jannny said:
Simplify (I'll use @ for theta)
(sin @) sin( 180 - @) - sin(90 - @) cos(180 - @)
the final answer is 1.
and with the solutions how did
sin (180 - @) = sin @?
sin ( 90 - @) = cos @ ?
cos ( 180 - @) = - cos @?
thanks, cheers.
sin (180 - θ) = sin θ
&
cos (180 - θ) = - cos θ
You know the four quadrants ASTC ?
1st. quad:
All trig ratios are positive (θ)
2nd. quad:
Sin only is positive (180 - θ)
3rd. quad:
Tan only is positive (180 + θ)
4th. quad:
Cos only is positive (360 - θ)
E.g
If θ were 45
0
cos 45
0 (1st quadrant; All trig ratios are positive)
cos 135
0 = cos (180
0 - 45
0) =
- cos 45
0 (2nd quadrant; cos is negative)
cos 225
0 = cos (180
0 + 45
0) =
- cos 45 (3rd quadrant; cos is negative)
cos 315
0 = cos (360
0 - 45
0) =
cos 45
0 (4th quadrant; cos is positive)
sin ( 90
0 - θ) = cos θ
The best way I can explain is:
sin 90 = cos 0 (cos 0 = 1 & sin 90 = 1)
sin 30 = cos 60 (cos 60 = 0.5 & sin 30 = 0.5)
It's from the Preliminary Maths course.
Remember:
sin ( 90
0 - θ) = cos θ
cos ( 90
0 - θ) = sin θ
tan ( 90
0 - θ) = cot θ
sec ( 90
0 - θ) = cosec θ
cosec ( 90
0 - θ) = sec θ
cot ( 90
0 - θ) = tan θ
