Lời giải:
a)
\((\sin c+\cos c)^2+(\sin ^2c-\cos ^2c)^2\)
\(=(\sin c+\cos c)^2+[(\sin c-\cos c)(\sin c+\cos c)]^2\)
\(=(\sin c+\cos c)^2[1+(\sin c-\cos c)^2]\)
\(=(\sin ^2c+2\sin c\cos c+\cos ^2c)(1+\sin ^2c-2\sin c\cos c+\cos ^2c)\)
\(=(1+2\sin c\cos c)(2-2\sin c\cos c)\) (nhớ là \(\sin ^2c+\cos ^2c=1\) )
\(=(1+\sin 2c)(2-\sin 2c)\)
b)
\(\sin ^6c+\cos ^6c+3\sin ^2c\cos ^2c\)
\(=(\sin ^2c)^3+(\cos ^2c)^3+3\sin ^2c\cos ^2c\)
\(=(\sin ^2c+\cos ^2c)(\sin ^4c-\sin ^2c\cos ^2c+\cos ^4c)+3\sin ^2c\cos ^2c\)
\(=\sin ^4c-\sin ^2c\cos ^2c+\cos ^4c+3\sin ^2c\cos ^2c\)
\(=\sin ^4c+2\sin ^2c\cos ^2c+\cos ^4c=(\sin ^2c+\cos ^2c)^2=1^2=1\)
Vậy biểu thức không phụ thuộc vào c
