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Prove that if two vectors have the same magnitude vv and make an angle θ\theta, their sum has a magnitude S=2vcos12θS=2v\cos\frac{1}{2}\theta and their difference is D=2vsin12θD=2v\sin\frac{1}{2}\theta.

#Mechanics #Sam

Note by Samuel Ayinde
6 years, 11 months ago

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Comments

If you have 2 vectors of magnitudes aa and bb with an angle θ\theta between them, then their resultant has a magnitude of a2+b2+2ab×cosθ\sqrt{a^2+b^2+2ab\times cos\theta} . (Try to prove this using the triangle law of vector addition).

Thus in your 1st case, the resultant is v2+v2+2v2cosθ=2v2(1+cosθ)=v2+2cosθ=v4cos2θ2=2v×cosθ2\sqrt{v^2+v^2 + 2v^2 cos\theta} = \sqrt{2v^2(1+cos\theta)} = v \sqrt{2+2cos \theta} = v\sqrt{4 cos^2\frac{\theta}{2}} = 2v\times cos\frac{\theta}{2} .....

we used the result " cos(θ2)=1+cosθ2cos\bigl( \frac{\theta}{2}\bigr) = \sqrt{\frac{1+cos\theta}{2}} ", which is easy to prove using cosθ=cos(θ2+θ2)=cos2(θ2)sin2(θ2)=2cos2(θ2)1cos\theta = \cos(\frac{\theta}{2}+\frac{\theta}{2})= cos^2 (\frac{\theta}{2}) -sin^2(\frac{\theta}{2})=2cos^2(\frac{\theta}{2})-1

In the seconds case, only difference occurring is θ\theta becomes 180θ180^\circ -\theta hence in the half angle thing, it will become 90cos(θ2)90^\circ - cos(\frac{\theta}{2}) and that will become sin(θ2)sin(\frac{\theta}{2}) , giving the final result as
2u×sin(θ2) 2 u \times sin(\frac{\theta}{2})

Aditya Raut - 6 years, 11 months ago

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oh thanks for your contribution aditya!

samuel ayinde - 6 years, 11 months ago

go for vector addition

S=\sqrt { v^{ 2 }+v^{ 2 }+2v.v.\cos \theta } =\sqrt { 2v^{ 2 }+2v^{ 2 }\cos \theta } =\sqrt { 2v^{ 2 }(1+\cos \theta ) } =\sqrt { 2v^{ 2 }(2\cos ^{ 2 } \theta /2) } ............................changing1+\cos \theta to half angle form =\sqrt { 4v^{ 2 }\cos ^{ 2 } \theta /2 } =2v\cos \theta /2

change this to text. And do same for difference.

Rajeev Sharma - 6 years, 11 months ago

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ok i'll try it. thanks @Rajeev sharma

samuel ayinde - 6 years, 11 months ago
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