giải giúp em vs ạ.
\(a,\sin^243^0+\sin^244^0+\sin^245^0+\sin^246^0+\sin^247^0\\ =\cos^247^0+\cos^246^0+\sin^245^0+\sin^246^0+\sin^247^0\\ =\left(\cos^247^0+\sin^247^0\right)+\left(\cos^246^0+\sin^246^0\right)+\sin^245^0=1+1+0,5=2,5\)
\(b,=\sin^289^0+\sin^288^0+...+\sin^246^0+\cos^245^0+\cos^246^0+...+\cos^288^0+\cos^289^0\\ =\left(\sin^289^0+\cos^289^0\right)+\left(\sin^288^0+\cos^288^0\right)+...+\left(\sin^246^0+\cos^246^0\right)+\cos^245^0\\ =1+1+..+1+0,5=44,5\)
\(c,=2\sqrt{\dfrac{\cot68^0}{\cot68^0}}+\cot34^0\cdot\cot56^0=2+1=3\)
a)=\(sin^243+sin^244+sin^245+cos^244+cos^243\)
=\(2+sin^245=\dfrac{5}{2}\)
b)=\(cos^21+cos^22+...+cos^245+sin^244+sin^243+...+sin^21\)
=\(\left(cos^21+sin^21\right)+\left(cos^22+sin^22\right)+...+cos^245\)
=44+\(cos^245\)=44,5
c)=\(\sqrt{\dfrac{4cot68}{cot68}}+1=\sqrt{4}+1=2+1=3\)
