§2. Giá trị lượng giác của một cung

Có \(a\) thuộc góc phần tư thứ III -> sin\(a\) < 0

+) sin\(a\)=-\(\sqrt{1-cos^2a}\)=-\(\sqrt{1-\left(\dfrac{-12}{13}\right)^2}\)=\(\dfrac{-5}{13}\)

\(cos2a=cos^2a-sin^2a\)=\(\left(\dfrac{-12}{13}\right)^2-\left(\dfrac{-5}{13}\right)^2=\dfrac{119}{169}\)

(điều kiện: sin \(\ne\) cos)

\(\dfrac{2cos^2-1}{sin+\cos}=\dfrac{2cos^2-\left(\sin^2+\cos^2\right)}{sin+\cos}=\dfrac{\cos^2-\sin^2}{sin+\cos}\)\(=\dfrac{\left(\cos-\sin\right).\left(\sin+\cos\right)}{sin+\cos}\)

= Cos - sin

\(\dfrac{sin^2x}{1+cotx}-\dfrac{cos^2x}{1+tanx}=\dfrac{sin^2x}{1+\dfrac{cosx}{sinx}}-\dfrac{cos^2x}{1+\dfrac{sinx}{cosx}}=\dfrac{sin^2x}{\dfrac{sinx+cosx}{sinx}}-\dfrac{cos^2x}{\dfrac{cosx+sinx}{cosx}}=\dfrac{sin^3x}{sinx+cosx}-\dfrac{cos^3x}{sinx+cosx}=\dfrac{\left(sinx-cosx\right)\left(sin^2x-sinx\cdot cosx+cos^2x\right)}{sinx+cosx}=\dfrac{\left(sinx-cosx\right)\left(1-sinx\cdot cosx\right)}{sinx+cosx}\)???

Đáp án A

\(=\cos\left(\Pi+\dfrac{\Pi}{2}-a\right)-\sin\left(\Pi+\dfrac{\Pi}{2}-a\right)+\sin a\)

\(=-\cos\left(\dfrac{\Pi}{2}-a\right)+\sin\left(\dfrac{\Pi}{2}-a\right)+\sin a\)

\(=-\sin a+\cos a+\sin a=\cos a\)

\(A=\cos\left(\text{π}-\dfrac{x}{2}\right)-\sin\left(\text{π}-x\right)\)

\(=\sin x+\sin x=2\cdot\sin x\)

\(B=\cos\left(2\text{π}+\dfrac{\text{π}}{2}-x\right)+\sin\left(4\text{π}+\dfrac{\text{π}}{2}-x\right)-\cos\left(6\text{π}+\dfrac{3}{2}\text{π}+x\right)-\sin\left(16\text{π}+\dfrac{3}{2}\text{π}+x\right)\)

\(=\sin x+\cos x-\cos\left(\dfrac{3}{2}\text{π}+x\right)-\sin\left(\dfrac{3}{2}\text{π}+x\right)\)

\(=\sin x+\cos x-\cos\left(\text{π}+\dfrac{\text{π}}{2}+x\right)-\sin\left(\text{π}+\dfrac{\text{π}}{2}+x\right)\)

\(=\cos x+\sin x+\cos\left(\dfrac{1}{2}\text{π}+x\right)+\sin\left(\dfrac{1}{2}\text{π}+x\right)\)

\(=\cos x+\sin x-\sin x+\cos x=2\cos x\)