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October 20, 2014

October 20, 2014

Posted by **Emily** on Thursday, August 22, 2013 at 4:16pm.

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)

- Math -
**Reiny**, Thursday, August 22, 2013 at 5:25pmYou 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.

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