ĐGTD ĐH Bách khoa - Tư duy Toán học - Công thức biến đổi lượng giác

$\tan \left(\frac{\pi }{2}-\alpha \right)=\cot \alpha$

$\tan \left(-\alpha \right)=-\tan \alpha$

$\tan \left(\pi +\alpha \right)=-\tan \alpha$

$\tan \left(\pi -\alpha \right)=-\tan \alpha$

$\cos a+\cos b=2\cos \frac{a+b}{2}.\cos \frac{a-b}{2}$

$\sin a-\sin b=2\cos \frac{a+b}{2}.sin\frac{a-b}{2}$

$\cos a-\cos b=2\sin \frac{a+b}{2}.sin\frac{a-b}{2}$

$\sin a+\sin b=2\sin \frac{a+b}{2}.\cos \frac{a-b}{2}$

sin 2a = 2 sina

cos 2a = cos⁴a – sin⁴a

(sin a + cos a)² = 1 + 2 sin 2a

cos 2a = 1 – 2cos²a

$\frac{1+m}{2}$

$\frac{1-m}{2}$

$\frac{1-m}{5}$

m

$\frac{\sqrt{23}}{4}$

$\pm \frac{\sqrt{23}}{4}$

$\frac{-\sqrt{30}}{4}$

$\frac{-\sqrt{23}}{4}$

Sinx

−sinx

−sin9x

sin9x

$\sin \alpha =\frac{\sqrt{7}}{4};\cos 2\alpha =\frac{1}{8}$

$\sin \alpha =-\frac{\sqrt{7}}{4};\cos 2\alpha =\frac{1}{8}$

$\sin \alpha =\frac{\sqrt{5}}{4};\cos 2\alpha =\frac{\sqrt{11}}{4}$

$\sin \alpha =\frac{\sqrt{7}}{4};\cos 2\alpha =-\frac{1}{8}$

$-\frac{3\sqrt{7}}{8}$

$-\frac{\sqrt{7}}{8}$

$\frac{3\sqrt{7}}{8}$

$\frac{\sqrt{7}}{8}$

$\cot \alpha$

$\tan \frac{\alpha }{2}$

$\sin 2\alpha$

$\tan \alpha$

${\sin }^2\left(\alpha +\beta \right)$

${\cos }^2\left(\alpha +\beta \right)$

$ta{n}^2\left(\alpha +\beta \right)$

$\sin \left(\alpha +\beta \right)$

$\frac{9}{7}$

1

$\frac{1}{2}$

$\frac{3}{5}$

sinα

3

1

Cosα

tan(α + β)

cot(α + β)

sin(α + β)

cos(α + β)

2

3

0

4

S = 1

$S=\frac{5}{4}$

S = 2

$S=\frac{7}{4}$

Tanα

Cotα

2sinα

2cosα

Tam giác đều

Tam giác cân

Tam giác vuông

Tam giác thường

$\sin \alpha +\sin \beta$

2 tanα

2

2 cotα

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{6}$

$\frac{1}{8}$