Bài tập nâng cao · Bài 5
Bài tập nâng cao Giá trị lượng giác của góc lượng giác · Bài 5
Rút gọn các biểu thức sau:
a) $A=\frac{\sin\left( 328^{\circ} \right)\sin 958^{\circ}}{\cot 572^{\circ}}-\frac{\cos\left( -508^{\circ} \right)\cos\left( -1022^{\circ} \right)}{\tan\left( -212^{\circ} \right)}$.
b) $B=\frac{\sin\left( -234^{\circ} \right)-\cos 216^{\circ}}{\sin 144^{\circ}-\cos 126^{\circ}}\tan 36^{\circ}$.
c) $C=\cos 20^{\circ}+\cos 40^{\circ}+\cos 60^{\circ}+\ldots+\cos 160^{\circ}+\cos 180^{\circ}$.
d) $D=\cos^{2}10^{\circ}+\cos^{2}20^{\circ}+\cos^{2}30^{\circ}+\ldots+\cos^{2}180^{\circ}$.
e) $E=\sin 20^{\circ}+\sin 40^{\circ}+\sin 60^{\circ}+\ldots+\sin 340^{\circ}+\sin 360^{\circ}$.
a) $A=\frac{\sin\left( 328^{\circ} \right)\sin 958^{\circ}}{\cot 572^{\circ}}-\frac{\cos\left( -508^{\circ} \right)\cos\left( -1022^{\circ} \right)}{\tan\left( -212^{\circ} \right)}$.
b) $B=\frac{\sin\left( -234^{\circ} \right)-\cos 216^{\circ}}{\sin 144^{\circ}-\cos 126^{\circ}}\tan 36^{\circ}$.
c) $C=\cos 20^{\circ}+\cos 40^{\circ}+\cos 60^{\circ}+\ldots+\cos 160^{\circ}+\cos 180^{\circ}$.
d) $D=\cos^{2}10^{\circ}+\cos^{2}20^{\circ}+\cos^{2}30^{\circ}+\ldots+\cos^{2}180^{\circ}$.
e) $E=\sin 20^{\circ}+\sin 40^{\circ}+\sin 60^{\circ}+\ldots+\sin 340^{\circ}+\sin 360^{\circ}$.
Xem lời giải
Lời giải
a)Ta có:
$\sin 328^{\circ}=\sin\left( 360^{\circ}-32^{\circ} \right)=-\sin 32^{\circ}$; $\sin 958^{\circ}=\sin\left( 3.360^{\circ}-122^{\circ} \right)=-\sin 122^{\circ}=-\sin\left( 90^{\circ}+32^{\circ} \right)=-\cos 32^{\circ}$; $\cot 572^{\circ}=\cot\left( 2.360^{\circ}-148^{\circ} \right)=-\cot 148^{\circ}=-\cot\left( 180^{\circ}-32^{\circ} \right)=\cot 32^{\circ}$; $\cos\left( -508^{\circ} \right)=\cos 508^{\circ}=\cos\left( 360^{\circ}+148^{\circ} \right)=\cos 148^{\circ}=\cos\left( 180^{\circ}-32^{\circ} \right)=-\cos 32^{\circ};$$\cos 1022^{\circ}=\cos\left( 3.360^{\circ}-58^{\circ} \right)=\cos 58^{\circ}=\cos\left( 90^{\circ}-32^{\circ} \right)=\sin 32^{\circ}$; $\tan\left( -212^{\circ} \right)=-\tan\left( 180^{\circ}+32^{\circ} \right)=-\tan 32^{\circ}$.
Khi đó: $A=\frac{\sin 32^{\circ}\cos 32^{\circ}}{\cot 32^{\circ}}-\frac{\cos 32^{\circ}\sin 32^{\circ}}{\tan 32^{\circ}}=\sin^{2}32^{\circ}-\cos^{2}32^{\circ}=-\cos 64^{\circ}$.
b)Ta có:
$\sin\left( -234^{\circ} \right)=\sin\left( 126^{\circ}-360^{\circ} \right)=\sin 126^{\circ}=\sin\left( 90^{\circ}+36^{\circ} \right)=\cos 36^{\circ}$; $\cos 216^{\circ}=\cos\left( 180^{\circ}+36^{\circ} \right)=-\cos 36^{\circ}$; $\sin 144^{\circ}=\sin\left( 180^{\circ}-36^{\circ} \right)=\sin 36^{\circ}$; $\cos 126^{\circ}=\cos\left( 90^{\circ}+36^{\circ} \right)=-\sin 36^{\circ}$.
Khi đó: $B=\frac{\cos 36^{\circ}+\cos 36^{\circ}}{\sin 36^{\circ}+\sin 36^{\circ}}\tan 36^{\circ}=1$.
c)Ta có:
$C=\left( \cos 20^{\circ}+\cos 160^{\circ} \right)+\left( \cos 40^{\circ}+\cos 140^{\circ} \right)+\left( \cos 60^{\circ}+\cos 120^{\circ} \right)+\left( \cos 80^{\circ}+\cos 100^{\circ} \right)-1$
=$\left( \cos 20^{\circ}-\cos 20^{\circ} \right)+\left( \cos 40^{\circ}-\cos 40^{\circ} \right)+\left( \cos 60^{\circ}-\cos 60^{\circ} \right)+\left( \cos 80^{\circ}-\cos 80^{\circ} \right)-1$
=$-1$.
d)Ta có:
$D=\left( \cos^{2}10^{\circ}+\cos^{2}170^{\circ} \right)+\left( \cos^{2}20^{\circ}+\cos^{2}160^{\circ} \right)+\ldots+\left( \cos^{2}80^{\circ}+\cos^{2}100^{\circ} \right)+\cos^{2}90^{\circ}+1$
=$\left( \cos^{2}10^{\circ}+\cos^{2}10^{\circ} \right)+\left( \cos^{2}20^{\circ}+\cos^{2}20^{\circ} \right)+\ldots+\left( \cos^{2}80^{\circ}+\cos^{2}80^{\circ} \right)+1$
=$2\left( \cos^{2}10^{\circ}+\cos^{2}20^{\circ}+\cos^{2}30^{\circ}+\ldots+\cos^{2}80^{\circ} \right)+1$
=$2\left[ \left( \cos^{2}10^{\circ}+\cos^{2}80^{\circ} \right)+\left( \cos^{2}20^{\circ}+\cos^{2}70^{\circ} \right)+\ldots+\left( \cos^{2}40^{\circ}+\cos^{2}50^{\circ} \right) \right]+1$
=$2\left[ \left( \cos^{2}10^{\circ}+\sin^{2}10^{\circ} \right)+\left( \cos^{2}20^{\circ}+\sin^{2}20^{\circ} \right)+\ldots+\left( \cos^{2}40^{\circ}+\sin^{2}40^{\circ} \right) \right]+1$
= 2.4+1=9.
e)Ta có:
$E=\sin 20^{\circ}+\sin 40^{\circ}+\sin 60^{\circ}+\ldots+\sin 340^{\circ}+\sin 360^{\circ}$
=$\left( \sin 20^{\circ}+\sin 340^{\circ} \right)+\left( \sin 40^{\circ}+\sin 320^{\circ} \right)+\ldots+\left( \sin 160^{\circ}+\sin 200^{\circ} \right)+\sin 180^{\circ}+\sin 360^{\circ}$
=$2\sin 180^{\circ}\cos 160^{\circ}+2\sin 180^{\circ}\cos 140^{\circ}+\ldots+2\sin 180^{\circ}\cos 20^{\circ}$
= 0.
$\sin 328^{\circ}=\sin\left( 360^{\circ}-32^{\circ} \right)=-\sin 32^{\circ}$; $\sin 958^{\circ}=\sin\left( 3.360^{\circ}-122^{\circ} \right)=-\sin 122^{\circ}=-\sin\left( 90^{\circ}+32^{\circ} \right)=-\cos 32^{\circ}$; $\cot 572^{\circ}=\cot\left( 2.360^{\circ}-148^{\circ} \right)=-\cot 148^{\circ}=-\cot\left( 180^{\circ}-32^{\circ} \right)=\cot 32^{\circ}$; $\cos\left( -508^{\circ} \right)=\cos 508^{\circ}=\cos\left( 360^{\circ}+148^{\circ} \right)=\cos 148^{\circ}=\cos\left( 180^{\circ}-32^{\circ} \right)=-\cos 32^{\circ};$$\cos 1022^{\circ}=\cos\left( 3.360^{\circ}-58^{\circ} \right)=\cos 58^{\circ}=\cos\left( 90^{\circ}-32^{\circ} \right)=\sin 32^{\circ}$; $\tan\left( -212^{\circ} \right)=-\tan\left( 180^{\circ}+32^{\circ} \right)=-\tan 32^{\circ}$.
Khi đó: $A=\frac{\sin 32^{\circ}\cos 32^{\circ}}{\cot 32^{\circ}}-\frac{\cos 32^{\circ}\sin 32^{\circ}}{\tan 32^{\circ}}=\sin^{2}32^{\circ}-\cos^{2}32^{\circ}=-\cos 64^{\circ}$.
b)Ta có:
$\sin\left( -234^{\circ} \right)=\sin\left( 126^{\circ}-360^{\circ} \right)=\sin 126^{\circ}=\sin\left( 90^{\circ}+36^{\circ} \right)=\cos 36^{\circ}$; $\cos 216^{\circ}=\cos\left( 180^{\circ}+36^{\circ} \right)=-\cos 36^{\circ}$; $\sin 144^{\circ}=\sin\left( 180^{\circ}-36^{\circ} \right)=\sin 36^{\circ}$; $\cos 126^{\circ}=\cos\left( 90^{\circ}+36^{\circ} \right)=-\sin 36^{\circ}$.
Khi đó: $B=\frac{\cos 36^{\circ}+\cos 36^{\circ}}{\sin 36^{\circ}+\sin 36^{\circ}}\tan 36^{\circ}=1$.
c)Ta có:
$C=\left( \cos 20^{\circ}+\cos 160^{\circ} \right)+\left( \cos 40^{\circ}+\cos 140^{\circ} \right)+\left( \cos 60^{\circ}+\cos 120^{\circ} \right)+\left( \cos 80^{\circ}+\cos 100^{\circ} \right)-1$
=$\left( \cos 20^{\circ}-\cos 20^{\circ} \right)+\left( \cos 40^{\circ}-\cos 40^{\circ} \right)+\left( \cos 60^{\circ}-\cos 60^{\circ} \right)+\left( \cos 80^{\circ}-\cos 80^{\circ} \right)-1$
=$-1$.
d)Ta có:
$D=\left( \cos^{2}10^{\circ}+\cos^{2}170^{\circ} \right)+\left( \cos^{2}20^{\circ}+\cos^{2}160^{\circ} \right)+\ldots+\left( \cos^{2}80^{\circ}+\cos^{2}100^{\circ} \right)+\cos^{2}90^{\circ}+1$
=$\left( \cos^{2}10^{\circ}+\cos^{2}10^{\circ} \right)+\left( \cos^{2}20^{\circ}+\cos^{2}20^{\circ} \right)+\ldots+\left( \cos^{2}80^{\circ}+\cos^{2}80^{\circ} \right)+1$
=$2\left( \cos^{2}10^{\circ}+\cos^{2}20^{\circ}+\cos^{2}30^{\circ}+\ldots+\cos^{2}80^{\circ} \right)+1$
=$2\left[ \left( \cos^{2}10^{\circ}+\cos^{2}80^{\circ} \right)+\left( \cos^{2}20^{\circ}+\cos^{2}70^{\circ} \right)+\ldots+\left( \cos^{2}40^{\circ}+\cos^{2}50^{\circ} \right) \right]+1$
=$2\left[ \left( \cos^{2}10^{\circ}+\sin^{2}10^{\circ} \right)+\left( \cos^{2}20^{\circ}+\sin^{2}20^{\circ} \right)+\ldots+\left( \cos^{2}40^{\circ}+\sin^{2}40^{\circ} \right) \right]+1$
= 2.4+1=9.
e)Ta có:
$E=\sin 20^{\circ}+\sin 40^{\circ}+\sin 60^{\circ}+\ldots+\sin 340^{\circ}+\sin 360^{\circ}$
=$\left( \sin 20^{\circ}+\sin 340^{\circ} \right)+\left( \sin 40^{\circ}+\sin 320^{\circ} \right)+\ldots+\left( \sin 160^{\circ}+\sin 200^{\circ} \right)+\sin 180^{\circ}+\sin 360^{\circ}$
=$2\sin 180^{\circ}\cos 160^{\circ}+2\sin 180^{\circ}\cos 140^{\circ}+\ldots+2\sin 180^{\circ}\cos 20^{\circ}$
= 0.