Trig. Find cot x/2, given tan x= -√5/2 with 90° having a lot of trouble with this problem. Find cot x/2, given tan x= -√5/2 with 90°
Trig help? find csc 13pi/8 using the half-angle identity.?
csc(13pi/8)= 1/sin(13pi/8)= -1/sin(5pi/8)-------(1) sin(5pi/4)=2sin(5pi/8)cos(5pi/8)=> -sqrt(2)/4= -sin(5pi/8)sqrt[1-sin^2(5pi/8)]=> 1/8=sin^2(5pi/8)[1-sin^2(5pi/8)]=> Let x=sin(5pi/8), then 1/8=x^2[1-x^2]=> 1/8=x^2-x^4=> 8x^4-8x^2+1=0=> x^2=[2+sqrt(2)]/4 (take the bigger value) x=sqrt[2+sqrt(2)]/2 (take +ve value) =>csc(13pi/8)=-1/sqrt[2+sqrt(2)]/2=> csc(13pi/8)=-sqrt(2+sqrt(2))(2-sqrt(2)...
Trig help? Given csc x calculate half angles?
you are given that sin Θ = - √35 / 6 and Θ is in 3rd quadrant ; cos Θ = - 1 / 6 use cos Θ = 2 cos² ( Θ / 2 ) - 1 ≡ 1 - 2 sin² (Θ / 2 ) to find the sine and cosine of Θ / 2 Note : Θ / 2 is in the 2nd quadrant
Help me answer Algebra 3-4/Pre-Calculus problems?
I'll start with number (5) and work backwards, to see how bored I get with the idea of answering 15 questions. (5) Since cot x = (cos x)/(sin x), we have sin x + (cos^2 x)/(sin x) = 1 / f(x) sin^2(x) + cos^2(x) = (sin x) / f(x) 1 = (sin x) / f(x) f(x) = sin x (4) sin (x + pi/3) = cos (pi/3) sin(x) + sin (pi/3) cos(x) = (1/2) sin(x) + {[sqrt(3)]/2} cos(x) A radical such as sqrt(3) is considered an "exact" number because it hasn't been rounded off. (3) For this one, I'm going to give answers without showing steps a. sin(x) b. tan^2(x) c. sec^2(x) d. 1 e. tan^2(x) (2) I'll just do "d" and you can try the others (d) cos (-7 pi/12) = cos (7 pi/ 12) = - cos (5 pi/ 12) = - sin (pi/ 12) Let's drop the minus sign for a moment while figuring out sin(pi/ 12). sin(pi/ 12) = sin[ (pi/3) - (pi/4) ] = sin (pi/3) cos (pi/4) - sin (pi/4) cos(pi/3) = (sqrt(3)/2) (sqrt(2)/2) - (sqrt(2)/2) (1/2) = [sqrt(6) - sqrt(2)] / 4 So finally, cos (-7 pi/12) = [ sqrt(2) - sqrt(6) ] / 4 Note that for 2a and 2b you'll have to use "half-angle" formulas, "difference" formulas won't do it. (1) I'll do (a), you try the others First note that sin(t) = sqrt(81-49)/9 = 4 sqrt(2) / 9 (a) cos (2t) = cos^2 (t) - sin^2 (t) = (7/9)^2 - 32/81 = 17/81
I need help in precalculus problems about trig formulas?
First let us write down our double angle formulae: sin 2u = 2sinu cosu cos 2u = 2cos^2 u -1 (we use this one since we know cos u) Now for sin 2u, we don't know sin u, but we can rewrite sin u as cosu*tanu (as tan u = sin u/cos u) Now tan u = 1/cot u So sin 2u = 2 sinu cosu = 2 cos^2 u / cot u = 2 (-2/3)^2 / -4 = -2/9 cos 2u = 2cos^2 u - 1 = 2(-2/3)^2 - 1 = 8/9 - 1 = -1/9 Half angle identities are below: Let t = tan(1/2 x) sin x = 2t/(1+t^2) cos x = (1-t^2)/(1+t^2) Let's concentrate on questions: We want sin165. Problem is 1/2 of 165 is not useful. So we need to think of other methods. One is the 'reverse' double angle formulae: cos330 = 2cos^2 (165) - 1 using 2x = 330 Now cos330 = cos(360-30) = cos(30) = 1/2 sqrt(3) so 1/2 sqrt(3) = 2cos^2(165) - 1 1/2 sqrt(3) + 1 = 2cos^2 (165) Then 1/4 sqrt(3) + 1/2 = cos^2 (165) cos(165) = - sqrt((1/4 sqrt(3) + 1/2)) = - 1/2 sqrt(sqrt(3) + 2) We can do the same for sin(165) using cos330 = 1 - 2 sin^2 (165)