Question

If $\sin A+\sin 2A=x$ and $\cos A+\cos 2A=y$ then $({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$is equal to
A . 2y
B . y
C . 3y
D . None of these

Answer 1

Hint: In this question, we have given that $\sin A+\sin 2A=x$and $\cos A+\cos 2A=y$and we have to find the value of $({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$. For this, first we solve $({{x}^{2}}+{{y}^{2}})$ by putting the values of x and y in the equation and after solving the equations and using the trigonometric identities, we are able to find the value and choose our correct option.

Formula Used:
In this question, we use the identities as follow:-
${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
And ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
And $2{{\cos }^{2}}A-1=\cos 2A$

Complete Step- by- step Solution:
Given $\sin A+\sin 2A=x$
And $\cos A+\cos 2A=y$
We have to find the value of $({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$
First we find the value of $({{x}^{2}}+{{y}^{2}})$
${{x}^{2}}+{{y}^{2}}={{(\sin A+\sin 2A)}^{2}}+{{(\cos A+\cos 2A)}^{2}}$
Now by using the identity ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we expand equation (1) as
${{x}^{2}}+{{y}^{2}}={{\sin }^{2}}A+{{\sin }^{2}}(2A)+2\sin A\sin 2A+{{\cos }^{2}}A+{{\cos }^{2}}(2A)+2\cos A\cos 2A$
Now we combine the terms as follow:-
${{x}^{2}}+{{y}^{2}}=({{\sin }^{2}}A+{{\cos }^{2}}A)+({{\sin }^{2}}(2A)+{{\cos }^{2}}(2A)+2(\sin A\sin 2A+\cos A\cos 2A)$
We know the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
And $\cos (x-y)=\cos x\cos y+\sin x\sin y$
Using the above identity, equation becomes
${{x}^{2}}+{{y}^{2}}=1+1+2\cos (2A-A)$
Then ${{x}^{2}}+{{y}^{2}}=2+2\cos (A)$
Now we take 2 common from R.H.S, we get
${{x}^{2}}+{{y}^{2}}=2(1+\cos A)$
Now the value of $({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$ = $2(1+\cos A)(2(1+\cos A)-3)$
On multiplying the equations, we get
$({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$= $4{{(1+\cos A)}^{2}}-6(1+\cos A)$
Now by using the identity ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, we get
$({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$ = $4(1+{{\cos }^{2}}A+2\cos A)-6(1+\cos A)$
On opening the brackets, we get
$({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$= $4+4{{\cos }^{2}}A+8\cos A-6-6\cos A$
$({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$= $4{{\cos }^{2}}A+2\cos A-2$
On simplifying the above equation, we get
$({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$ = $2(2{{\cos }^{2}}A-1)+2\cos A$
Now by using the identity $2{{\cos }^{2}}A-1=\cos 2A$, we get
$({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$= $2\cos 2A+2\cos A$
Now tale 2 common from R.H.S, we get
$({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$= $2(\cos 2A+\cos A)$
As $\cos A+\cos 2A=y$ is given in the equation
On putting the value, we get
$({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$= $2y$
Hence the value of $({{x}^{2}}+{{y}^{2}})({{x}^{2}}+{{y}^{2}}-3)$ is $2y$

Thus, Option (A) is correct.

Note: In these questions, students started to solve the complete equation. By solving the whole equation at one time makes us confused and it takes plenty of time. Always try to solve the questions in parts so that we cannot get confused and able to solve it in lesser time.