\(F=cos\left(\frac{\pi}{4}+a\right)\cdot cos\left(\frac{\pi}{4}-a\right)\)

\(=\frac12\cdot\left\lbrack cos\left(\frac{\pi}{4}+a-\frac{\pi}{4}+a\right)+cos\left(\frac{\pi}{4}+a+\frac{\pi}{4}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack cos\left(2a\right)+cos\left(\frac{\pi}{2}\right)\right\rbrack=\frac12\cdot cos2a\)

\(G=\sin\left(\frac{\pi}{3}+a\right)\cdot cos\left(\frac{\pi}{3}-a\right)\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{3}+a+\frac{\pi}{3}-a\right)+\sin\left(\frac{\pi}{3}+a-\frac{\pi}{3}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac23\pi\right)+\sin2a\right\rbrack=\frac12\cdot\left\lbrack\frac12+\sin2a\right\rbrack\)

\(H=cos\left(\frac{\pi}{2}-a\right)\cdot\sin\left(\frac{\pi}{2}+a\right)\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{2}+a+\frac{\pi}{2}-a\right)+\sin\left(\frac{\pi}{2}+a-\frac{\pi}{2}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack\sin\left(\pi\right)+\sin2a\right\rbrack=\frac12\left\lbrack2\cdot\sin a\cdot cosa\right\rbrack=\sin a\cdot cosa\)

\(I=\sin\left(\frac{\pi}{4}+a\right)-cos\left(\frac{\pi}{4}-a\right)\)

\(=\sin\left(\frac{\pi}{4}+a\right)-\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+a\right)=\sin\left(\frac{\pi}{4}+a\right)-\sin\left(\frac{\pi}{4}+a\right)\)

=0

\(K=cos\left(\frac{\pi}{6}-x\right)-\sin\left(\frac{\pi}{3}+x\right)\)

\(=\sin\left(\frac{\pi}{2}-\frac{\pi}{6}+x\right)-\sin\left(\frac{\pi}{3}+x\right)=\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}+x\right)\)

=0

`A=sin(π-α)+cos(π+α)+cos(-α)`

`= sinα-cosα+cosα=sinα=3/5`

Chọn A.

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: \(cos\left(a-\dfrac{pi}{6}\right)\)

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

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

Chọn D.

Vì  ⇒ sin⁡α > 0, cos⁡α < 0.

Từ sin⁡α + 2cos⁡α = -1 ⇒ sin⁡α = -1 - 2cos⁡α.

Ta có:

(-1 - 2cos⁡α ) 2  +  cos 2 α  = 1

⇔ 1 + 4cos⁡α + 4 cos 2 α  +  cos 2 α  = 1

⇔ 5cos2⁡α + 4cos⁡α = 0

⇔ cos⁡α.(5cos⁡α + 4) = 0

Chọn D.

Ta có:

sin 2 α +  cos 2 α  = 1

sin⁡α + 2cos⁡α = -1 ⇔ sin⁡α = -1 - 2cos⁡α

⇔ (-1 - 2cos⁡α ) 2  +  cos 2 α  = 1

⇔ 1 + 4cos⁡α + 4 cos 2 α  +  cos 2 α  = 1

⇔ 5 cos 2 α  + 4cos⁡α = 0

Vì π/2 < α < π ⇒ cos⁡α < 0. Do đó, cos⁡ α = -4/5

Ta lại có:

Vì \(\dfrac{\pi}{2}< \alpha< \pi\) \(\Rightarrow\) cos \(\alpha\) < 0

\(\Rightarrow\) cos \(\alpha\) = \(-\sqrt{1-sin^2\alpha}\) = \(-\dfrac{2\sqrt{2}}{3}\)

\(\Rightarrow\) tan \(\alpha\) = \(\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\sqrt{2}}{4}\)

\(\Rightarrow\) cot \(\alpha\) = \(\dfrac{1}{tan\alpha}\) = \(-2\sqrt{2}\)

Chúc bn học tốt!

Chọn D.

Ta có: sin⁡2α = a ⇒ 2sin⁡α.cos⁡α = a với 0 < α < 90o.

sin 2 α +  cos 2 α  = 1

⇔  sin 2 α  +  cos 2 α  + 2sin⁡αcos⁡α - 2sin⁡αcos⁡α = 1

⇔ (sin⁡α + cosα ) 2  - 2sin⁡αcos⁡α = 1

⇔ (sin⁡α + cosα ) 2  = 1 + 2sin⁡αcos⁡α

⇔ (sin⁡α + cosα ) 2  = 1 + a

Chọn C.

Áp dụng công thức: cos2α = 1 - 2 sin 2 α = 1 - 2.(0,6 ) 2  = 0,28