Question

If $a\cos \theta -b\sin \theta =x$ and$a\sin \theta +b\cos \theta =y$, then prove that ${{a}^{2}}+{{b}^{2}}={{x}^{2}}+{{y}^{2}}$.

Answer 1

Hint: Using the square identity , square both the given expression and add them, Now use the fact that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ .

Complete step-by-step answer:
Let us consider the given two trigonometric expressions,
$a\cos \theta -b\sin \theta =x...........(1)$
$a\sin \theta +b\cos \theta =y...........(2)$
Squaring the equation (1), we get
${{\left( a\cos \theta -b\sin \theta \right)}^{2}}={{x}^{2}}$
The square of difference of the terms is expanded as the subtraction of two times the product of two terms from the sum of the squares of the terms.
Take $p=a\cos \theta $ and $q=b\sin \theta $ , and replace them in the expansion of the formula for calculating the value in mathematical form ${{(p-q)}^{2}}={{p}^{2}}+{{q}^{2}}-2pq$.
${{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2(a\cos \theta )(b\sin \theta )={{x}^{2}}$
${{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2ab\cos \theta \sin \theta ={{x}^{2}}............(3)$
Squaring the equation (2), we get
${{\left( a\sin \theta +b\cos \theta \right)}^{2}}={{y}^{2}}$
The square of the sum of the terms is expanded as the sum of two times the product of two terms from the sum of the squares of the terms.
Take $p=a\sin \theta $ and $q=b\cos \theta $ , and replace them in the expansion of the formula for calculating the value in mathematical form ${{(p+q)}^{2}}={{p}^{2}}+{{q}^{2}}+2pq$.
${{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta -2(a\sin \theta )(b\cos \theta )={{y}^{2}}$
${{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta -2ab\cos \theta \sin \theta ={{y}^{2}}............(4)$
Adding equation (3) and equation (4), we get
$\left( {{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2ab\cos \theta \sin \theta \right)+\left( {{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta -2ab\cos \theta \sin \theta \right)={{x}^{2}}+{{y}^{2}}$
Cancelling the term$2ab\cos \theta \sin \theta $, we get
$\left( {{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta \right)+\left( {{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta \right)={{x}^{2}}+{{y}^{2}}$
If ${{a}^{2}}$ and ${{b}^{2}}$ are the common terms, then we have
${{a}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)+{{b}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)={{x}^{2}}+{{y}^{2}}$
We know that, the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
${{a}^{2}}\left( 1 \right)+{{b}^{2}}\left( 1 \right)={{x}^{2}}+{{y}^{2}}$
${{a}^{2}}+{{b}^{2}}={{x}^{2}}+{{y}^{2}}$

Note: The sum of squares or sum of difference formulas is used to calculate the sum or difference of two or more squares in an expression.