Tính giá trị lượng giác và biểu thức lượng giác

Tính các giá trị lượng giác sau: \(\cos 795^{\circ}, \sin 18^{\circ}, \tan \frac{7 \pi}{12}, \cot \frac{5 \pi}{8}\).

Lời giải

•  Vì \(795^{\circ}=75^{\circ}+2.360^{\circ}=30^{\circ}+45^{\circ}+2.360^{\circ}\) nên

\(\cos 795^{\circ}=\cos 75^{\circ}=\cos 30^{\circ} \cos 45^{\circ}-\sin 30^{\circ} \sin 45^{\circ}\)

\(=\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}-\frac{1}{2} \cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\)

•  Vì \(54^{\circ}+36^{\circ}=90^{\circ}\) nên \(\sin 54^{\circ}=\cos 36^{\circ}\)

Do đó \(3 \sin 18^{\circ}-4 \sin ^{8} 18^{\circ}=1-2 \sin ^{2} 18^{\circ}\)

\(\Leftrightarrow\left(\sin 18^{\circ}-1\right)\left(4 \sin ^{2} 18^{\circ}+2 \sin 18^{\circ}-1\right)=0\)

\(\Leftrightarrow \sin 18^{\circ}=1\) hoặc \(\sin 18^{\circ}=\frac{\sqrt{5}-1}{2}\) hoặc \(\sin 18^{\circ}=\frac{\sqrt{5}+1}{2}\) 

Vì \(0\lt \sin 18^{\circ}\lt 1\) nên \(\sin 18^{\circ}=\frac{\sqrt{5}-1}{2}\).

•  \(\tan \frac{7 \pi}{12}=\tan \left(\frac{\pi}{3}+\frac{\pi}{4}\right)=\frac{\tan \frac{\pi}{3}+\tan \frac{\pi}{4}}{1-\tan \frac{\pi}{3} \tan \frac{\pi}{4}}=\frac{\sqrt{3}+1}{1-\sqrt{3}}=-2-\sqrt{3}\)
•  \(\cot \frac{5 \pi}{8}=\cot \left(\frac{\pi}{2}+\frac{\pi}{8}\right)=-\tan \frac{\pi}{8}\)

Ta lại có \(1=\tan \frac{\pi}{4}=\tan \left(2 \cdot \frac{\pi}{8}\right)=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^{2} \frac{\pi}{8}}\) suy ra

\(\begin{array}{l}1-\tan ^{2} \frac{\pi}{8}=2 \tan \frac{\pi}{8} \Leftrightarrow \tan ^{2} \frac{\pi}{8}+2 \tan \frac{\pi}{8}-1=0 \\ \Leftrightarrow \tan \frac{\pi}{8}=-1-\sqrt{2} \text { hoặc } \tan \frac{\pi}{8}=-1+\sqrt{2}\end{array}\)

Do \(\tan \frac{\pi}{8}\gt 0\) nên \(\tan \frac{\pi}{8}=-1+\sqrt{2}\)

Vậy \(\cot \frac{5 \pi}{8}=1-\sqrt{2}\)

Tính giá trị biểu thức lượng giác sau:

a) \(A=\sin 22^{\circ} 30^{\prime} \cos 202^{\circ} 30^{\prime}\)

b) \(B=4 \sin ^{4} \frac{\pi}{16}+2 \cos \frac{\pi}{8}\)

c) \(O=\frac{\sin \frac{\pi}{5}-\sin \frac{2 \pi}{15}}{\cos \frac{\pi}{5}-\cos \frac{2 \pi}{15}}\)

d) \(D=\sin \frac{\pi}{9}-\sin \frac{5 \pi}{9}+\sin \frac{7 \pi}{9}\)

Lời giải

a) Cách 1: Ta có \(\cos 202^{\circ} 30^{\prime}=\cos \left(180^{\circ}+22^{\circ} 30^{\prime}\right)=-\cos 22^{\circ} 30^{\prime}\)

Do đó \(A=-\sin 22^{\circ} 30^{\prime} \cos 22^{\circ} 30^{\prime}=-\frac{1}{2} \sin 45^{\circ}=-\frac{\sqrt{2}}{4}\)

Cách 2: \(A=\frac{1}{2}\left[\sin \left(22^{\circ} 30^{\prime}+202^{\circ} 30^{\prime}\right)+\sin \left(22^{\circ} 30^{\prime}-202^{\circ} 30^{\prime}\right)\right]\).

\(=\frac{1}{2}\left[\sin 225^{\circ}+\sin \left(-180^{\circ}\right)\right]\)

\(=\frac{1}{2}\left[\sin \left(180^{\circ}+45^{\circ}\right)-\sin 180^{\circ}\right]=-\frac{1}{2} \sin 45^{\circ}=-\frac{\sqrt{2}}{4}\)

b) \(B=\left(2 \sin ^{2} \frac{\pi}{16}\right)^{2}+2 \cos \frac{\pi}{8}=\left[1-\cos \left(2 \cdot \frac{\pi}{16}\right)\right]^{2}+2 \cos \frac{\pi}{8}\)

\(=1-2 \cos \frac{\pi}{8}+\cos ^{2} \frac{\pi}{8}+2 \cos \frac{\pi}{8}=1+\frac{1+\cos \frac{\pi}{4}}{2}=1+\frac{1+\frac{\sqrt{2}}{2}}{2}=\frac{6+\sqrt{2}}{4}\)

c) \(C=\frac{\sin \frac{\pi}{5}-\sin \frac{2 \pi}{15}}{\cos \frac{\pi}{5}-\cos \frac{2 \pi}{15}}=\frac{2 \cos \frac{1}{2}\left(\frac{\pi}{5}+\frac{2 \pi}{15}\right) \sin \frac{1}{2}\left(\frac{\pi}{5}-\frac{2 \pi}{15}\right)}{-2 \sin \frac{1}{2}\left(\frac{\pi}{5}+\frac{2 \pi}{15}\right) \sin \frac{1}{2}\left(\frac{\pi}{5}-\frac{2 \pi}{15}\right)}\)

\(=-\frac{\cos \frac{\pi}{6}}{\sin \frac{\pi}{6}}=-\cot \frac{\pi}{6}=-\sqrt{3}\)

d) \(D=\left(\sin \frac{\pi}{9}+\sin \frac{7 \pi}{9}\right)-\sin \frac{5 \pi}{9}=2 \sin \frac{4 \pi}{9} \cdot \cos \frac{\pi}{3}-\sin \frac{5 \pi}{9}=\sin \frac{4 \pi}{9}-\sin \frac{5 \pi}{9}=0\)