Đáp án:
$1)sin\alpha =\frac{\sqrt{15}}{4}\\
tan\alpha =\sqrt{15}\\
cot\alpha =\frac{\sqrt{15}}{15}$
2)
$ A=tanx$
Giải thích các bước giải:
$1) cos^2\alpha +sin^2\alpha =1\\
\Rightarrow sin^2\alpha =1-cos^2\alpha =1-\frac{1}{4}^2=\frac{15}{16}\\
\Rightarrow sin\alpha =\pm \sqrt{\frac{15}{16}}=\pm \frac{\sqrt{15}}{4}$
Vì $0<\alpha <\frac{\pi}{2}\Rightarrow sin\alpha =\frac{\sqrt{15}}{4}\\
tan\alpha =\frac{sin\alpha }{cos\alpha }=\frac{\frac{\sqrt{15}}{4}}{\frac{1}{4}}=\sqrt{15}\\
cot\alpha =\frac{1}{tan\alpha }=\frac{\sqrt{15}}{15}$
2)
$A=\frac{sin(x+\frac{\pi}{4})-cos(x+\frac{\pi}{4})}{sin(x+\frac{\pi}{4}+cos(x+\frac{\pi}{4}))} \\
=\frac{\frac{1}{cos(x+\frac{\pi}{4})}\left [sin(x+\frac{\pi}{4})-cos(x+\frac{\pi}{4}) \right ]}{\frac{1}{cos(x+\frac{\pi}{4})}.\left [sin(x+\frac{\pi}{4})+cos(x+\frac{\pi}{4}) \right ]} \\
=\frac{tan(x+\frac{\pi}{4})-1}{tan(x+\frac{\pi}{4})+1}\\
+)
tan(x+\frac{\pi}{4})+1\\
=\frac{tanx+tan\frac{\pi}{4}}{1-tanxtan\frac{\pi}{4}}+1\\
=\frac{tanx+1}{1-tanx}+1\\
=\frac{tanx+1+1-tanx}{1-tanx}\\
=\frac{2}{1-tanx}\\
+)
tan(x+\frac{\pi}{4})-1\\
=\frac{tanx+tan\frac{\pi}{4}}{1-tanxtan\frac{\pi}{4}}-1\\
=\frac{tanx+1}{1-tanx}-1\\
=\frac{tanx+1-1+tanx}{1-tanx}\\
=\frac{2tanx}{1-tanx}\\
\Rightarrow A=\frac{\frac{2tanx}{1-tanx}}{\frac{2}{1-tanx}}\\
=\frac{2tanx}{1-tanx}.\frac{1-tanx}{2}\\
=tanx$
3)
$VT=4sin(x+\frac{\pi}{3}).sin(x-\frac{\pi}{3})\\
=4.\left [ \frac{1}{2}\left ( cos\frac{2\pi}{3}-cos2x \right ) \right ]\\
=2.(\frac{-1}{2}-cos2x)\\
=-1-2cos2x\\
=-1-2(1-2sin^2x)\\
=-1-2+4sin^2\\
=4sin^2x-3=VP$
