14 câu Trắc nghiệm Ôn tập Toán 10 Chương 6 có đáp án

Đáp án A

Ta có:

cotα=3⇒cosαsinα=3⇔cosα=3sinα

Thay vào biểu thức đề bài, ta được:

3sinα−2.3sinα12sin3α+43sinα3=−3sinα120sin3α=−140.1sin2α=−1401+cot2α=−1401+32=−14

Đáp án D

Ta có:

tan2α2=sin2α2cos2α2=1−cosα21+cosα2=1−cosα1+cosα⇔1−cosα=tan2α21+cosα

Đặt tanα2=t thì cosα=1−t21+t2

Với t = 2 ⇒cosα=1−41+4=−35

Suy ra B=1+5−353−2−35=−2215=−1021

Đáp án B

A=tan2π24+cot2π24=1cos2π24−1+1sin2π24−1=1cos2π24.sin2π24−2=4sin2π12−2=81−cosπ6−2=81−32−2=162−3−2=16−4+232−3=12+232−3

Đáp án B

Ta có: sina−cosa=34 suy ra (sina−cosa)2=916

⇔sin2a+cos2a−2sinacosa=916⇔1−sin2a=916⇔sin2a=1−916⇔sin2a=716

Đáp án B

sin4a−cos4a=sin2a−cos2asin2a+cos2a=sin2a−cos2a.1=−cos2a

nên A sai

2sin4a+cos4a=2sin2a+cos2a2−2sin2a.cos2a=21−2.14sin22a=2−sin22a

nên B đúng

sina−cosa2=1−2sina.cosa=1−sin2anên C sai

sin2a+cos2a3=1 và 1+2sin4a.cos4a=1+2.12sin2a4=1+18sin42a nên D sai

Đáp án A

G=cos2π6+cos22π6+...+cos25π6+cos2π=cos2π6+cos22π6+cos24π6+cos25π6+cos2π2+cos2π=cos2π6+cos2π3+cos22π6+cos2π6+1=2cos2π6+cos22π6+1=2cos2π6+cos2π6+1=3

Đáp án A

E=sinπ5+sin2π5+...+sin9π5=sinπ5+sin9π5+sin2π5+sin8π5+...+sin4π5+sin6π5+sin5π5=sinπ5+sin2π−π5+sin2π5+sin2π−2π5+...+sin4π5+sin2π−4π5+sinπ=sinπ5+sin−π5+sin2π5+sin−2π5+...+sin4π5+sin−4π5+0=sinπ5−sinπ5+sin2π5−sin2π5+...+sin4π5−sin4π5= 0 + 0 + … + 0 = 0

Đáp án A

sin8x+cos8x=sin4x+cos4x2−2sin4x.cos4x=1−2sin2x.cos2x−18sin42x=1−12sin22x.2−18sin42x=1−sin22x+18sin42x=1−1−cos4x2+181−cos4x22=1−1−cos4x2+1321−2cos4x+1+cos8x22=3564+716cos4x+164cos8x⇒a=35,b=7,c=1⇒a−5b+c=1

Đáp án C

Ta có:

0<α<900⇔0<α2<450⇒0<sinα2<22⇔0<x−12x<22⇔x>0sin2α2+cos2α2=1⇒cosα2=1−sin2α2vì 0<α2<450⇔cosα2=x+12x⇒tanα2=x−1x+1tanα=2tanα21−tan2α2=2x−1x+11−x−1x+1=x2−1⇒cotα=1tanα=1x2−1=x2−1x2−1

Đáp án C

Ta có:

cosA+cosB+cosC+1=2cosA+B2cosA−B2+2sin2C2=2cosπ2−C2cosA−B2+2sin2C2=2sinC2cosA−B2+2sin2C2=2sinC2cosA−B2+2sinC2=2sinC2cosA−B2+cosA+B2=2sinC2.2cosA2cosB2=4cosA2cosB2sinC2

sinA+sinB+sinC=2sinA+B2cosA−B2+2sinC2cosC2=2sinπ2−C2cosA−B2+2sinC2cosC2=2cosC2cosA−B2+2sinC2cosC2=2cosC2cosA−B2+sinπ2−A+B2=2cosC2cosA−B2+cosA+B2=2cosC2.2cosA2cosB2=4cosA2cosB2cosC2⇒cosA+cosB+cosC+1sinA+sinB+sinC=4cosA2cosB2sinC24cosA2cosB2cosC2=tanC2⇒tanC2=1⇔C2=450⇔C=900

⇒ΔABC vuông tại C

Đáp án C

+)sinxcos3x−cosxsin3x=sinx3cosx+cos3x4−cosx.3sinx−sin3x4=34sinxcosx+14sinxcos3x−34sinxcosx+14sin3xcosx=14sinxcos3x+sin3xcosx=14sinx+3x=sin4x4

+)sin4x+cos4x=sin2x+cos2x2−2sin2xcos2x=1−12sin22x=1−12−1−cos4x2=3+cos4x4

+)cot2x+tan2x=cos2xsin2x+sin2xcos2x=cos4x+sin4xsin2xcos2x=3+cos4x414sin22x=3+cos4x121−cos4x=2cos4x+61−cos4x

Đáp án C

sina−cosa=15⇒sina−cosa2=125⇔sin2a−2sinacosa+cos2a=125⇔1−sin2a=125⇔sin2a=2425

Ta có:

sin22a+cos22a=1⇒24252+cos22a=1⇔cos22a=49625⇔cos2a=±725Mà 1350<a<1800⇔2700<2a<3600⇒cos2a>0⇒cos2a=725tan2a=sin2acos2a=2425725=247

Đáp án A

2cos2x−14tanπ4−xsin2π4+x=cos2x4.sinπ4−xcosπ4−x.1−cosπ2+2x2=cos2x2.2cosx−sinx2cosx+sinx.1+sin2x=cos2x2.cosx−sinxcosx+sinx.sinx+cosx2=cos2x2cosx−sinxsinx+cosx=cos2x2cos2x−sin2x=cos2x2cos2x=12

Đáp án B

B=sin3a3+3sin3a32+32sin3a33+...+3n−1sin3a3n=3sina3−sina4+3.3sina32−sina34+32.3sina33−sina324+...+3n−1.3sina3n−sina3n−14=14.−sina+3sina3−3sina3+32sina32−32sina32+33sina33−...−3n−1sina3n−1+3nsina3n=143nsina3n−sina=3nsina3n−sina4