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How do you verify that the given trigonometric equation is correct and satisfy the condition, the given function is \[\left( {\cos (x)} \right)\left( {\cos (x)} \right) = 1 - {\sin ^2}x\]?

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Last updated date: 20th Jun 2024
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Answer
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Hint: To prove a trigonometric function either we can use direct properties available for the function to achieve our result or you can satisfy the equation by assuming any angle and get that left the hand side of the equation is equal to the right hand side, both ways are correct, accurate and will give the result.

Formulae Used:
\[ \Rightarrow {\sin ^2}x + {\cos ^2}x = 1\]

Complete step by step solution:
The given question is \[\left( {\cos (x)} \right)\left( {\cos (x)} \right) = 1
- {\sin ^2}x\]
Here we will solve this question first by using the trigonometric property that is:
\[ \Rightarrow {\sin ^2}x + {\cos ^2}x = 1\]
By rearranging the splitting the above property we get:
\[
\Rightarrow {\sin ^2}x + {\cos ^2}x = 1 \\
\Rightarrow {\cos ^2}x = 1 - {\sin ^2}x \\
\]
Here we obtain our result.
Now lets satisfy the given equation at zero degree, on solving we get:
\[
\Rightarrow \left( {\cos (x)} \right)\left( {\cos (x)} \right) = 1 - {\sin ^2}x \\
\Rightarrow \cos (0) = 1,\,and\,\sin (0) = 0 \\\]
Putting the values, we get:
\[\Rightarrow \cos (0)\cos (0) = 1 - {\sin ^2}(0) \\
\Rightarrow 1 \times 1 = 1 - {\left( 0 \right)^2} \\
\Rightarrow 1 = 1 - 0 \\
\Rightarrow 1 = 1 \\
\]
Here also we obtain that the left hand side and right hand side of the given equation gives the same value on the assumed angle, and since it's already a trigonometric property hence will satisfy for every angle.

Additional Information: Here you can solve after transforming every term into a single function then you will get that the given function of “sin” and “cos” will become one and one minus one will give you zero which is present on the right side of the equation.

Note: To prove any trigonometric equation you can also simplify the equation and since every trigonometric function is inter relatable and accordingly you can solve for it, with this method you can convert every given function into one trigonometric identity and then you can prove for left hand side of equation equals to right hand side.