`sin^2 25^o + sin^2 65^o`

`=cos^2 65^o + sin^2 65^o`

=1`

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`***` Áp dụng công thức lượng giác: `sin^2 \alpha +cos^2 \alpha =1`

Đề cs sai 0 nhỉ phải là `sin^2 25^o + sin^2 65^o`

 Hoặc `sin^2 35^o + sin^2 55^o` chứ

b: \(\cos30^0=\dfrac{\sqrt{3}}{2}\)

Mà\(\left\{{}\begin{matrix}sin32^o=cos58^o\\sin^210^o=cos^280^o\\tan43^o=cot47^0\end{matrix}\right.\)

=> sin 32^o - cos 58^o + 5(sin²10^o + sin²80^o) + 2.\(\dfrac{tan43^o}{cot47^o}\) 

       = cos 58^o - cos 58^o + 5(cos²80^o + sin²80^o) + 2. \(\dfrac{cot47^o}{cot47^o}\)

Mà cos²80^o + sin²80^o = 1

=> cos 58^o - cos 58^o + 5(cos²80^o + sin²80^o) + 2. \(\dfrac{cot47^o}{cot47^o}\)

      = 5.1 + 2.1

      = 5 + 2

      = 7

Vậy sin 32^o - cos 58^o + 5(sin²10^o + sin²80^o) + 2.\(\dfrac{tan43^o}{cot47^o}\) = 7

Ta có: \(\sin {70^o} = \cos {20^o};\;\cos {110^o} =  - \cos {70^o} =  - \sin {20^o}\)

\(\begin{array}{l} \Rightarrow A = {(\sin {20^o} + \cos {20^o})^2} + {(\cos {20^o} - \sin {20^o})^2}\\ = ({\sin ^2}{20^o} + {\cos ^2}{20^o} + 2\sin {20^o}\cos {20^o}) + ({\cos ^2}{20^o} + {\sin ^2}{20^o} - 2\sin {20^o}\cos {20^o})\\ = 2({\sin ^2}{20^o} + {\cos ^2}{20^o})\\ = 2\end{array}\)

Ta có: \(\tan {110^o} =  - \tan {70^o} =  - \cot {20^o};\;\cot {110^o} =  - \cot {70^o} =  - \tan {20^o}.\)

\( \Rightarrow B = \tan {20^o} + \cot {20^o} + ( - \cot {20^o}) + ( - \tan {20^o}) = 0\)

\(=\left(\sin^212^0+\sin^278^0\right)+\left(\sin^270^0+\sin^220^0\right)-\left(\sin^235^0+\sin^255^0\right)+\sin^230^0\)

\(=1+1-1+\dfrac{1}{4}=1+\dfrac{1}{4}=\dfrac{5}{4}\)

A=(sin²10+sin²80)+(sin²20+sin70)+(sin²30+sin²60)+(sin²40+sin²50)

Lại có: sin80=cos10; sin70=cos20; sin60=cos30; sin50=cos40

=> sin²80=cos²10; sin²70=cos²20; sin²60=cos²30; sin²50=cos²40

=>A=(sin²10+cos²10)+(sin²20+cos²20)+(sin²30+cos²30)+(sin²40+cos²40)

=>A=1+1+1+1=4

A = 4

học tốt nha

A=4

HỌC TỐT!!!!