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# Help!help!!help!!!

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 $$D=2v\sin\frac{1}{2}\theta$$.

Note by Samuel Ayinde
3 years, 9 months ago

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If 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
$$2 u \times sin(\frac{\theta}{2})$$

- 3 years, 9 months ago

- 3 years, 9 months ago

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.

- 3 years, 9 months ago

ok i'll try it. thanks @Rajeev sharma

- 3 years, 9 months ago