Question

How do you simplify the expression \[\dfrac{{{\cos }^{3}}t+{{\sin }^{3}}t}{{{\left( \cos t+\sin t \right)}^{2}}}\]?

Answer 1

Hint: Assume the given trigonometric expression as E. Now, leave the denominator as it is and simplify the numerator by using the algebraic identity: - \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\]. Cancel the common terms and use the identity: - \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to get the final simplified form and answer.

Complete step by step solution:
Here, we have been provided with the expression \[\dfrac{{{\cos }^{3}}t+{{\sin }^{3}}t}{{{\left( \cos t+\sin t \right)}^{2}}}\] and we are asked to simplify it. Let us assume the given trigonometric expression as E and use some algebraic and trigonometric formulas to solve this question. So, we have,
\[\Rightarrow E=\dfrac{{{\cos }^{3}}t+{{\sin }^{3}}t}{{{\left( \cos t+\sin t \right)}^{2}}}\]
Now, leaving the denominator as it is and simplifying the numerator by using the algebraic identity: - \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\], we get,
\[\Rightarrow E=\dfrac{\left( \cos t+\sin t \right)\left( {{\cos }^{2}}t-{{\sin }^{2}}t-\sin t\cos t \right)}{{{\left( \cos t+\sin t \right)}^{2}}}\]
Cancelling the common term which is \[\left( \cos t+\sin t \right)\] from the numerator and denominator, we get,
\[\Rightarrow E=\dfrac{\left( {{\cos }^{2}}t-{{\sin }^{2}}t-\sin t\cos t \right)}{\left( \cos t+\sin t \right)}\]
Using the trigonometric identity: - \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to simplify the numerator, we get,
\[\Rightarrow E=\dfrac{\left( 1-\sin t\cos t \right)}{\left( \cos t+\sin t \right)}\]
Hence, the above expression is the required simplified form.

Note: One may note that we can write the obtained simplified form of the expression in terms of other trigonometric functions also. For example: - we can divide the numerator and the denominator with the term \[\sin t\cos t\] and in this way we will get the simplified form as \[\left( \dfrac{\sec t\csc t-1}{\sec t+\csc t} \right)\] which can also be our answer. The main things you have to remember are certain algebraic identities and trigonometric identities like the basic three identities: - \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], \[1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta \] and \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \]. Remember the algebraic identities: - \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\] and \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)\]. You can also remember the formulas: - \[2\sin \theta \cos \theta =\sin 2\theta \] and \[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\cos 2\theta \], which are used in higher trigonometry.