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Câu hỏi: Chứng minh đẳng thức lượng giác:
\(\begin{array}{l}a) sin(\alpha + \beta ).sin(\alpha - \beta ) = si{n^2}\alpha - si{n^2}\beta \\b) co{s^4}\alpha - co{s^4}\left( {\alpha - \frac{\pi }{2}} \right) = cos2\alpha \end{array}\)
\(\begin{array}{l}a) sin(\alpha + \beta ).sin(\alpha - \beta ) = si{n^2}\alpha - si{n^2}\beta \\b) co{s^4}\alpha - co{s^4}\left( {\alpha - \frac{\pi }{2}} \right) = cos2\alpha \end{array}\)
Phương pháp giải
Áp dụng:
a, \(\sin a\sin b = \frac{1}{2}\left[ {\cos \left( {a - b} \right) - \cos \left( {a + b} \right)} \right]\)
b, \(\cos \left( {\frac{\pi }{2} - \alpha } \right) = \sin \alpha \)
Lời giải chi tiết
\(a) sin(\alpha + \beta ).sin(\alpha - \beta ) = \frac{1}{2}.\left[ {cos\left( {\alpha + \beta - \alpha + \beta } \right) - cos\left( {\alpha + \beta + \alpha - \beta } \right)} \right]\)
\(\begin{array}{l} = \frac{1}{2}.(cos2\beta - cos2\alpha ) = \frac{1}{2}.(1 - 2si{n^2}\beta - 1 + 2si{n^2}\alpha )\\ = si{n^2}\alpha - si{n^2}\beta \end{array}\)
\(\begin{array}{l}b) co{s^4}\alpha - co{s^4}\left( {\alpha - \frac{\pi }{2}} \right) = co{s^4}\alpha - si{n^4}\alpha \\ = (co{s^2}\alpha + si{n^2}\alpha )(co{s^2}\alpha - si{n^2}\alpha )\\ = co{s^2}\alpha -si{n^2}\alpha = cos2\alpha .\end{array}\)
Áp dụng:
a, \(\sin a\sin b = \frac{1}{2}\left[ {\cos \left( {a - b} \right) - \cos \left( {a + b} \right)} \right]\)
b, \(\cos \left( {\frac{\pi }{2} - \alpha } \right) = \sin \alpha \)
Lời giải chi tiết
\(a) sin(\alpha + \beta ).sin(\alpha - \beta ) = \frac{1}{2}.\left[ {cos\left( {\alpha + \beta - \alpha + \beta } \right) - cos\left( {\alpha + \beta + \alpha - \beta } \right)} \right]\)
\(\begin{array}{l} = \frac{1}{2}.(cos2\beta - cos2\alpha ) = \frac{1}{2}.(1 - 2si{n^2}\beta - 1 + 2si{n^2}\alpha )\\ = si{n^2}\alpha - si{n^2}\beta \end{array}\)
\(\begin{array}{l}b) co{s^4}\alpha - co{s^4}\left( {\alpha - \frac{\pi }{2}} \right) = co{s^4}\alpha - si{n^4}\alpha \\ = (co{s^2}\alpha + si{n^2}\alpha )(co{s^2}\alpha - si{n^2}\alpha )\\ = co{s^2}\alpha -si{n^2}\alpha = cos2\alpha .\end{array}\)