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Question: How do you rewrite \(2\sin \left( x \right)-2\cos \left( x \right)\) as \(A\sin \left( x+\phi \right...

How do you rewrite 2sin⁡(x)−2cos⁡(x)2\sin \left( x \right)-2\cos \left( x \right) as Asin⁡(x+ϕ)?x and ϕ=?A\sin \left( x+\phi \right)?x\ and\ \phi =?

Explanation

Solution

For this question, we have to use basic trigonometric identities in order to obtain the solution. We write the equation of the trigonometric identity and then compare it with the equation given in question. We then obtain the desired form as Asin⁡(x+ϕ).A\sin \left( x+\phi \right). Using this equation, we can get the values of the constant A and angle ϕ.\phi .

Complete step by step solution:
We know the basic trigonometric identity for addition is given as,
sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B……(1)\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\ldots \ldots \left( 1 \right)
Here, let us assume A=xA=x and B=ϕ.B=\phi . We can now rewrite this equation (1) as follows.
⇒sin⁡(x+ϕ)=sin⁡xcos⁡ϕ+cos⁡xsin⁡ϕ\Rightarrow \sin \left( x+\phi \right)=\sin x\cos \phi +\cos x\sin \phi
We now multiply both sides of the above given equation by a constant term A.
⇒Asin⁡(x+ϕ)=Asin⁡xcos⁡ϕ+Acos⁡xsin⁡ϕ……(2)\Rightarrow A\sin \left( x+\phi \right)=A\sin x\cos \phi +A\cos x\sin \phi \ldots \ldots \left( 2 \right)
We need to compare this equation to the one given in the question 2sin⁡(x)−2cos⁡(x).2\sin \left( x \right)-2\cos \left( x \right).
We need to compare the right-hand side of equation (2) to the one in question.
Comparing these equations,
⇒Asin⁡xcos⁡ϕ+Acos⁡xsin⁡ϕ=2sin⁡(x)−2cos⁡(x)\Rightarrow A\sin x\cos \phi +A\cos x\sin \phi =2\sin \left( x \right)-2\cos \left( x \right)
Comparing both the terms individually, taking the first term first,
⇒Asin⁡xcos⁡ϕ=2sin⁡x\Rightarrow A\sin x\cos \phi =2\sin x
Cancelling out the sin⁡x\sin x terms on both sides,
⇒Acos⁡ϕ=2……(3)\Rightarrow A\cos \phi =2\ldots \ldots \left( 3 \right)
Comparing the second term now,
⇒Acos⁡xsin⁡ϕ=−2cos⁡x\Rightarrow A\cos x\sin \phi =-2\cos x
Cancelling out the cos⁡x\cos x terms on both sides,
⇒Asin⁡ϕ=−2……(4)\Rightarrow A\sin \phi =-2\ldots \ldots \left( 4 \right)
Dividing equation (4) by equation (3),
⇒Asin⁡ϕAcos⁡ϕ=−22\Rightarrow \dfrac{A\sin \phi }{A\cos \phi }=\dfrac{-2}{2}
Cancelling A term in the numerator and denominator of left-hand side. We also know that sin⁡xcos⁡x=tan⁡x.\dfrac{\sin x}{\cos x}=\tan x.
Therefore, LHS simplifies to
⇒tan⁡ϕ=−1\Rightarrow \tan \phi =-1
We can calculate ϕ\phi by taking the tan function as tan inverse on the right-hand side.
⇒ϕ=tan⁡−1(−1)\Rightarrow \phi ={{\tan }^{-1}}\left( -1 \right)
Referring to the table for tan angle values, we know that tan⁡−1(−1)=−π4.{{\tan }^{-1}}\left( -1 \right)=-\dfrac{\pi }{4}.
Therefore, we get the value of ϕ\phi as,
⇒ϕ=−π4\Rightarrow \phi =-\dfrac{\pi }{4}
Using equation (4) and substituting the value of ϕ,\phi ,
⇒Asin⁡(−π4)=−2\Rightarrow A\sin \left( -\dfrac{\pi }{4} \right)=-2
We know that sin⁡(−π4)=−sin⁡(π4)=−12.\sin \left( -\dfrac{\pi }{4} \right)=-\sin \left( \dfrac{\pi }{4} \right)=-\dfrac{1}{\sqrt{2}}. Substituting this in the above equation,
⇒A(−12)=−2\Rightarrow A\left( -\dfrac{1}{\sqrt{2}} \right)=-2
Cross multiplying the term (−12)\left( -\dfrac{1}{\sqrt{2}} \right) to the RHS,
⇒A=−2×−2\Rightarrow A=-2\times -\sqrt{2}
⇒A=22\Rightarrow A=2\sqrt{2}
Taking the RHS of equation (2), and substituting the values of A and ϕ\phi obtained from above,
⇒Asin⁡(x+ϕ)=22sin⁡(x−π4)\Rightarrow A\sin \left( x+\phi \right)=2\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right)
Hence, we rewrite 2sin⁡(x)−2cos⁡(x)2\sin \left( x \right)-2\cos \left( x \right) as 22sin⁡(x−π4),2\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right), where A=22A=2\sqrt{2} and ϕ=−π4.\phi =-\dfrac{\pi }{4}.

Note: We can even solve this question using the identity for sin⁡(A−B)\sin \left( A-B \right) similarly. All the steps remain the same. We need to have a good knowledge of trigonometric identities to solve such questions.