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\(=\left(\sin^212^0+\sin^278^0\right)+\left(\sin^270^0+\sin^220^0\right)-\left(\sin^235^0+\sin^255^0\right)+\sin^230^0\)
\(=1+1-1+\dfrac{1}{4}=1+\dfrac{1}{4}=\dfrac{5}{4}\)
\(ADCT:\sin^2\alpha+\cos^2\alpha=1\)
\(A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0\)
\(A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}\)
\(A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)
Câu b lm tương tự
a, \(\cos^215+\cos^225+\cos^235+\cos^245+\sin^235+\sin^225+\sin^215\)
=\(\left(\cos^215+\sin^215\right)+\left(\cos^225+\sin^225\right)+\left(\cos^235+\sin^235\right)+\cos^245\)
=\(1+1+1+\frac{1}{2}=\frac{7}{2}\)
b.\(\sin^210-\sin^220-\sin^230-\sin^240-\cos^240-\cos^220+\cos^210\)
=\(\left(\sin^210+\cos^210\right)-\left(\sin^220+\cos^220\right)-\left(\sin^240+\cos^240\right)-\sin^230\)
=\(1-1-1-\frac{1}{4}=-\frac{5}{4}\)
c,\(\sin15+\sin75-\sin75-\cos15+\sin30=\sin30=\frac{1}{2}\)
Áp dụng tính chất 2 góc phụ nhau nha bạn.
\(\sin^235^0+\tan22^0-\dfrac{\cot13^0}{\tan77^0}-\cot68^0+\sin^255\)
\(=\left(\sin^235^0+\sin^255^0\right)+\left(\tan22^0-\cot68^0\right)-\dfrac{\cot13^0}{\tan77^0}\)
\(=\left(\sin^235^0+cos^235^0\right)+\left(\tan22^0-\tan22^0\right)-\dfrac{\cot13^0}{\cot13^0}\)
\(=1+0-1=0\)
Ta có \(\sin x=\cos\left(90^0-x\right)\)
\(\Rightarrow M=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin^245^0\)
\(=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\sin^245^0\)
\(=1+1+1+\left(\frac{\sqrt{2}}{2}\right)^2=3+\frac{1}{2}=\frac{7}{2}\)
a/ \(\tan40.\cot40+\frac{\sin50}{\cos40}\)
\(=1+\frac{\cos40}{\cos40}=1+1=2\)
\(=cos^2\left(90^o-35^o\right)+sin^255^o+cot\left(90^o-17^o\right)-cot73^o-\frac{tan\left(90^o-47^o\right)}{tan53^o}\)
\(=cos^255^o+sin^255^o+cot73^o-cot73^o-\frac{tan53^o}{tan53^o}\)
\(=1-1=0\)