find tan 7π/4
- Determine quadrant – since 7π/4 is greater than 3π/2 and less than 2π, the angle is in the fourth quadrant.
- Draw altitude.
- Determine the acute angle which is: π/4
- cos π/4 = sqrt(2)/2 as is sin π/4.
- Determine sign: since the angle is in the fourth quadrant and x is positive and y is negative, so cos 7π/4 is positive and sin 7π/4 is negative.
- tan 7π/4 = (sin 7π/4 )/ (cos 7π/4) = -1
Qadrants: 30, 700, 5π/3 and - 3π/5
- 30: Quadrant 1
- 700 : Q IV
- 5π/3 : Q IV
- 3π/5: Q III
sin(30 )
1/2
sin(45)
sqrt(2) /2
sin 60
sqrt(3) / 2
sin 90
1
sin 120
sqrt(3) / 2
sin 135
sqrt(2) / 2
sin 150
1/2
sin 180
0
sin 210
- 1/2
sin 225
-sqrt(2) / 2
sin 240
- sqrt (3) / 2
sin 270
-1
sin 300
- sqrt(3) / 2
sin 315
- sqrt(2) / 2
sin 330
- 1/2
sin 360
= sin 0 = 0
cos 0
1
cos 30
sqrt(3) / 2
cos 45
sqrt(2) / 2
cos (60)
1/2
cos 90
0
cos 120
- 1/2
cos 135
- sqrt(2) / 2
cos 150
- sqrt(3) / 2
cos 180
-1
cos 210
- sqrt(3) / 2
cos (225)
- sqrt(2) / 2
cos(240)
- 1 /2
cos 270
0
cos 300
1/2
cos 315
sqrt(2) / 2
cos (330)
sqrt(3) / 2
cos 360
= cos 0 = 1
pi/6 radians
30 degrees
pi/3 radians
60 degrees
π / 4 radians
45o
π / 2 radians
90o
2 π/3
120 degrees
Determine sin 300
Quadrant ==> IV
- sqrt(3) / 2
csc 150
1 / sin
==> 1 / (sin 150)
== 1 / sin(30)
== 2
cot 5π/3
Quadrant IV
cot(-π/3)
cot (- π/3)
- sqrt(3) / 3
cos(π/3) = 1/2
cot(π/3) [=] 1/3sqrt(3)
csc(π/3) = 2/3sqrt(3)
(7)
[sec(pi/3)] [=] [2]
(8)
[sin(pi/3)] [=] [1/2sqrt(3)]
(9)
[tan(pi/3)] [=] [sqrt(3).]
signs of cos theta
and sin thetha
given theta?
cos is positive in I,IV, negative in II, III
sin is positive in I, II, negative in III, IV
find sin(105)
What is the exact value of sin(105º)?
We can use a sum angle formula noticing that 105º = 45º + 60º.
We have sin(105º) = sin(45º + 60º) = sin(45º )cos(60º) + cos(45º )sin(60º).
We know the exact values of trig functions for 60º and 45º.
Therefore, sin(45º )cos(60º) + cos(45º )sin(60º)
(sqrt(2) + sqrt(6))/4