Question

Prove the following trigonometric equation :
${{\text{sin}}^2}{\text{A = co}}{{\text{s}}^2}({\text{A - B}}) + {\cos ^2}{\text{B}} - 2\cos ({\text{A - B}})\cos {\text{AcosB}}$

Answer 1

Hint: In order to solve such types of problems, we must keep one thing in our mind that how can we arrange the terms so that we can apply available trigonometric formulas then expression will automatically start to get reduced.

Complete step-by-step answer:
$\Rightarrow {\text{si}}{{\text{n}}^2}{\text{A = co}}{{\text{s}}^2}({\text{A - B}}) + {\cos ^2}{\text{B}} - 2\cos ({\text{A - B}})\cos {\text{AcosB}}$
On writing ${\text{co}}{{\text{s}}^2}{\text{B}}$ term first in RHS as shown below, because through such type rearrangement we can proceed to solve by taking ${\text{cos}}({\text{A - B}})$ term common
$\Rightarrow {\text{RHS = co}}{{\text{s}}^2}({\text{A - B}}) + {\cos ^2}{\text{B}} - 2\cos ({\text{A - B}})\cos {\text{AcosB}}$
On taking ${\text{cos}}({\text{A - B}})$ term common,
$\Rightarrow {\text{ }}{\cos ^2}{\text{B + cos}}({\text{A - B}})\left( {(\cos ({\text{A - B}}) - 2\cos {\text{AcosB)}}} \right)$ --- (1)
We know that.
$\Rightarrow {\text{ cos}}({\text{A - B}}) = {\text{ cosAcosB + sinAsinB}}$
So on putting the value of ${\text{cos}}({\text{A - B}})$ in expression (1)
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( {(\cos {\text{AcosB + sinAsinB}} - 2\cos {\text{AcosB)}}} \right)$
On subtracting ${\text{cos}}({\text{A)cos (B)}}$ term we get
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( {({\text{sinAsinB}} - \cos {\text{AcosB)}}} \right)$
On rearranging minus sign, try to make any formula of cos
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( {( - \cos {\text{AcosB + sinAsinB)}}} \right)$
On taking minus sign common from 2nd bracket
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( { - (\cos {\text{AcosB - sinAsinB)}}} \right)$
We can take this minus sign out from 2nd bracket
$\Rightarrow {\cos ^2}B{\text{ - cos}}({\text{A - B}})\left( {\cos {\text{AcosB - sinAsinB}}} \right)$ --- (2)
We know the formula of ${\text{cos}}({\text{A + B}})$ so here we can use that
${\text{ cos}}({\text{A + B}}) = {\text{ cosAcosB - sinAsinB}}$
So using the formula of ${\text{cos}}({\text{A + B}})$ in expression (2)
$\Rightarrow {\cos ^2}B{\text{ - cos}}(A - B)(\cos ({\text{A + B}}){\text{)}}$ --- (3)
We know that.
${\text{ cos}}({\text{A - B}}) = {\text{ cosAcosB + sinAsinB}}$
${\text{ cos}}({\text{A + B}}) = {\text{ cosAcosB - sinAsinB}}$
On using above results of {\text{cos}}({\text{A - B}}){\text{ & cos}}({\text{A - B}}) in expression (3)
$\Rightarrow {\cos ^2}B{\text{ - }}\left( {{\text{(cosAcosB + sinAsinB)}}(\cos {\text{AcosB - sinAsinB)}}} \right)$
In algebra, there is a formula known as the Difference of two squares:$({{\text{m}}^2}{\text{ - }}{{\text{n}}^2}) = ({\text{m + n}})({\text{m - n}})$
Here, ${\text{m = cosAcosB}}$ {\text{ & n = sinAsinB}}
So on using Difference of two squares formula
$\Rightarrow {\cos ^2}B{\text{ - }}\left( {{\text{(co}}{{\text{s}}^{^2}}{\text{Aco}}{{\text{s}}^2}{\text{B - si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}} \right)$
On further solving
$\Rightarrow {\cos ^2}B{\text{ - co}}{{\text{s}}^{^2}}{\text{Aco}}{{\text{s}}^2}{\text{B + si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}$
On taking the ${\text{co}}{{\text{s}}^2}({\text{B}})$ term common
$\Rightarrow {\cos ^2}B{\text{ (1 - co}}{{\text{s}}^{^2}}{\text{A) + si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}$
We know the formula of ${\text{si}}{{\text{n}}^2}{\text{A + co}}{{\text{s}}^2}{\text{A = 1 }}$ so here we can use
${\text{ (1 - co}}{{\text{s}}^{^2}}{\text{A) = si}}{{\text{n}}^2}{\text{A}}$ in above expression
$\Rightarrow {\cos ^2}B{\text{ (si}}{{\text{n}}^{^2}}{\text{A) + si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}$
On taking the ${\sin ^2}({\text{A}})$ term common
$\Rightarrow {\text{(si}}{{\text{n}}^{^2}}{\text{A) }}\left( {{{\cos }^2}B{\text{ + si}}{{\text{n}}^2}{\text{B}}} \right)$
Again using ${\text{si}}{{\text{n}}^2}{\text{B + co}}{{\text{s}}^2}{\text{B = 1 }}$
$\Rightarrow {\text{(si}}{{\text{n}}^{^2}}{\text{A) }} \times \left( 1 \right)$
$\Rightarrow {\text{si}}{{\text{n}}^{^2}}{\text{A = LHS}}$

Note: Whenever we face such a type of problem always remember the trigonometry identities which are written above then simplify the given statements using these identities. we will get the required answer.