You should memorize the following diagrams:
1. the 30°-60°-90° triangle with matching sides
1 -- √3 -- 2 (angles in radians, π/6 π/3 π/2)
2. the 45° -- 45° -- 90° triangle with corresponding sides 1 -- 1 -- √2
So sin π/4 or sin 45° = 1/√2
arccos(-√3/2) = .....
I know from looking at my triangle that cos60° or cosπ/3 = √3/2
Also I know that the cosine is negative in II and III
so arccos(-√3/2) = 180-60 = 120° or 2π/3
cscØ = 2
sinØ = 1/2
Ø = 30° or π/6 , looking at my triangle
With identities, it is usually a good idea to change all trig ratios to sines and cosines, and let the chips fall as they may.
I will do the first one:
LS = cotx cosx + sinx
= (cosx/sinx)(cosx) + sinx
= (cos^2 x + sin^2 x)/sinx
= 1/sinx
= csc x
= RS
try the others using that concept, let me know if you were successful.
1) evaluate without a calculator: a)sin(3.14/4) b) cos(-3(3.14)/4) c) tan(4(3.14)/3) d) arccos(- square root of three/2) e) csctheata=2
2) verify the following identities: a) cotxcosx+sinx=cscx b)[(1+sinx)/ cosx] + [cosx/ (1+sinx)]= 2secx c) (sin^2x + 2sinx+1)/cos^2x= (1+sinx)/(1-sinx)
1 answer
1 answer