Question

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

Answer 1

Hint: 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 $A\sin \left( x+\phi \right).$ Using this equation, we can get the values of the constant A and angle $\phi .$

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

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