Ta có α + β = π nên sinα = sin(π – α) = sinβ, suy ra sin²α = sin²β.

a) A = sin²α + cos²β = sin²β + cos²β = 1.

b) Ta có α + β = π nên cosα = – cos(π – α) = – cosβ.

Khi đó, B = (sinα + cosβ)² + (cosα + sinβ)²

= (sinβ + cosβ)² + (– cosβ + sinβ)²

= (sinβ + cosβ)² + (sinβ – cosβ )²

= sin²β + 2sinβ cosβ + cos²β + sin²β – 2sinβ cosβ + cos²β

= 2(sin²β + cos²β)

= 2 . 1 = 2.

Chọn đáp án B.

* Ta có  sin 2 α + cos 2 α = 1 ⇒ sin 2 α + 16 25 = 1 ⇒ sin 2 α = 9 25

Mà  sin α < 0 ⇒ sin α = - 3 5

* Vì  sin 2 β + cos 2 β = 1 ⇒ 9 16 + cos 2 β = 1 ⇒ c o s 2 β = 7 16

  cos β > 0 ⇒ cos β = 7 4

*  sin α + β = sin α . cos β + c o s α . sin β = - 3 5 . 7 4 + 4 5 . 3 4 = 12 - 3 7 20

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)

Đáp án B

`A=sin(π-α)+cos(π+α)+cos(-α)`

`= sinα-cosα+cosα=sinα=3/5`

\(tanb-4cotb=3\)

=>\(tanb-\dfrac{4}{tanb}=3\)

=>\(tan^2b-4=3tanb\)

=>(tanb-4)(tanb+1)=0

=>tan b=-1 hoặc tan b=4

0<=b<=90

=>tan b ko thể bằng -1 được

=>tan b=4

1+tan^2b=1/cos^2b

=>1/cos^2b=17

=>cosb=1/căn 17

=>sin b=4/căn 17

\(P=\left(\dfrac{1}{\sqrt{17}}+\dfrac{4}{\sqrt{17}}\right)\cdot\sqrt{17}=5\)