wishboy SG got a few qns i cannot solve or solved halfway Prove the following identity. ( 3 - 6 cos² x ) / ( sin x - cos x ) = 3 ( sin x + cos x ) Prove that (a) cot x + tan x = cosec x sec x (b) cosec² x + sec² x = cosec² x sec² x Hence, deduce that (cot x + tan x )² = cosec² x + sec² x 5. Find all the angles between 0° and 360° which satisfy the following equations. (d) cos ( 2x - 40° ) = 0.8 this qn i found 2 answers which are 38.4° and 218.4°, but there are 2 more answers which are 1.6° and 181.6°. i dunno how to find those 2. (e) cot ( 2x + 10° ) = - 0.5 this qn i get the answers all wrong de my 1st step was tan ( 2x + 10° ) = - 2 answers: 53.3°, 143.3°, 233.3°, 323.3° will appreciate any help
eagle SG ( 3 - 6 cos² x ) = 3 - 3cos² x -3cos² x = 3( sin² x - cos² x ) = 3 (sin x + cos x) (sin x - cos x) after divide u will get answer a) cos x / sin x + sin x / cos x = ( cos²x + sin²x ) / (sin x cos x) = 1 / (sin x cos x) = cosec x sec x b) 1 / sin² x + 1 / cos² x = ( cos² x + sin² x ) / (sin² x cos² x) = 1 / (sin² x cos² x) = cosec² x sec² x (part a)² = (cot x + tan x )² = cosec² x sec² x = cosec² x + sec² x (deduced)
^tamago^ SG Prove: (3-6 cos²x)/(sin x-cos x) = 3(sin x+cos x) LHS = (3-6 cos²x)/(sin x-cos x) = 3(1-2cos²x)/(sin x-cos x) = 3(sin²x-cos²x)/(sin x-cos x) = 3(sin x+cos x) = RHS (proven)
eagle SG U must know 0 < x < 360 0 < 2x < 720 -40 < 2x - 40 < 680 hence, 2x - 40 = -36.87, 36.87, 323.13, 396.87 x = 1.6, 38.4, 181.6, 218.4 understand part d, then try part e by yourself
^tamago^ SG cos(2x-40°) = 0.8 For -40° 2x-40°=-36.86°, 36.86°, 360-36.86°, 360+36.86° = -36.86°, 36.86°, 323.14°, 396.86° 2x=3.14°, 76.86°, 363.14°, 436.86° x=1.57°, 38.43°, 181.57°, 218.43° = 1.6°, 38.4°, 181.6°, 218.4°
^tamago^ SG cot(2x+10°) = -0.5 tan(2x+10°) = -2 For 10° 2x+10°= 180-63.43°,360-63.43°, 540-63.43°, 720-63.43° = 116.57°, 296.56°, 476.57°, 656.56° 2x= 106.57°, 286.56°, 466.57°, 646.56° x= 53.29°, 143.28°, 233.29°, 323.28° = 53.3°, 143.3°, 233.3°, 323.3°
