Question

What is the whole square of \[({a^2} - {b^2})\sin \theta + 2ab\cos \theta \]?

Answer 1

Hint: We have to find the whole square of \[({a^2} - {b^2})\sin \theta + 2ab\cos \theta \]. We have addition sign in between and so we will use the identity \[{(x + y)^2} = {x^2} + {y^2} + 2xy\] considering the first term as ‘a’ and the second term as ‘b’. Then, again the first term has minus sign in between the two terms. So, when we square this term , we will again use the identity \[{(x - y)^2} = {x^2} + {y^2} - 2xy\]. And, also we need to remember the exponential properties.
We know, \[{(abc)^2} = {a^2}{b^2}{c^2}\]

Complete step-by-step solution:
We need to find the whole square of \[({a^2} - {b^2})\sin \theta + 2ab\cos \theta \]. i.e. we need to find
\[{[({a^2} - {b^2})\sin \theta + 2ab\cos \theta ]^2}\]
Considering \[({a^2} - {b^2})\sin \theta \] as \[x\] and \[2ab\cos \theta \] as \[y\] in \[{(x + y)^2} = {x^2} + {y^2} + 2xy\], we get
\[{[({a^2} - {b^2})\sin \theta + 2ab\cos \theta ]^2} = {(({a^2} - {b^2})\sin \theta )^2} + {(2ab\cos \theta )^2} + 2 \times (({a^2} - {b^2})\sin \theta ) \times (2ab\cos \theta )\]
\[ = {({a^2} - {b^2})^2}{\sin ^2}\theta + {2^2}{a^2}{b^2}{\cos ^2}\theta + 4 \times [({a^2} - {b^2})\sin \theta ] \times (ab\cos \theta )\]
\[ = {({a^2} - {b^2})^2}{\sin ^2}\theta + {2^2}{a^2}{b^2}{\cos ^2}\theta + 4ab({a^2} - {b^2})\sin \theta \cos \theta \] --(1)
Now, using \[{(x - y)^2} = {x^2} + {y^2} - 2xy\]on \[{({a^2} - {b^2})^2}\], we get
\[{({a^2} - {b^2})^2} = {({a^2})^2} + {({b^2})^2} - 2{a^2}{b^2}\]
\[ = {a^4} + {b^4} - 2{a^2}{b^2}\] (\[{({x^2})^2} = {x^{2 \times 2}} = {x^4}\]) ----(2)
Now, substituting (2) in (1), we get
\[{[({a^2} - {b^2})\sin \theta + 2ab\cos \theta ]^2} = ({a^4} + {b^4} - 2{a^2}{b^2}){\sin ^2}\theta + {2^2}{a^2}{b^2}{\cos ^2}\theta + 4ab({a^2} - {b^2})\sin \theta \cos \theta \]
\[ = ({a^4} + {b^4} - 2{a^2}{b^2}){\sin ^2}\theta + 4{a^2}{b^2}{\cos ^2}\theta + 4ab({a^2} - {b^2})\sin \theta \cos \theta \]
\[ = {a^4}{\sin ^2}\theta + {b^4}{\sin ^2}\theta - 2{a^2}{b^2}{\sin ^2}\theta + 4{a^2}{b^2}{\cos ^2}\theta + 4{a^3}b\sin \theta \cos \theta - 4a{b^3}\sin \theta \cos \theta \] (Opening the brackets)
Hence , we get
\[{[({a^2} - {b^2})\sin \theta + 2ab\cos \theta ]^2} = {a^4}{\sin ^2}\theta + {b^4}{\sin ^2}\theta - 2{a^2}{b^2}{\sin ^2}\theta + 4{a^2}{b^2}{\cos ^2}\theta + 4{a^3}b\sin \theta \cos \theta - 4a{b^3}\sin \theta \cos \theta \]

Note: We can even further solve this problem and simplify it but we can leave it here as well. We need to be very thorough with the identities for \[{(x + y)^2}\] and \[{(x - y)^2}\]. Also, we need to take care of squaring trigonometric terms. i.e. \[{(\sin x)^2} = {\sin ^2}x\]not \[{(\sin x)^2} = {\sin ^2}{x^2}\]. And, two trigonometric terms can be added if they have same exponential powers. And while solving, we need to be very careful with the terms i.e. not forgetting any term or carrying the wrong term forward or not changing the sign while shifting or squaring.