1) 4- csc? 0 = 0 %3D 3) sec? 0-2 = 0 %3D 5) -3 = -3csc² 0 + 1
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- 2cosθ+1=0 solution set cotθ-1=0 solution set 4secθ+1=9 solution set sec30/2=2 solution set sinθ= sqrt3/2 first 6 solutions sin(2θ)=sqrt2/2 first 6 solutions sinθ/2=-1/2 first 6 solutions sin^2θ-cos^2θ=1+cosθ solution set sin^2θ=6(cos(-θ)-1 colution set 5cosθ=5sin(-θ) solution setif cos alpha + 2 cos beta + 3 cos gamma = 0 ; sin alpha + 2 Sin beta + 3 sin gamma = 0 then cos 3 alpha + 8 cos 3 beta + 27 cos 3 gamma = ? cos (2alpha - beta - gamma ) + 8 cos (2beta - gamma - alpha) + 27 cos (2y - alpha - beta) = ? sin 3 alpha + 8 din 3 beta + sin 3 gamma = ? sin (2 alpha - beta - gamma) + 8 din (2beta - gamma - alpha) + 27 sin (2gamma - alpha - beta) =?Which equation out of the seven is used to find the value of y for the figure shown and which equation out of the seven is true for: if tan 22.5° =w/12 Equation 1: Sin 36= 10/y Equation 2: Tan 36°= 10/y Equation 3: Cos 36°= 10/y Equation 4: Tan 36°= y/10 Equation 5: W=12 tan 22.5° Equation 6: W= tan 22.5°/12 Equation 7: W= 12/tan 22.5°
- If (Sec A + tan A ) (SecB + tanB) (Sec C + tanC) = (Sec A -tanA) (Sec B - tan B) (Sec C - tan C ) . Prove that each of the side is equal to ±1 • Best Of Luck !!• Spammers Stay Away !!Find all solutions to 2sin(θ)=√32sin(θ)=radical 3 on the interval 0≤θ<2π?Show how it derived to 4sin(npi)-4npi cos(npi)-2n^2pi^2 sin(npi)/pin^3
- why is ((1/pi)((cosnpi))/(1+2n))+((cos(npi)/(1-2n))=2*(-1)^(n+1)/pi(4n^2-1)Find the areas of the regions . 1. Inside one leaf of the four-leaved rose r = cos 2u 2. Inside one leaf of the three-leaved rose r = cos 3uThe points A(1, 0, 2), B(2, 0, 1) and C(1, 2, k) from a triangle of area √6. Determine the value(s) of K for which this is true