Bài tập nâng cao · Bài 2
Dạng 1. Công thức cộng
Bài tập nâng cao Công thức lượng giác · Bài 2
Chứng minh rằng:
a) $\sin x+\cos x=\sqrt{2}\sin\left( x+\frac{\pi}{4} \right)$;
b) $\sin\left( a+b \right)\sin\left( a-b \right)=\sin^{2}a-\sin^{2}b=\cos^{2}b-\cos^{2}a$;
c) $4\sin\left( x+\frac{\pi}{3} \right)\sin\left( x-\frac{\pi}{3} \right)=4\sin^{2}x-3$;
d) $\sin\left( x+\frac{\pi}{4} \right)-\sin\left( x-\frac{\pi}{4} \right)=\sqrt{2}\cos x$.
a) $\sin x+\cos x=\sqrt{2}\sin\left( x+\frac{\pi}{4} \right)$;
b) $\sin\left( a+b \right)\sin\left( a-b \right)=\sin^{2}a-\sin^{2}b=\cos^{2}b-\cos^{2}a$;
c) $4\sin\left( x+\frac{\pi}{3} \right)\sin\left( x-\frac{\pi}{3} \right)=4\sin^{2}x-3$;
d) $\sin\left( x+\frac{\pi}{4} \right)-\sin\left( x-\frac{\pi}{4} \right)=\sqrt{2}\cos x$.
Xem lời giải
Lời giải
a) $\sin x+\cos x=\sqrt{2}\left( \frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x \right)=\sqrt{2}\left( \frac{\sqrt{2}}{2}\sin x+\frac{\sqrt{2}}{2}\cos x \right)$
$=\sqrt{2}\left( \cos\frac{\pi}{4}\sin x+\sin\frac{\pi}{4}\cos x \right)=\sqrt{2}\sin(x+\frac{\pi}{4})$.
b) $\sin\left( a+b \right)\sin\left( a-b \right)=\left( \sin a\cos b+\cos a\sin b \right)\left( \sin a\cos b-\cos a\sin b \right)$
$=\sin^{2}a\cos^{2}b-\cos^{2}a\sin^{2}b=\sin^{2}a(1-\sin^{2}b)-(1-\sin^{2}a)\sin^{2}b$
$\begin{array}{l}=\sin^{2}a-\sin^{2}a\sin^{2}b-\sin^{2}b+\sin^{2}a\sin^{2}b\\=\sin^{2}a-\sin^{2}b=1-\cos^{2}a-\left( 1-\cos^{2}b \right)\end{array}$
$=\cos^{2}b-\cos^{2}a$.
Cách 2
$\begin{array}{l}\sin\left( a+b \right)\sin\left( a-b \right)=-\frac{1}{2}\left[ \cos 2a-\cos 2b \right]\\=-\frac{1}{2}\left[ \left( 2\cos^{2}a-1 \right)-\left( 2\cos^{2}b-1 \right) \right]=\cos^{2}b-\cos^{2}a\\=-\frac{1}{2}\left[ \left( 1-2\sin^{2}a \right)-\left( 1-2\sin^{2}b \right) \right]=\sin^{2}a-\sin^{2}b\end{array}$
c) Áp dụng ý b) ở trên và $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, ta được
$4\sin\left( x+\frac{\pi}{3} \right)\sin\left( x-\frac{\pi}{3} \right)=4\left( \sin^{2}x-\sin^{2}\frac{\pi}{3} \right)=4\left( \sin^{2}x-\frac{3}{4} \right)$
$=4\sin^{2}x-3$.
d) $\sin\left( x+\frac{\pi}{4} \right)-\sin\left( x-\frac{\pi}{4} \right)=\sin x\cos\frac{\pi}{4}+\cos x\sin\frac{\pi}{4}-\left( \sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4} \right)$
$=2\cos x\sin\frac{\pi}{4}=2\frac{\sqrt{2}}{2}\cos x=\sqrt{2}\cos x$.
$=\sqrt{2}\left( \cos\frac{\pi}{4}\sin x+\sin\frac{\pi}{4}\cos x \right)=\sqrt{2}\sin(x+\frac{\pi}{4})$.
b) $\sin\left( a+b \right)\sin\left( a-b \right)=\left( \sin a\cos b+\cos a\sin b \right)\left( \sin a\cos b-\cos a\sin b \right)$
$=\sin^{2}a\cos^{2}b-\cos^{2}a\sin^{2}b=\sin^{2}a(1-\sin^{2}b)-(1-\sin^{2}a)\sin^{2}b$
$\begin{array}{l}=\sin^{2}a-\sin^{2}a\sin^{2}b-\sin^{2}b+\sin^{2}a\sin^{2}b\\=\sin^{2}a-\sin^{2}b=1-\cos^{2}a-\left( 1-\cos^{2}b \right)\end{array}$
$=\cos^{2}b-\cos^{2}a$.
Cách 2
$\begin{array}{l}\sin\left( a+b \right)\sin\left( a-b \right)=-\frac{1}{2}\left[ \cos 2a-\cos 2b \right]\\=-\frac{1}{2}\left[ \left( 2\cos^{2}a-1 \right)-\left( 2\cos^{2}b-1 \right) \right]=\cos^{2}b-\cos^{2}a\\=-\frac{1}{2}\left[ \left( 1-2\sin^{2}a \right)-\left( 1-2\sin^{2}b \right) \right]=\sin^{2}a-\sin^{2}b\end{array}$
c) Áp dụng ý b) ở trên và $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, ta được
$4\sin\left( x+\frac{\pi}{3} \right)\sin\left( x-\frac{\pi}{3} \right)=4\left( \sin^{2}x-\sin^{2}\frac{\pi}{3} \right)=4\left( \sin^{2}x-\frac{3}{4} \right)$
$=4\sin^{2}x-3$.
d) $\sin\left( x+\frac{\pi}{4} \right)-\sin\left( x-\frac{\pi}{4} \right)=\sin x\cos\frac{\pi}{4}+\cos x\sin\frac{\pi}{4}-\left( \sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4} \right)$
$=2\cos x\sin\frac{\pi}{4}=2\frac{\sqrt{2}}{2}\cos x=\sqrt{2}\cos x$.