Question

What does $cosx\ sinx$ equal?

Answer 1

Hint: This kind of question is solved based on the concept of trigonometry. We should be aware of trigonometry identities to solve the given problem. We are going to solve this given function by multiplying the given function with 2 and after that writing that particular function into separate terms, by using trigonometric identities of relevant equations, the resultant would be out.

Complete step by step solution:
The trigonometric identities are the equations which are accurate in the case of right angles triangles.
A few of them are:
\[\begin{align}
  & {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \\
 & {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\
 & {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \\
\end{align}\]
The above equations can also be written in the below form
\[\begin{align}
  & {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 \\
 & {{\cot }^{2}}\theta +1={{\csc }^{2}}\theta \\
 & {{\cot }^{2}}\theta ={{\csc }^{2}}\theta -1 \\
\end{align}\]
 Other than the basic trigonometric identities, we can compute the below equations from the basic trigonometric identities, they are
\[\begin{align}
  & \sin 2\theta =2\sin \theta \cos \theta \\
 & \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\
 & \tan 2\theta =2\tan \theta /\left( 1-{{\tan }^{2}}\theta \right) \\
 & \cot 2\theta =\left( {{\cot }^{2}}\theta -1 \right)/2\cot \theta \\
\end{align}\]
Above all equations are also known as Pythagorean identities.
Let us solve the given question
Given problem,
\[\cos x\sin x\]
Let us consider the given function as
\[f\left( x \right)=\cos x\sin x\]
To the above function we are multiplying 2 to the given function
\[\Rightarrow 2\cos x\sin x\]
Which also can be written as separate terms as below,
\[\Rightarrow \cos x\sin x+\cos x\sin x\]
Thus, it is the double angle formula for the sine.
\[\Rightarrow \cos x\sin x+\cos x\sin x=\sin \left( 2x \right)\]
We know that the formula of \[\sin 2x\] is
\[2\cos x\sin x=\sin \left( 2x \right)\]
So, we can simplify as
\[\Rightarrow \cos x\sin x=\dfrac{\sin \left( 2x \right)}{2}\]
Therefore the \[\cos x\sin x\] is equal to \[\dfrac{\sin \left( 2x \right)}{2}\] .

Note: Students do not understand how to approach a given trigonometric problem from the concept. Students make errors in applying identities and also do errors in calculations. Most of the mistakes occur especially during multiplications. Most students make errors at the manipulation of trig ratios using formula. Understanding the trigonometric based question is also a time taking process.