1.a) \(4cos\dfrac{\alpha}{2}.cos\dfrac{\beta}{2}.cos\dfrac{f}{2}\)

\(=\dfrac{1}{2}.4\left[cos\left(\dfrac{\alpha-\beta}{2}\right)+cos\left(\dfrac{\alpha+\beta}{2}\right)\right].cos\dfrac{f}{2}\)

\(=2.cos\left(\dfrac{\alpha-\beta}{2}\right)cos\dfrac{f}{2}+2.cos\left(\dfrac{\alpha+\beta}{2}\right).cos\dfrac{f}{2}\)

\(=cos\left(\dfrac{\alpha-\left(\beta+f\right)}{2}\right)+cos\left(\dfrac{\alpha-\beta+f}{2}\right)+cos\left(\dfrac{\alpha+\beta-f}{2}\right)+cos\left(\dfrac{\alpha+\beta+f}{2}\right)\)

\(=cos\left(\dfrac{2\alpha-\pi}{2}\right)+cos\left(\dfrac{\pi-2\beta}{2}\right)+cos\left(\dfrac{\pi-2f}{2}\right)+cos\left(\dfrac{\pi}{2}\right)\)

\(=cos\left(-\dfrac{\pi}{2}+\alpha\right)+cos\left(\dfrac{\pi}{2}-\beta\right)+cos\left(\dfrac{\pi}{2}-f\right)\)

\(=sin\alpha+sin\beta+sinf\) (đpcm)

a2) \(1+4sin\dfrac{\alpha}{2}.sin\dfrac{\beta}{2}.sin\dfrac{f}{2}\)

\(=1+2\left[cos\left(\dfrac{\alpha-\beta}{2}\right)-cos\left(\dfrac{\alpha+\beta}{2}\right)\right].sin\dfrac{f}{2}\)

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

\(=1+sin\left(\dfrac{f-\alpha+\beta}{2}\right)+sin\left(\dfrac{a-\beta+f}{2}\right)-sin\left(\dfrac{f-\left(\alpha+\beta\right)}{2}\right)-sin\left(\dfrac{\alpha+\beta+f}{2}\right)\)

\(=1+sin\left(\dfrac{\pi-2\alpha}{2}\right)+sin\left(\dfrac{\pi-2\beta}{2}\right)-sin\left(\dfrac{2f-\pi}{2}\right)-sin\left(\dfrac{\pi}{2}\right)\)

\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+sin\left(\dfrac{\pi}{2}-\beta\right)+sin\left(\dfrac{\pi}{2}-f\right)\)

\(=cos\alpha+cos\beta+cosf\) (đpcm)

a: \(\cos\alpha=\dfrac{1}{2}\)

\(\tan\alpha=\sqrt{3}\)

\(\cot\alpha=\dfrac{\sqrt{3}}{3}\)

b) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\cos^2\alpha=\dfrac{16}{25}\)

hay \(\cos\alpha=\dfrac{4}{5}\)

Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)

\(=5\cdot\left(\dfrac{3}{5}\right)^2+6\cdot\left(\dfrac{4}{5}\right)^2\)

\(=5\cdot\dfrac{9}{25}+6\cdot\dfrac{16}{25}\)

\(=\dfrac{141}{25}\)

c) Ta có: \(\tan\alpha=\dfrac{1}{\cot\alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\)

\(D=\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)

\(=\dfrac{\dfrac{9}{16}+\dfrac{16}{9}}{\dfrac{9}{16}-\dfrac{16}{9}}=-\dfrac{337}{175}\)

a: pi/2<a<pi

=>sin a>0

\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)

\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)

\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)

b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

c: \(sin\left(a-\dfrac{pi}{3}\right)\)

\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)

\(=\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}\)

d: \(cos\left(a-\dfrac{pi}{6}\right)\)

\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)

\(sin\alpha^2+cos\alpha^2=1\Rightarrow sin\alpha^2=1-cos\alpha^2=1-\dfrac{1}{25}=\dfrac{24}{25}\Rightarrow sin\alpha=\dfrac{2\sqrt{6}}{5}\)

\(\Rightarrow cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{1}{5}:\dfrac{2\sqrt{6}}{5}=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{24}\)

\(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\sin^2\alpha=1-\dfrac{1}{25}=\dfrac{24}{25}\)

hay \(\sin\alpha=\dfrac{2\sqrt{6}}{5}\)

\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)

\(\cot\alpha=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)

a: \(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\)

\(\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)(đúng)

b: Ta có: \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)

\(=\dfrac{4\cdot\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)

=4

Để giải bài toán này, ta sẽ sử dụng các công thức và quy tắc trong lượng giác để tính toán.

Trước hết, ta có: sin(α+β) = sinα.cosβ + cosα.sinβ cos(α+β) = cosα.cosβ - sinα.sinβ

Đề bài cho α+β = 1313 và tanα = -2tanβ. Ta có thể suy ra các thông tin sau: tanα = -2tanβ => sinα/cosα = -2sinβ/cosβ => sinα.cosβ = -2sinβ.cosα

Bài toán yêu cầu tính A = sin(α+3π/8) . cos(α+π/8) + sin(β-5π/12) . sin(β-π/12)

Để tính A, ta sẽ thay các giá trị đã biết vào công thức trên:

A = sin(α+3π/8) . cos(α+π/8) + sin(β-5π/12) . sin(β-π/12) = (sinα . cos(3π/8) + cosα . sin(3π/8)) . (cosα . cos(π/8) - sinα . sin(π/8)) + (sinβ . cos(5π/12) - cosβ . sin(5π/12)) . (cosβ . cos(π/12) + sinβ . sin(π/12)) = (sinα . cos(3π/8) + cosα . sin(3π/8)) . (cosα . cos(π/8) - sinα . sin(π/8)) + (sinβ . cos(5π/12) - cosβ . sin(5π/12)) . (cosβ . cos(π/12) + sinβ . sin(π/12)) = (sinα . cos(3π/8) + cosα . sin(3π/8)) . (cosα . cos(π/8) - sinα . sin(π/8)) + (sinβ . cos(5π/12) - cosβ . sin(5π/12)) . (cosβ . cos(π/12) + sinβ . sin(π/12)) = (sinα . cos(3π/8) + cosα . sin(3π/8)) . (cosα . cos(π/8) - sinα . sin(π/8)) + (sinβ . cos(5π/12) - cosβ . sin(5π/12)) . (cosβ . cos(π/12) + sinβ . sin(π/12))

Tuy nhiên, để tính giá trị chính xác của A, cần biết thêm giá trị cụ thể của α và β. Trong câu hỏi của bạn, không có thông tin về α và β, do đó không thể tính toán giá trị của A.