Question

If \[\sin A - \cos A = k,\] then ${\sin ^3}A - {\cos ^3}A = $
A.$3k + {k^3}$
B.$3k - {k^3}$
C.$\dfrac{{3k + {k^3}}}{2}$
D.$\dfrac{{3k - {k^3}}}{2}$

Answer 1

Hint: Here we will use different identities using the formula for the square of the difference of two terms and cube of the difference between two terms. Also, use the identity of the square of sine and cosine functions.

Complete step-by-step answer:
Take the given expression –
\[\sin A - \cos A = k\] .... (i)
Take square on both sides of the equation. Use the identity of difference of cube formula: ${(a - b)^2} = {a^2} - 2ab + {b^2}$
${\sin ^2}A - 2\sin A\cos A + {\cos ^2}A = {k^2}$
Also, the identity states that ${\sin ^2}A + {\cos ^2}A = 1$ . Place the values in the above equation.
$
  \operatorname{s} \underline {i{n^2}A + {{\cos }^2}A} - 2\sin A\cos A = {k^2} \\
   \Rightarrow 1 - 2\sin A\cos A = {k^2} \\
 $
Make sine and cosine the subject and take rest on one side of the equation-
$ \Rightarrow 1 - {k^2} = 2\sin A\cos A$
When the term multiplicative at one side changes its side, then it goes in the division.
$ \Rightarrow \sin A\cos A = \dfrac{{1 - {k^2}}}{2}$ ...... (ii)
Again take the given expression
\[\sin A - \cos A = k\]
Take the cube of the given equation on both sides of the equation.
\[{\left( {\sin A - \cos A} \right)^3} = {k^3}\]
Use the identity of difference of cube formula: ${(a - b)^3} = {a^3} - {b^3} - 3ab(a - b)$
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A - 3\cos A\sin A(\operatorname{Sin} A - \cos A) = {k^3}\]
Place the value in the above equation from the equation (i)
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A - 3\cos A\sin A(k) = {k^3}\]
Take all the terms on one side and the required terms on the left hand side of the equation. When the term is moved from one side to another of the equation, then the sign of the term is also changed. Positive terms become negative and vice-versa.
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A = {k^3} + 3\cos A\sin A(k)\]
Take the common multiple on the right hand side of the equation –
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A = k({k^2} + 3\cos A\sin A)\]
Place the values in the above equation from equation (ii)
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A = k({k^2} + 3\left( {\dfrac{{1 - {k^2}}}{2}} \right))\]
Simplify the above equation –
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A = k({k^2} + \left( {\dfrac{{3 - 3{k^2}}}{2}} \right))\]
Take LCM on the right hand side of the equation-
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A = \dfrac{k}{2}(2{k^2} + 3 - 3{k^2})\]
Add and subtract among like terms –
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A = \dfrac{k}{2}(\underline {2{k^2} - 3{k^2}} + 3)\]
Remember when you do subtraction and addition among the terms do subtraction and give signs of greater numbers.
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A = \dfrac{k}{2}( - {k^2} + 3)\]
Above equation can be re-written as –
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A = (\dfrac{{3k - {k^3}}}{2})\]
So, the correct answer is “\[ {\sin ^3}A - {\cos ^3}A = (\dfrac{{3k - {k^3}}}{2})\]”.

Note: Always remember the basic standard formulas of the algebraic expressions to expand the terms. Place the value using the relevant trigonometric function and reduce the resultant expansion. Remember and apply the basic trigonometric identities.