who can help in doing this?

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

who can help in doing this?

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

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

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

we used the result" \(cos\bigl( \frac{\theta}{2}\bigr) = \sqrt{\frac{1+cos\theta}{2}}\) ", which is easy to prove using \(cos\theta = \cos(\frac{\theta}{2}+\frac{\theta}{2})= cos^2 (\frac{\theta}{2}) -sin^2(\frac{\theta}{2})=2cos^2(\frac{\theta}{2})-1 \)In the seconds case, only difference occurring is \(\theta\) becomes \(180^\circ -\theta\) hence in the half angle thing, it will become \(90^\circ - cos(\frac{\theta}{2})\) and that will become \(sin(\frac{\theta}{2})\) , giving the final result as

\( 2 u \times sin(\frac{\theta}{2})\) – Aditya Raut · 3 years, 3 months ago

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– Samuel Ayinde · 3 years, 3 months ago

oh thanks for your contribution aditya!Log in to reply

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 · 3 years, 3 months ago

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– Samuel Ayinde · 3 years, 3 months ago

ok i'll try it. thanks @Rajeev sharmaLog in to reply