If ( + )=a and ( + )=b, then prove that 2( - )-4ab ( - )=1-2a^2-2b^2 [Hint: Express ( - )= (( + )-( + ))]

We have ( + )=a...(i) ( + )=b...(ii) ( + )= 1-a^2 and ( + )= 1-b^2 ( - )= [( + )-( + )] = ( + ) ( + )+ ( + ) ( + ) = 1-a^2 1-b^2 +ab=a b+ 1-a^2-b^2+a^2b^2 2( - )-4ab ( - ) =2 ^2( - )-1-4ab ( - ) =2 ( - )[ ( - )-2ab]-1 =2 (ab+ 1-a^2-b^2+a^2b^2 ) (ab+ 1-a^2-b^2+a^2b^2 -2ab )-1 =2 [ ( 1-a^2-b^2+a^2b^2+ab ) ( 1-a^2-b^2+a^2b^2 -ab ) ]-1 =2[1-a^2-b^2+a^2b^2-a^2b^2]-1 =2-2a^2-2b^2-1=1-2a^2-2b^2

If ( + )=a and ( + )=b, then prove that 2( - )-4ab ( - )=1-2a^2-2…

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Question

If $\sin(\theta+\alpha)=\text{a}$ and $\sin(\theta+\beta)=\text{b},$ then prove that $\cos2(\alpha-\beta)-4\text{ab}\cos(\alpha-\beta)=1-2\text{a}^2-2\text{b}^2$
[Hint: Express $\cos(\alpha-\beta)=\cos((\theta+\alpha)-(\theta+\beta))$]

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Answer

We have $\sin(\theta+\alpha)=\text{a}...(\text{i})$
$\sin(\theta+\beta)=\text{b}...(\text{ii})$
$\therefore\cos(\theta+\alpha)=\sqrt{1-\text{a}^2}$ and $\cos(\theta+\beta)=\sqrt{1-\text{b}^2}$
$\therefore\cos(\alpha-\beta)=\cos[(\theta+\alpha)-(\theta+\beta)]$
$=\cos(\theta+\beta)\cos(\theta+\alpha)+\sin(\theta+\alpha)\sin(\theta+\beta)$
$=\sqrt{1-\text{a}^2}\sqrt{1-\text{b}^2}+\text{ab}=\text{a b}+\sqrt{1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2}$
$\Rightarrow\cos2(\alpha-\beta)-4\text{ab}\cos(\alpha-\beta)$
$=2\cos^2(\alpha-\beta)-1-4\text{ab}\cos(\alpha-\beta)$
$=2\cos(\alpha-\beta)[\cos(\alpha-\beta)-2\text{ab}]-1$
$=2\big(\text{ab}+\sqrt{1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2}\big)\\\big(\text{ab}+\sqrt{1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2}-2\text{ab}\big)-1$
$=2\big[\big(\sqrt{1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2+\text{ab}}\big)\\\big(\sqrt{1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2}-\text{ab}\big)\big]-1$
$=2[1-\text{a}^2-\text{b}^2+\text{a}^2\text{b}^2-\text{a}^2\text{b}^2]-1\\=2-2\text{a}^2-2\text{b}^2-1=1-2\text{a}^2-2\text{b}^2$

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