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Prove by the vector method , cos$(A+B)=$ cos$A$ cos$B$ - sin$A$ sin$B$

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  • $ \overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b | \cos \theta$ $ \therefore \cos \theta = \large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a||\overrightarrow b|} \Rightarrow \theta = \cos^{-1} \large\frac{\overrightarrow a.\overrightarrow b}{|\overrightarrow a||\overrightarrow b|}$
Let $OL, OL'$ be the radii of the unit circle with centre $O$ such that $ <{xOL} = A$ and $ <{L'Ox}=B. \: \therefore <{L'OL} = A+B$. Let $\overrightarrow i, \overrightarrow j$ be the unit vectors along $ Ox \: and\: Oy.$ Draw $LM, L'M' \perp \: to \: Ox$ We now have $ \overrightarrow {OL} = \overrightarrow {OM}+\overrightarrow {ML}=OM\overrightarrow i+ML\overrightarrow j$
$\overrightarrow {OL'}=\overrightarrow {OM'}+\overrightarrow {M'L'}=OM'\overrightarrow i-M'L'\overrightarrow j$
Consider the scalar product $ \overrightarrow {OL}.\overrightarrow {OL'}$
By definition $ \overrightarrow {OL}.\overrightarrow {OL'} = (OL)(OL') \cos(A+B)$
$ = \cos (A+B)$ (i) since $OL=OL'=1$
But $\overrightarrow {OL}=OM\overrightarrow i+ML\overrightarrow j\: and \: \overrightarrow {OL'}=OM'\overrightarrow i-M'L'\overrightarrow j$
$ \therefore \overrightarrow {OL}.\overrightarrow {OL'}=(OM\overrightarrow i+ML\overrightarrow j).(OM'\overrightarrow i-M'L'\overrightarrow j)$
$ = (OM)(OM')-(ML)(M'L') $ (ii)
Now in $ \Delta \: OML, \large\frac{OM}{OL} = \cos A\: or OM = OL\cos A = \cos A \: ( since\: OL = 1)$
Also $ \large\frac{ML}{OL} = \sin A\: or \: ML = OL \sin A = \sin A$
Similarly, from $ \Delta OM'L'$ we have $ OL'=\cos B\: and M'L'=\sin B$
(ii) reduces to $ \overrightarrow {OL}.\overrightarrow {OL'} = \cos A \cos B. \sin A \sin B (iii)$
From (i) and (iii) , we have $ \cos (A+B) = \cos A \cos B = \sin A \sin B$ Hence proved


answered Jun 4, 2013 by thanvigandhi_1
edited Jun 20, 2013 by thanvigandhi_1

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