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7 tháng 4 2019

Chọn C.

Ta có A =  sin 2a.tan2a + 4sin2a - tan2a + 3cos2a

= 3sin2a + 3cos2a = 3.

18 tháng 4 2021

\(A=\dfrac{sin\left(a-b\right)}{cosa.cosb}+\dfrac{sin\left(b-c\right)}{cosb.cosc}+\dfrac{sin\left(c-a\right)}{cosc.cosa}\)

\(=\dfrac{sina.cosb-cosa.sinb}{cosa.cosb}+\dfrac{sinb.cosc-cosb.sinc}{cosb.cosc}+\dfrac{sinc.cosa-cosc.sina}{cosc.cosa}\)

\(=\dfrac{sina}{cosa}-\dfrac{sinb}{cosb}+\dfrac{sinb}{cosb}-\dfrac{sinc}{cosc}+\dfrac{sinc}{cosc}-\dfrac{sina}{cosa}\)

\(=0\)

30 tháng 4 2019

Chọn C.

Ta có:

25 tháng 2 2018

Chọn C.

Ta có: 

A = sin( a-160) .cos( a + 140) – sin( a + 140) .cos(a - 160) = sin[ ( a - 170) – (a + 130) ] = sin( -300) = -0,5.

16 tháng 5 2021

\(A=\dfrac{sin2\alpha+sin5\alpha-sin3\alpha}{1+cos\alpha-2sin^22\alpha}\)

\(=\dfrac{2sin\alpha.cos\alpha+2.cos4\alpha.sin\alpha}{cos4\alpha+cos\alpha}\)

\(=\dfrac{2sin\alpha.\left(cos\alpha+cos4\alpha\right)}{cos4\alpha+cos\alpha}=2sin\alpha\)

NV
21 tháng 4 2021

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

b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)

c/\(P=sin\left(30+60\right)=sin90=1\)

d/

\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)

\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)

\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)

e/

\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)

\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)

21 tháng 4 2021

sina - 1 = sina - sin\(\dfrac{\pi}{2}\)

 

6 tháng 7 2021

1.Ý A

\(P=cos^4x-sin^4x=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=cos2x\)

2. Ý B

\(D=sin\left(\dfrac{5\pi}{2}-\alpha\right)+cos\left(13\pi+\alpha\right)-3sin\left(\alpha-5\pi\right)\)

\(=sin\left(2\pi+\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha+\pi-6\pi\right)\)

\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha+\pi\right)\)

\(=cos\alpha-cos\alpha+3sin\alpha=3sin\alpha\)

13 tháng 5 2022

`P=sin(\pi/2 - \alpha)+cos(\alpha+5\pi)`

`P=cos \alpha+cos(\alpha+\pi)`

`P=cos \alpha-cos \alpha=0`

       `->A`

15 tháng 1 2018

Chọn C.

Ta có