Bài tập nâng cao · Bài 4
Bài tập nâng cao Giá trị lượng giác của góc lượng giác · Bài 4
Tính giá trị các biểu thức sau:
a) $A=\sin\frac{7\pi}{6}+\cos 9\pi+\tan(-\frac{5\pi}{4})+\cot\frac{7\pi}{2}$
b) $B=\frac{1}{\tan 368^{\circ}}+\frac{2\sin 2550^{\circ}\cos(-188^{\circ})}{2\cos 638^{\circ}+\cos 98^{\circ}}$
c) $C=\sin^{2}25^{\circ}+\sin^{2}45^{\circ}+\sin^{2}60^{\circ}+\sin^{2}65^{\circ}$
d) $D=\tan^{2}\frac{\pi}{8}.\tan\frac{3\pi}{8}.\tan\frac{5\pi}{8}$
a) $A=\sin\frac{7\pi}{6}+\cos 9\pi+\tan(-\frac{5\pi}{4})+\cot\frac{7\pi}{2}$
b) $B=\frac{1}{\tan 368^{\circ}}+\frac{2\sin 2550^{\circ}\cos(-188^{\circ})}{2\cos 638^{\circ}+\cos 98^{\circ}}$
c) $C=\sin^{2}25^{\circ}+\sin^{2}45^{\circ}+\sin^{2}60^{\circ}+\sin^{2}65^{\circ}$
d) $D=\tan^{2}\frac{\pi}{8}.\tan\frac{3\pi}{8}.\tan\frac{5\pi}{8}$
Xem lời giải
Lời giải
a) Ta có $A=\sin\left( \pi+\frac{\pi}{6} \right)+\cos\left( \pi+4.2\pi \right)-\tan\left( \pi+\frac{\pi}{4} \right)+\cot\left( \frac{\pi}{2}+3\pi \right)$
$\Rightarrow A=-\sin\frac{\pi}{6}+\cos\pi-\tan\frac{\pi}{4}+\cot\frac{\pi}{2}=-\frac{1}{2}-1-1+0=-\frac{5}{2}$
b) Ta có $B=\frac{1}{\tan\left( 8^{\circ}+360^{\circ} \right)}+\frac{2\sin\left( 30^{\circ}+7.360^{\circ} \right)\cos(8^{\circ}+180^{\circ})}{2\cos\left( -90^{\circ}+8^{\circ}+2.360^{\circ} \right)+\cos\left( 90^{\circ}+8^{\circ} \right)}$
$\begin{array}{l} B=\frac{1}{\tan 8^{\circ}}+\frac{2\sin 30^{\circ}\left( -\cos 8^{\circ} \right)}{2\cos\left( 8^{\circ}-90^{\circ} \right)-\sin 8^{\circ}}=\frac{1}{\tan 8^{\circ}}+\frac{2.\frac{1}{2}\left( -\cos 8^{\circ} \right)}{2\cos\left( 90^{\circ}-8^{\circ} \right)-\sin 8^{\circ}}= \\ \,\,\,\,=\frac{1}{\tan 8^{\circ}}-\frac{\cos 8^{\circ}}{2\sin 8^{\circ}-\sin 8^{\circ}}=\frac{1}{\tan 8^{\circ}}-\frac{\cos 8^{\circ}}{\sin 8^{\circ}}=0 \end{array}$
c) Vì $25^{\circ}+65^{\circ}=90^{\circ}\Rightarrow \sin 65^{\circ}=\cos 25^{\circ}$ do đó
$C=\left( \sin^{2}25^{\circ}+\cos^{2}25^{\circ} \right)+\sin^{2}45^{\circ}+\sin^{2}60^{\circ}=1+\left( \frac{\sqrt{2}}{2} \right)^{2}+\left( \frac{\sqrt{3}}{2} \right)^{2}$
Suy ra $C=\frac{9}{4}$.
d) $D=-\left( \tan\frac{\pi}{8}.\tan\frac{3\pi}{8} \right).\left[ \tan\left( -\frac{\pi}{8} \right)\tan\frac{5\pi}{8} \right]$
Mà $\frac{\pi}{8}+\frac{3\pi}{8}=\frac{\pi}{2},\,-\frac{\pi}{8}+\frac{5\pi}{8}=\frac{\pi}{2}\Rightarrow \tan\frac{3\pi}{8}=\cot\frac{\pi}{8},\,\tan\frac{5\pi}{8}=\cot\left( -\frac{\pi}{8} \right)$
Nên $D=-\left( \tan\frac{\pi}{8}.\cot\frac{\pi}{8} \right).\left[ \tan\left( -\frac{\pi}{8} \right)\cot\left( -\frac{\pi}{8} \right) \right]=-1$.
$\Rightarrow A=-\sin\frac{\pi}{6}+\cos\pi-\tan\frac{\pi}{4}+\cot\frac{\pi}{2}=-\frac{1}{2}-1-1+0=-\frac{5}{2}$
b) Ta có $B=\frac{1}{\tan\left( 8^{\circ}+360^{\circ} \right)}+\frac{2\sin\left( 30^{\circ}+7.360^{\circ} \right)\cos(8^{\circ}+180^{\circ})}{2\cos\left( -90^{\circ}+8^{\circ}+2.360^{\circ} \right)+\cos\left( 90^{\circ}+8^{\circ} \right)}$
$\begin{array}{l} B=\frac{1}{\tan 8^{\circ}}+\frac{2\sin 30^{\circ}\left( -\cos 8^{\circ} \right)}{2\cos\left( 8^{\circ}-90^{\circ} \right)-\sin 8^{\circ}}=\frac{1}{\tan 8^{\circ}}+\frac{2.\frac{1}{2}\left( -\cos 8^{\circ} \right)}{2\cos\left( 90^{\circ}-8^{\circ} \right)-\sin 8^{\circ}}= \\ \,\,\,\,=\frac{1}{\tan 8^{\circ}}-\frac{\cos 8^{\circ}}{2\sin 8^{\circ}-\sin 8^{\circ}}=\frac{1}{\tan 8^{\circ}}-\frac{\cos 8^{\circ}}{\sin 8^{\circ}}=0 \end{array}$
c) Vì $25^{\circ}+65^{\circ}=90^{\circ}\Rightarrow \sin 65^{\circ}=\cos 25^{\circ}$ do đó
$C=\left( \sin^{2}25^{\circ}+\cos^{2}25^{\circ} \right)+\sin^{2}45^{\circ}+\sin^{2}60^{\circ}=1+\left( \frac{\sqrt{2}}{2} \right)^{2}+\left( \frac{\sqrt{3}}{2} \right)^{2}$
Suy ra $C=\frac{9}{4}$.
d) $D=-\left( \tan\frac{\pi}{8}.\tan\frac{3\pi}{8} \right).\left[ \tan\left( -\frac{\pi}{8} \right)\tan\frac{5\pi}{8} \right]$
Mà $\frac{\pi}{8}+\frac{3\pi}{8}=\frac{\pi}{2},\,-\frac{\pi}{8}+\frac{5\pi}{8}=\frac{\pi}{2}\Rightarrow \tan\frac{3\pi}{8}=\cot\frac{\pi}{8},\,\tan\frac{5\pi}{8}=\cot\left( -\frac{\pi}{8} \right)$
Nên $D=-\left( \tan\frac{\pi}{8}.\cot\frac{\pi}{8} \right).\left[ \tan\left( -\frac{\pi}{8} \right)\cot\left( -\frac{\pi}{8} \right) \right]=-1$.