Question

Solve the following equation \[\sin \theta +\cos \theta =1\].

Answer 1

Hint: In the given equation we will multiply it by \[\dfrac{1}{\sqrt{2}}\] to both sides of the equality and then we will use the trigonometric identity to solve it further which is given as follows:

Complete step-by-step answer:

\[\operatorname{sinA}\operatorname{cosB}+\cos A\sin B=\sin \left( A+B \right)\]

We have been given the equation \[\sin \theta +\cos \theta =1\].

Now we will multiply the equation by \[\dfrac{1}{\sqrt{2}}\] and we get as follows:

\[\dfrac{1}{\sqrt{2}}\sin \theta +\dfrac{1}{\sqrt{2}}\cos \theta =\dfrac{1}{\sqrt{2}}\]

As we know that the value of \[\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\]

So by using the above values in the equation, we get as follows:

\[\cos \dfrac{\pi }{4}\sin \theta +\sin \dfrac{\pi }{4}\cos \theta =\dfrac{1}{\sqrt{2}}\]

We also know the trigonometric identity as follows:

\[\operatorname{sinA}\operatorname{cosB}+\cos A\sin B=\sin \left( A+B \right)\]

So by using this trigonometric identity in the above equation, we get as follows:

\[\begin{align}

  & \cos \dfrac{\pi }{4}\sin \theta +\sin \dfrac{\pi }{4}\cos \theta =\dfrac{1}{\sqrt{2}} \\

 & \sin \left( \theta +\dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right) \\

\end{align}\]

As we know that the general solution for \[\sin \theta =\sin \alpha \] is given by \[\theta =n\pi +{{(-1)}^{n}}\alpha \] where ‘n’ is any integer.

\[\begin{align}

  & \sin \left( \theta +\dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right) \\

 & \theta +\dfrac{\pi }{4}=n\pi +{{(-1)}^{n}}\dfrac{\pi }{4} \\

\end{align}\]

On taking \[\dfrac{\pi }{4}\] to the right hand side of the equation, we get as follows:

\[\theta =n\pi +{{(-1)}^{n}}\dfrac{\pi }{4}-\dfrac{\pi }{4}\]

On taking \[\dfrac{\pi }{4}\] as common, we get as follows:

\[\theta =n\pi +\left( {{\left( -1 \right)}^{n}}-1 \right)\dfrac{\pi }{4}\]

On substituting the obtained value of ‘\[\theta \]’in the given equation we can observe that the given equation does not satisfy for n = 0.

Therefore, the solution of the given equation is given by \[\theta =n\pi +\left( {{\left( -1

\right)}^{n}}-1 \right)\dfrac{\pi }{4}\] where \[n\in Z-\left\{ 0 \right\}\].

Here, Z means integer.


Note: Check the solution once by substituting the values in the equation by squaring both sides of the equation as it might give you the correct answer some time but it is not a correct method to solve.

We can also use the trigonometric identity \[\cos \left( A+B \right)\] instead of \[\sin (A+B)\], we will get the answer.