I know someone asked for calc help but now i rlyyy need some trig help... For Question 8, I can't seem to understand what she wants. Notice that sin = -1/2... but sin is in Q2, right? ISn't tat impossible? Did she make a mistake? For Question 9, I can't figure out how you know which quadrant 2theta is in? Because if you do 90<x<180 you get 180<2x<360... which is Q 3 and 4.... I mean, hmm... idk...
By the way, I didn't get to use a calculator on this test. So I don't get how you do it! My friend helped me and we do the inverse of tangent to get it in Q4, but when you do it like 2tanx / 1-tan^2(x) you get a positive?? :s ---------- Post added 15th Nov 2011 at 10:19 PM ---------- Is it working now?
Picture is showing now yeah. Haven't taken trig since freshman year in high-school. Gonna take me a little while to remember it, but I'll try and dig through my notes.
B is correct. sin (theta) = 20/29 (theta) = sin^-1 (20/29) (theta) = 43.6 degrees. HOWEVER, angle is in second quadrant. i.e. 136.4 degrees. And go from there. East stuff, I do Maths C. Ladies and men, please form a y = mx + c.
What I do not understand is, if you find tangent based on sin=20/29 (like she showed us how to do), you get tanx = 20/21... which when you plug into 2tanx/(1-tan^2(x)) Is 2(20/21) / (1-(20/21)^2) Which is.... (40/21) / (441/441 - 400/441) = (840/441) / (41/441) = 840/41 But that's postive... Wait..... tanx should be -20/29.... which makes it negative... okay... i get it :S But for number 8 she made an error right?
There's a typo in Question 8. 3(pi)/2 < A < (pi) Means: 3(pi)/2 < (pi) Which is obviously wrong. Assuming they meant this: (pi) < A < 3(pi)/2, you get the angle of 210. (Working in degrees here, radians look ugly). B is around 165.52 degrees (cba exact value), but all in all you get sin 44.48 = 0.7006. Subtract this from D, and it's pretty much correct.
Oh! Thanks for pointing that out... for some reason I kept getting 3pi/2 as 135 ugg idk why hahaha. So it's 270? So it's in Quadrant 3..... i think i asked her this when we were taking the test, I said "Is this right...?" And she said yes but guess she didn't get what I was saying lol. So apparently I was not thinking for either of these two problems. I guess I'll have to redo that... not sure what u mean by "Subtract this from D, and it's pretty much correct." Thanks so much for the help guys!
How'd you do? On this test, the average was a 54... The highest was an 88. I think/thought I did decent. She said she's not giving us our scores back until thursday, and if we want we can complete the test for an extra 15 points if we get them all correct when re-doing it. :/ Haha thanks anyways! I'll let you know!
I'm talking about Option D in multiple choice (i.e. getting an exact value without having to deal with surds etc.)
Not sure if you learnt this but ... sin (A+B) = sinAcosB + cosAsinB sin (A-B) = sinAcosB - cosAsinB cos (A+B) = cosAcosB - sinAsinB cos (A-B) = cosAcosB + sinAsinB tan (A+B) = (tan A + tan B)/(1-tanAtanB) tan (A-B) = (tan A - tan B)/(1+tanAtanB) Exactly how these identities are derived I cannot remember anymore haha. But using sin(A-B) = sinAcosB - cosAsinB and using a right angled triangle and assuming that it is 3pi/2 > A > pi, sin(A-B) = (-1/2)(-sqrt 15 /4) - (-sqrt 3 /2)(1/4) = sqrt 15 /8 + sqrt 3 /8 = (sqrt 15 + sqrt 3) / 8 which is option D
