Bi tp cung v gc lng gic

Bi tp cung v gc lng gicPhn 1: Bin i lng gicBi 1: CM cc ng thc sau:

a, sin4x + cos4x = 1- 2sin2xcos2x = 1 sin22xb, sin6x + cos6x = 1-3sin2xcos2x = 1- sin22x

c,

Bi 2: Rt gn biu thc

Bi 3: Tnh gi tr cc biu thc sau:a, Cho sinx + cosx = 5/4. Tnh A = sinxcosx

B = sinx cosxC= sin3x cos3x

b, Cho tanx cotx = m. Tnh A = tan2x cot2xB= tan2x + cot2xC= tan3x + cot3x

Bi 4: CMR cc biu thc sau khng ph thuc vo x

Bi 5: Rt gn

Bi 6: Tnh gi tr cc biu thc:

Phn 2: H thc lng trong tam gicBi 1: CMR trong tam gic ta lun c:

a, sinA + sinB + sinC = 4 cos(A/2) cos(B/2) cos(C/2) b, cosA+cosB+cosC = 1+4sin(A/2)sin(B/2)sin(C/2)

c, sin2A+sin2B+sin2C = 2+ cosAcosBcosCd, tanA + tanB + tanC = tanAtanBtanC < tam gic ko vung>

e, tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2) = 1

f, cotAcotB + cotBcotC + cotCcotA = 1g,

Bi 2: CMR iu kin cn v tam gic ABC vung l:

a, cos2A + cos2B + cos2C = -1b, sinA + sinB + sinC + 1 = cosA + cosB + cosC

c, sinB + sinC = cosB + cosC

d, sin2B + sin2C = 4 sinBsinCe,

Bi 3: CMR tam gic ABC cn nu:

a, c = 2a.cosB

b, tanA + 2tanB = tanA.tan2B

c, sinC = 2sinAsinB.tan(C/2)

d, asin(B-C) + bsin(C-A) = 0

e, tanA + tanB = 2cot(C/2)

Bi 4: CMR : Nu 0x,y ( th

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