Question

How do you simplify the expression \[{{\sin }^{3}}x+\sin x{{\cos }^{2}}x\]?

Answer 1

Hint: Take \[\sin x\] common from both the terms and write the remaining terms inside the bracket. Now, use the trigonometric identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to simplify the terms inside the bracket and get the answer.

Complete step by step solution:
Here, we have been provided with the trigonometric expression \[{{\sin }^{3}}x+\sin x{{\cos }^{2}}x\] and we are asked to simplify this expression. Here, we are going to use some basic trigonometric identity involving the sine and cosine function.
Now, let us assume this expression as E, so we have,
 \[\Rightarrow E={{\sin }^{3}}x+\sin x{{\cos }^{2}}x\]
Clearly, we can see that we have \[\sin x\] common in both the terms, so taking this sine function common, we get,
\[\Rightarrow E=\sin x\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\]
Now, in a right-angle triangle we have, \[\sin x=\dfrac{p}{h}\] and \[\cos x=\dfrac{b}{h}\], where p = perpendicular, b = base and h = hypotenuse. So, on squaring the sine and cosine function and adding them, we get,
\[\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=\dfrac{{{p}^{2}}}{{{h}^{2}}}+\dfrac{{{b}^{2}}}{{{h}^{2}}}\]
Taking L.C.M. in the R.H.S. and simplifying, we get,
\[\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=\left( \dfrac{{{p}^{2}}+{{b}^{2}}}{{{h}^{2}}} \right)\]
In a right-angle triangle we know that the sum of squares of perpendicular and base is equal to the square of hypotenuse of the triangle, so mathematically we have,
\[\Rightarrow {{p}^{2}}+{{b}^{2}}={{h}^{2}}\]
Substituting this value in the above expression, we get,
\[\begin{align}
  & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=\dfrac{{{h}^{2}}}{{{h}^{2}}} \\
 & \Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
\end{align}\]
So, in general we have \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and this result is valid for any value of \[\theta \], so applying this identity we get the expression ‘E’ as: -
\[\begin{align}
  & \Rightarrow E=\sin x\times 1 \\
 & \Rightarrow E=\sin x \\
\end{align}\]

Note: You must remember the three basic trigonometric identities which are: - \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], \[1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta \] and \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \]. Without using the first identity mentioned here it will be difficult to solve the given question. You may check the answer by substituting some particular angular value of x like: - x can be \[{{0}^{\circ }},{{30}^{\circ }},{{45}^{\circ }},{{60}^{\circ }},{{90}^{\circ }}\] etc. The trigonometric values of these angles are known to us so they can be used as a proof.