Quay lại đáp án

sinA" role="presentation" style="font-size: 13.696px; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">sinA−−−−√sinA , cosB2" role="presentation" style="font-size: 13.696px; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">cosB2−−−−−√cosB2 , sinC" role="presentation" style="font-size: 13.696px; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">sinC−−−−√sinC => $\sqrt{sinA.sinC}=cos \frac{B}{2}$=> $sinA.sinC=cos^2 \frac{B}{2}$=> $-\frac{1}{2}(cos(A+C)-cos(A-C))=cos^2\frac{B}{2}$=> $cos(A-C)-cos(180-B)=1+cosB$=> $cos(A-C)+cosB=1+cosB$=> $cos(A-C)=1$ => $A=C$=> $\Delta ABC$ cân tại B

−−−> $\sqrt{sinA.sinC}=cos \frac{B}{2}$Do $\sqrt{sinA}, \sqrt{cos\frac{B}{2}}, \sqrt{sinC} $ lập thành CSN=> $sinA.sinC=cos^2 \frac{B}{2}$=> $-\frac{1}{2}(cos(A+C)-cos(A-C))=cos^2\frac{B}{2}$=> $cos(A-C)-cos(180-B)=1+cosB$=> $cos(A-C)+cosB=1+cosB$=> $cos(A-C)=1$ => $A=C$=> $\Delta ABC$ cân tại B

sinA" role="presentation" style="font-size: 13.696px; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">sinA−−−−√sinA , cosB2" role="presentation" style="font-size: 13.696px; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">cosB2−−−−−√cosB2 , sinC" role="presentation" style="font-size: 13.696px; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">sinC−−−−√sinC => $\sqrt{sinA.sinC}=cos \frac{B}{2}$=> $sinA.sinC=cos^2 \frac{B}{2}$=> $-\frac{1}{2}(cos(A+C)-cos(A-C))=cos^2\frac{B}{2}$=> $cos(A-C)-cos(180-B)=1+cosB$=> $cos(A-C)+cosB=1+cosB$=> $cos(A-C)=1$ => $A=C$

sinA" role="presentation" style="font-size: 13.696px; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">sinA−−−−√sinA , cosB2" role="presentation" style="font-size: 13.696px; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">cosB2−−−−−√cosB2 , sinC" role="presentation" style="font-size: 13.696px; display: inline; line-height: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; position: relative;">sinC−−−−√sinC => $\sqrt{sinA.sinC}=cos \frac{B}{2}$=> $sinA.sinC=cos^2 \frac{B}{2}$=> $-\frac{1}{2}(cos(A+C)-cos(A-C))=cos^2\frac{B}{2}$=> $cos(A-C)-cos(180-B)=1+cosB$=> $cos(A-C)+cosB=1+cosB$=> $cos(A-C)=1$ => $A=C$=> $\Delta ABC$ cân tại B