Suppose f(x) = sin(pi*cosx) On

Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
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hey, i would really appreciate some help solving for x when:sin2x=cosx Use the identity sin 2A = 2sinAcosA so: sin 2x = cos x
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I have a question relating to limits that I solvedlim(x-->0) (1-cosx)/2x^2 I multiplied the numerator and denominator by
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Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
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sin2x+cosx=0 , [-180,180)= 2sinxcosx+cosx=0 = cosx(2sinx+1)=0 cosx=0 x1=cos^-1(0) x1=90 x2=360-90 x2=270 270 doesn't fit in
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Solve this equation fo rx in the interval 0<=x<=3603sinxtanx=8 I would do it this way: sinxtanx = 8/3 sinx(sinx/cosx)=8/3
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Suppose that X is an angle in standard position and that X' is it's reference angle. If cosX = -4/ squareroot of 17. find the
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Suppose f(x) = sin(pi*cosx) On any interval where the inverse function y = f ^–1(x) exists, the derivative of f ^–1(x) with
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2. asked by Anonymous
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Suppose f(x) = sin(pi*cosx) On any interval where the inverse function y = f –1(x) exists, the derivative of f –1(x) with
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I can't seem to prove these trig identities and would really appreciate help:1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1:
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2. asked by Heather
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