Nội dung bài giảng
Chứng minh rằng:
a) \(\sin \alpha + \cos \alpha = \sqrt 2 \sin (\alpha + {\pi \over 4})\)
b) \(\sin \alpha - \cos \alpha = \sqrt 2 \sin (\alpha - {\pi \over 4})\)
c) \(\tan ({\pi \over 4} - \alpha ) = {{1 - \tan \alpha } \over {1 + \tan \alpha }}\,\,(\alpha \ne {\pi \over 2} + k\pi ;\,\,\alpha \ne {{3\pi } \over 4} + k\pi )\)
d) \(\tan ({\pi \over 4} + \alpha ) = {{1 + \tan \alpha } \over {1 - \tan \alpha }}\,\,(\alpha \ne {\pi \over 2} + k\pi ;\,\,\alpha \ne {\pi \over 4} + k\pi )\)
Đáp án
a) Ta có:
\(\eqalign{
& \sqrt 2 \sin (\alpha + {\pi \over 4}) = \sqrt 2 (\sin \alpha \cos {\pi \over 4} + \sin {\pi \over 4}\cos \alpha ) \cr
& = \sqrt 2 (\sin \alpha {{\sqrt 2 } \over 2} + {{\sqrt 2 } \over 2}\cos \alpha ) \cr
& = \sin \alpha + \cos \alpha \cr} \)
b) Ta có:
\(\eqalign{
& \sqrt 2 \sin (\alpha - {\pi \over 4}) = \sqrt 2 (\sin \alpha \cos {\pi \over 4} - \sin {\pi \over 4}\cos \alpha ) \cr
& = \sin\alpha - \cos \alpha \cr} \)
c) Ta có:
\(\tan ({\pi \over 4} - \alpha ) = {{\tan {\pi \over 4} - \tan \alpha } \over {1 + \tan {\pi \over 4}\tan \alpha }} = {{1 - \tan \alpha } \over {1 + \tan \alpha }}\,\)
d) Ta có:
\(\tan ({\pi \over 4} + \alpha ) = {{\tan {\pi \over 4} + \tan \alpha } \over {1 - \tan {\pi \over 4}\tan \alpha }} = {{1 + \tan \alpha } \over {1 - \tan \alpha }}\,\,\)