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http://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.771795.html this might help
Well I think it is supposed to be solved using trig identities
cosx = 1- 2 sin^2 (x/2) cosx - 1 = -2sin^2(x/2) -0.5cosx +0.5 = sin^2(x/2) cosx = -4/5 (2nd quarter) sinx/2 = sqrt(-0.5 * -4/5 + 0.5) = 3sqrt(10)/10 |dw:1446760327411:dw| tanx/2 = -3sqrt(10)/sqrt(10) = -3 (2nd quarter) Any mistakes ?
I get tan x = -3/4 since cos= -4/5 and sin=3/5 so (3/5)/(-4/5)=-3/4
but then since its tan x/2 I think you divide that by two so tan x/2=(-3/4)/2 tanx/2=-3/8
sin(60) = sqrt(3)/2 so sin(30) = sqrt(3)/ 4 ??? of course not , you have to use half angle identity
cosx = 1- 2 sin^2 (x/2) cosx = 2cos^2(x/2) - 1 cosx = cos^2x/2 - sin^2x/2 sinx = 2 sinx/2 cosx/2
tanx = 2tanx/2 / 1- tan^2x/2
That's the double angle identities , you can derive the half angle ones from them.
Okay yeah I have those in my notes. But I dont understand where and how they are used when trying to solve these kind of questions.
??? liek I see that you plugged them in but how did you know that and when do you use this type of identity versus other ones?
It's easy to notice when to use them, if it's asking for ex for sin2x and the given is sinx so you have to use this identities , sin4x and you 're given sin2x or sinx or sin0.5x it asks for sin0.5x and you have sin4x and so on.
always think of them when the angles are double or half in either sin or cos or tan or cot or cosec or sec xD
okay well it looks like tan x = 3 not -3
I am still very confused with this math. I am going to try another question and see if it will help my understanding
im pretty sure its x=3
@RCCB , x is in quadrant 2, is in the second quarter so it's negative.
180 >= x >= 90 (divide the whole equation over 2) 90 >= x/2 >= 45 (first quarter ) so x/2 in the 1st , x in the 2nd