
How do you simplify the expression $\sin t - \sin t{\cos ^2}t$?
Answer
561k+ views
Hint: This problem deals with some basic trigonometric identities and basic trigonometric formulas. Along with this we also need to understand and should be able to solve simple mathematical equations.
Here the trigonometric identity which is used here is as given below:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$
Hence by rearranging the terms, the above expression becomes, as given below:
$ \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta $
Complete step-by-step answer:
Here considering the given expression of trigonometric function, as given below:
$ \Rightarrow \sin t - \sin t{\cos ^2}t$
We can observe that from the above expression, the term $\sin t$ is common, and hence taking this term common, as shown below:
$ \Rightarrow \sin t\left( {1 - {{\cos }^2}t} \right)$
We know that from basic trigonometric identity that ${\sin ^2}t + {\cos ^2}t = 1$, hence we can deduce that the expression $1 - {\cos ^2}t = {\sin ^2}t$, hence the expression $1 - {\cos ^2}t$ can be replaced with the term ${\sin ^2}t$, as shown below:
$ \Rightarrow \sin t\left( {1 - {{\cos }^2}t} \right)$
Substituting the expression of $1 - {\cos ^2}t$ as ${\sin ^2}t$, as given below:
$ \Rightarrow \sin t\left( {{{\sin }^2}t} \right)$
Now simplifying the above expression, as shown below:
$ \Rightarrow {\sin ^3}t$
So to simplify the expression $\sin t - \sin t{\cos ^2}t$ is equal to ${\sin ^3}t$.
$\therefore \sin t - \sin t{\cos ^2}t = {\sin ^3}t$
Final Answer: The simplification of the expression $\sin t - \sin t{\cos ^2}t$ is equal to ${\sin ^3}t$.
Note:
Please note that while solving the above problem, we used the basic trigonometric identity which is the sum of the squares of the sine and cosine trigonometric ratios is always equal to 1. Which is given by:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$
Similarly there are other two basic identities, where the difference of the squares of the secant and the tangent trigonometric ratios is equal to 1.
$ \Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1$
Also another identity:
$ \Rightarrow \cos e{c^2}\theta - {\cot ^2}\theta = 1$
Here the trigonometric identity which is used here is as given below:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$
Hence by rearranging the terms, the above expression becomes, as given below:
$ \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta $
Complete step-by-step answer:
Here considering the given expression of trigonometric function, as given below:
$ \Rightarrow \sin t - \sin t{\cos ^2}t$
We can observe that from the above expression, the term $\sin t$ is common, and hence taking this term common, as shown below:
$ \Rightarrow \sin t\left( {1 - {{\cos }^2}t} \right)$
We know that from basic trigonometric identity that ${\sin ^2}t + {\cos ^2}t = 1$, hence we can deduce that the expression $1 - {\cos ^2}t = {\sin ^2}t$, hence the expression $1 - {\cos ^2}t$ can be replaced with the term ${\sin ^2}t$, as shown below:
$ \Rightarrow \sin t\left( {1 - {{\cos }^2}t} \right)$
Substituting the expression of $1 - {\cos ^2}t$ as ${\sin ^2}t$, as given below:
$ \Rightarrow \sin t\left( {{{\sin }^2}t} \right)$
Now simplifying the above expression, as shown below:
$ \Rightarrow {\sin ^3}t$
So to simplify the expression $\sin t - \sin t{\cos ^2}t$ is equal to ${\sin ^3}t$.
$\therefore \sin t - \sin t{\cos ^2}t = {\sin ^3}t$
Final Answer: The simplification of the expression $\sin t - \sin t{\cos ^2}t$ is equal to ${\sin ^3}t$.
Note:
Please note that while solving the above problem, we used the basic trigonometric identity which is the sum of the squares of the sine and cosine trigonometric ratios is always equal to 1. Which is given by:
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$
Similarly there are other two basic identities, where the difference of the squares of the secant and the tangent trigonometric ratios is equal to 1.
$ \Rightarrow {\sec ^2}\theta - {\tan ^2}\theta = 1$
Also another identity:
$ \Rightarrow \cos e{c^2}\theta - {\cot ^2}\theta = 1$
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