Question

How do you simplify the expression $\sin t - \sin t{\cos ^2}t$?

Answer 1

Hint: We will first take sin t common from the given expression, then we will obtain such an expression in which we could use the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$.

Complete step-by-step solution:
We are given that we are required to simplify the given expression $\sin t - \sin t{\cos ^2}t$.
Let us assume this expression to be $I = \sin t - \sin t{\cos ^2}t$.
Now, since we can take sin t common out of the given expression, we can write the above given expression as following:-
$ \Rightarrow I = \sin t\left( {1 - {{\cos }^2}t} \right)$ …………..(1)
Since, we know that we have an identity which is given by the following expression:-
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1$ for any real number $\theta $.
Replacing $\theta $ by t in the above expression, we will then obtain the following expression:-
$ \Rightarrow {\sin ^2}t + {\cos ^2}t = 1$ for any real number t.
Taking the term with cosine of t from addition on the left hand side to the right hand side, we will then obtain the following equation:-
$ \Rightarrow {\sin ^2}t = 1 - {\cos ^2}t$ for any real number t.
Putting the above expression in equation number (1), we will then obtain the following equation with us:-
$ \Rightarrow I = \sin t\left( {{{\sin }^2}t} \right)$
Clubbing the like terms on the right hand side, we will then obtain the following expression with us:-
$ \Rightarrow I = {\sin ^3}t$

Thus, we have the required answer as $\sin t - \sin t{\cos ^2}t = {\sin ^3}t$.

Note: The students must note the identity which is given by the expression: ${\sin ^2}\theta + {\cos ^2}\theta = 1$ for any real number $\theta $. This can be proved as well.
Since, we know that the sine of any angle $\theta $ is given by the ratio of perpendicular and the hypotenuse. Therefore, $\sin \theta = \dfrac{P}{H}$ …………..(A)
We also know that the cosine of any angle $\theta $ is given by the ratio of base and the hypotenuse. Therefore, $\cos \theta = \dfrac{B}{H}$ …………..(B)
Squaring both the equations (A) and (B) and adding both of them to obtain the following expression:-
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = {\left( {\dfrac{P}{H}} \right)^2} + {\left( {\dfrac{B}{H}} \right)^2}$
Simplifying it, we will then obtain:-
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = \dfrac{{{P^2} + {B^2}}}{{{H^2}}}$
Since, we know that by using Pythagorean Theorem, we have: ${P^2} + {B^2} = {H^2}$. Putting this, we get:-
$ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = \dfrac{{{P^2} + {B^2}}}{{{P^2} + {B^2}}} = 1$