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How do you prove that $\sin x + \cos x = 1$ is not an identity by showing a
counterexample?

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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Hint: As we know that above question is related to trigonometric function since sine and cosine are trigonometric ratios. The trigonometric identity is represented with the help of an equation that consists of trigonometric ratios. These equations are true for Right angled triangles. We can show this with the help of a counterexample.

Complete step by step solution:
We know that to verify or to prove an identity means to prove that the equation is true by showing that both the left hand side and right hand side of the equation are equal. We have to use logical steps to show that one side of the equation can be transformed into the other side of the equation.
Here we have the equation $\sin x + \cos x = 1$ . By taking the left hand side and assuming $x = \dfrac{\pi }{4}$.
Now in the left hand side we have $\sin \left( {\dfrac{\pi }{4}} \right) + \cos \left( {\dfrac{\pi }{4}}
\right)$. We know that the value of $\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$.
By substituting the values we get $\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} \Rightarrow
\dfrac{2}{{\sqrt 2 }}$,
Therefore in the right hand side we get the value $\sqrt 2 $, It is not equal to the right hand side, $\sqrt 2 \ne 1$.
Hence it is proved that $\sin x + \cos x = 1$is not an identity with the help of a counterexample.

Note: We should be careful while solving this type of question and we have to be careful while putting the identities of trigonometric ratios in the wrong place. These can also be solved by using another counterexample. The trigonometric identities are equations that are true for Right angled triangles.
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