
How do you verify $\cos \left( {x + \dfrac{\pi }{2}} \right) = - \sin x$ ?
Answer
546.3k+ views
Hint: For solving this question, we will simplify the term on the left hand side and then prove that it is equal to the term on right hand side therefore, we will use the cosine sum formula, $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ by putting the values in the formula we will get term on the right hand side.
Complete step by step solution:
From the question, we know that, we have to prove the following equation –
$\cos \left( {x + \dfrac{\pi }{2}} \right) = - \sin x$
Now, from the above, we can conclude that, $\cos \left( {x + \dfrac{\pi }{2}} \right)$ is the left hand side term and $ - \sin x$ is the right hand side term. Therefore, we will take the term on left hand side because it will make easy to prove the equation by using the cosine sum formula, which is, $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
Hence, the left – hand side term is –
$ \Rightarrow \cos \left( {x + \dfrac{\pi }{2}} \right)$
Now, comparing the above term with the cosine sum formula, we get –
$
A = x \\
B = \dfrac{\pi }{2} \\
$
Therefore, using the cosine sum formula, we get –
$ \Rightarrow \cos \left( {x + \dfrac{\pi }{2}} \right) = \cos x\cos \dfrac{\pi }{2} - \sin x\sin \dfrac{\pi }{2}$
Now, we know that, $\cos \dfrac{\pi }{2} = 0$ and $\sin \dfrac{\pi }{2} = 1$ , putting these values in the above equation, we get –
$
\Rightarrow \cos \left( {x + \dfrac{\pi }{2}} \right) = \cos x \times 0 - \sin x \times 1 \\
\Rightarrow \cos \left( {x + \dfrac{\pi }{2}} \right) = - \sin x \\
$
Hence, $\cos \left( {x + \dfrac{\pi }{2}} \right) = - \sin x$, therefore, the left – hand side is equal to the term on the right hand side.
Hence, we verified the identity which was given to us in the question.
Note:
We should read the question very well and then move into the right direction, so that we get our required result. Don’t get confused between cosine sum identity and cosine difference identity. Use those identities according to the situation, when two angles are added we use a sum formula and if both are subtracted, we use a different formula. The identities of trigonometry should be kept in mind as they are required to solve any trigonometry question.
Complete step by step solution:
From the question, we know that, we have to prove the following equation –
$\cos \left( {x + \dfrac{\pi }{2}} \right) = - \sin x$
Now, from the above, we can conclude that, $\cos \left( {x + \dfrac{\pi }{2}} \right)$ is the left hand side term and $ - \sin x$ is the right hand side term. Therefore, we will take the term on left hand side because it will make easy to prove the equation by using the cosine sum formula, which is, $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
Hence, the left – hand side term is –
$ \Rightarrow \cos \left( {x + \dfrac{\pi }{2}} \right)$
Now, comparing the above term with the cosine sum formula, we get –
$
A = x \\
B = \dfrac{\pi }{2} \\
$
Therefore, using the cosine sum formula, we get –
$ \Rightarrow \cos \left( {x + \dfrac{\pi }{2}} \right) = \cos x\cos \dfrac{\pi }{2} - \sin x\sin \dfrac{\pi }{2}$
Now, we know that, $\cos \dfrac{\pi }{2} = 0$ and $\sin \dfrac{\pi }{2} = 1$ , putting these values in the above equation, we get –
$
\Rightarrow \cos \left( {x + \dfrac{\pi }{2}} \right) = \cos x \times 0 - \sin x \times 1 \\
\Rightarrow \cos \left( {x + \dfrac{\pi }{2}} \right) = - \sin x \\
$
Hence, $\cos \left( {x + \dfrac{\pi }{2}} \right) = - \sin x$, therefore, the left – hand side is equal to the term on the right hand side.
Hence, we verified the identity which was given to us in the question.
Note:
We should read the question very well and then move into the right direction, so that we get our required result. Don’t get confused between cosine sum identity and cosine difference identity. Use those identities according to the situation, when two angles are added we use a sum formula and if both are subtracted, we use a different formula. The identities of trigonometry should be kept in mind as they are required to solve any trigonometry question.
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