wishboy SG 5. Find all angles between 0° and 360° which satisfy the following equations. 3 sin x = 2 cos (x + 45°) my ans different from answer key correct ans is 17.8°, 197.8°.
eagle SG how u do? o levels or a levels? got learn how to expand cos(x + 45°) a not before? edit: I just realised topic says A maths. I think of an o level method that get back to u
eagle SG if u know the expansion, you can easily expand it then u make one side all sin x, one side all cos x then u divide the cos x over to get tan x, and divide the left side number to the right side. u should then get tan x = sqrt(2) / { 3 + sqrt(2) } which you should get x = 17.8 degrees. Add 180 degrees for the other answer becuz this is tangent. But I'm not sure if this expansion is in o level syllabus
wishboy SG (3 sin x) / 2 = 3/2 sin x EDIT: and since cos 45° = sin 45°, cos x cos 45° - sin x sin 45° = sin 45° (cos x - sin x) are they correct?
^tamago^ SG best to do by formulae to avoid doubt 3 sin x = 2 cos (x + 45°) 3 sin x = 2 cos x cos 45° - 2 sin x sin 45° 3 sin x = sqrt(2)(cos x-sin x) x = 17.76°, 180+17.76° (see ASTC) = 17.8°, 197.8°
wishboy SG lol somehow i got the answer, although the methods i used look so much alike thx for ur help
^tamago^ SG it's in the syllabus and in the textbook. try to memorise this table. it's helpful all the way to uni.
wishboy SG ooo so i learn liao... but forgot for my textbook they give sin 45° = 1/sqrt(2) value same so nvm another qn, 0° < x < 360° tan 5x = tan 2x find all values of x this qn got * meaning more difficult than normal ones
eagle SG i always teach my students this method sin 0 = sqrt(0)/2 sin 30 = sqrt(1)/2 sin 45 = sqrt(2)/2 sin 60 = sqrt(3)/2 sin 90 = sqrt(4)/2 cos just reverse everything but hor... everything is already second nature to me
eagle SG First 2 line RHS got problem and don't need do till so mah fan mah tan 5x = tan 2x 0 < 5x < 1800 0 < 2x < 720 0 < 5x - 2x < 1080 then 5x = 2x + 180, 2x + 360, 2x + 540, 2x + 720, 2x + 900 3x = 180, 360, 540, 720, 900 x = 60, 120, 180, 240, 300
