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Bài 8. Chứng minh các biểu thức sau không phụ thuộc vào \(x\)
a) \(A = \sin ({\pi \over 4} + x) – \cos ({\pi \over 4} – x)\)
b) \(B = \cos ({\pi \over 6} – x) – \sin ({\pi \over 3} + x)\)
c) \(C = {\sin ^2}x + \cos ({\pi \over 3} – x)cos({\pi \over 3} + x)\)
d) \(D = {{1 – \cos 2x + \sin 2x} \over {1 + \cos 2x + \sin 2x}}.\cot x\)
a)
\(\eqalign{
& A = \sin ({\pi \over 4} + x) – \cos ({\pi \over 4} – x) \cr
& = \sin {\pi \over 4}\cos x + \cos {\pi \over 4}\sin x – \cos x\cos {\pi \over 4} – \sin {\rm{x}}\cos {\pi \over 4} \cr
& = {{\sqrt 2 } \over 2}(\cos x + \sin x – \cos x – {\mathop{\rm s}\nolimits} {\rm{inx}}) = 0 \cr} \)
Không phụ thuộc vào \(x\)
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b)
\(\eqalign{
& B = \cos ({\pi \over 6} – x) – \sin ({\pi \over 3} + x) \cr
& = \cos {\pi \over 6}{\mathop{\rm cosx}\nolimits} + \sin {\pi \over 6}sinx – sin{\pi \over 3}\cos x – \cos {\pi \over 3}\sin x \cr
& = \cos x(\cos {\pi \over 6} – sin{\pi \over 3}) + {\mathop{\rm s}\nolimits} {\rm{inx}}(\sin {\pi \over 6} – \cos {\pi \over 3}) = 0 \cr} \)
c)
\(\eqalign{
& C = {\sin ^2}x + \cos ({\pi \over 3} – x)cos({\pi \over 3} + x) \cr
& = {\sin ^2}x + \left[ {\cos {\pi \over 3}\cos x + \sin {\pi \over 3}\sin x} \right]\left[ {\cos {\pi \over 3}\cos x – \sin {\pi \over 3}\sin x} \right] \cr
& = {\sin ^2}x + {\cos ^2}{\pi \over 3}{\cos ^2}x – {\sin ^2}{\pi \over 3}{\sin ^2}x \cr
& = {\sin ^2}x + {1 \over 4}{\cos ^2}x – {3 \over 4}{\sin ^2}x = {1 \over 4}({\cos ^2}x + {\sin ^2}x) = {1 \over 4} \cr} \)
d) \(D = {{2{{\sin }^2}x + 2\sin x\cos x} \over {2{{\cos }^2}x + 2\sin x\cos x}}\cot x = {{{\mathop{\rm s}\nolimits} {\rm{inx}}} \over {{\mathop{\rm cosx}\nolimits} }}.{{{\mathop{\rm cosx}\nolimits} } \over {{\mathop{\rm s}\nolimits} {\rm{inx}}}} = 1\)