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Given vector m=(a-sinθ,-1/2), vector n=(1/2, cosθ).(1)When a=√2/2, and m⊥n, find the value of sin2θ. (2) When a=0 and vector m‖ vector n, the value of tanθ is obtained.
(1) A=√2/2
→
M =(√2/2-sinθ,-1/2),
→ →
M
1/2(√2/2-Sinθ)-1/2×cosθ
=√2/4-1/2(Sinθ+ cosθ)=0
Sin cosθ=√2/2
(Sinθ+cosθ)2=1+sin2θ=1/2
S in2θ=-1/2
(2) A=0
→
M=(-sinθ,-1/2)
→ →
M‖n
-Sinθ cosθ=-1/4
Sinθcosθ=sinθcosθ/(sin ²θ+ cos2θ)
= Tanθ/(1+tan2θ)=1/4
Tanθ=2 3
Given vector m =(root 3sin (x/4),1), vector n =(cos (x/4), cos^2(x/4)) (1) If m.n=1, find the cos (2π/3-x) value. (2) Let f (x)=m.n, in △ABC, the opposite edge of ABC is abc, satisfying (2a-c) cosB=bccosC, find the value range of f (A). Given vector m =(root 3sin (x/4),1), vector n =(cos (x/4), cos^2(x/4)) (1) If m.n=1, get cos (2π/3-x) value. (2) Let f (x)=m.n, in △ABC, the opposite edge of ABC is abc, satisfying (2a-c) cosB=bccosC, find the value range of f (A).
Could you tell me if there is an extra c in the conditional equation? Should be "(2a-c) cosB=bcosC "? You can't do it! 1.M={√3s in (x/4),1}, n={ cos (x/4), cos^(x/4)} m*n=√3s in (x/4)* cos (x/4)+1* cos^(x/4)=(√3/2)* sin [2*(x/4)]+{1+ cos [2*(x/4)]}...
The known vector m =(sinA,1/2) is collinear with n =(3, sinA + root number 3 cosA), where A is the inner angle of the triangle ABC... If BC=2, find the maximum value of the area S of the triangle ABC, and judge the shape of the triangle ABC when S is the maximum value
From the meaning of the title
SinA (sinA 3 cosA)=3/2
(SinA)^2 3 sinAcosA=3/2
√3/2Sin2A-cos2A/2=1
Sin (2A-π/6)=1
Because A is the inner angle of the triangle
So 2A-π/6=π/2
So A=π/3
From Cosine Theorem
CosA = cos60=(b^2+c^2-a^2)/2bc =1/2
So b^2+ c^2-bc=4
Because b^2+ c^2>=2 bc
So bc >=4
S=bcsinA/2=√3bc/4>=√3
So maximum value of triangle ABC area S√3
Where bc =4 and b = c
So b=c=2
So it's an equilateral triangle.
Given that vector m =(sinA,1/2) is collinear with n =(3, sinA + root number 3 cosA), where A is the inner angle of triangle ABC, find the magnitude of angle A Given that the vector m=(sinA,1/2) is collinear with n=(3, sinA + root number 3 cosA), where A is the inner angle of the triangle ABC, find the magnitude of angle A
From the meaning of the title
SinA (sinA 3 cosA)=3/2
(SinA)^2 3 sinAcosA=3/2
√3/2Sin2A-cos2A/2=1
Sin (2A-π/6)=1
Because A is the inner angle of the triangle
So 2A-π/6=π/2
So A=π/3
From the meaning
SinA (sinA 3 cosA)=3/2
(SinA)^2 3 sinAcosA=3/2
√3/2Sin2A-cos2A/2=1
Sin (2A-π/6)=1
Because A is the inner angle of the triangle
So 2A-π/6=π/2
So A=π/3
Known vector M=(sinA, cosA), N=( 3,-1), M• N =1, and A is an acute angle. (1) Find the size of angle A; (2) Find the range of function f (x)=cos2x+4cosAsinx (x∈R). Known vector M =(sinA, cosA), N=( 3,-1), M• N =1, and A is an acute angle. (1) Find the size of angle A; (2) Find the range of function f (x)=cos2x+4cosAsinx (x∈R).
(1) From the title m•n=3sinA-cosA=1,2sin (A-π6)=1, sin (A-π6)=12, A-π6=π6, A=π3.(2) From (1) cosA=12, f (x)= cos2x+2sinx=1-2sin2x+2sinx=-2(sinx-12)2+32, because x∈R, so sinx∈[-1,...
Given vector m=(cosa-(root 2)/3,-1), n=(sina,1) The collinear vector of m and n and a belongs to [-π/2,0], find the value of (sin2a)/(sina-cosa) sina+cosa=(-root number 2)/3
√ Because m and n are collinear vectors
So (cosa-√2/3)/ sina=-1/1
I.e. sina+cosa=√2/3
(Sina+cosa)^2=1+2sinacosa=2/9
So sin2a=2sinacosa=-7/9
Because a belongs to [-π/2,0],
So sina0
Sina-cosa
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