Question

How do you prove ${\cos ^4}\left( x \right) - {\sin ^4}\left( x \right) = \cos \left( {2x} \right)$?

Answer 1

Hint : Here in this question, given a trigonometric equation. We have to prove left hand side is equal to right hand side $\left( {LHS = RHS} \right)$ of given trigonometric equation, left hand side solve by using a algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ and right hand solve by using a cosine sum rule $\cos \left( {a + b} \right) = \cos \left( a \right)\cos \left( b \right) - \sin \left( a \right)\sin \left( b \right)$. And on further simplification to get the required solution.

Complete step by step solution:
In trigonometry the functions sine (sin), cosine (cos) and tangent (tan) of an angle are sometimes referred to as the primary or basic trigonometric functions. The remaining trigonometric functions secant (sec), cosecant (csc), and cotangent (cot) are defined as the reciprocal functions of cosine, sine, and tangent, respectively.
Now, consider a given trigonometric equation:
$\Rightarrow {\cos ^4}\left( x \right) - {\sin ^4}\left( x \right) = \cos \left( {2x} \right)$
Here, we have to prove $LHS = RHS$
Consider left hand side of equation i.e., $LHS$
$\Rightarrow {\cos ^4}\left( x \right) - {\sin ^4}\left( x \right)$
Or it can be re-written as
$\Rightarrow {\left( {{{\cos }^2}\left( x \right)} \right)^2} - {\left( {{{\sin }^4}\left( x \right)} \right)^2}$
It’s looks similar as the algebraic identity: ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
Here, $a = \cos \left( x \right)$ and $b = \sin \left( x \right)$. Then on substituting we have
$\Rightarrow \left( {{{\cos }^2}\left( x \right) - {{\sin }^2}\left( x \right)} \right)\left( {{{\cos }^2}\left( x \right) + {{\sin }^2}\left( x \right)} \right)$
As by the standard trigonometric identity: ${\sin ^2}x + {\cos ^2}x = 1$, then
$\Rightarrow \left( {{{\cos }^2}\left( x \right) - {{\sin }^2}\left( x \right)} \right)\left( 1 \right)$
$\Rightarrow \,\,{\cos ^2}\left( x \right) - {\sin ^2}\left( x \right)$---------(1)
Next, consider right hand side of equation i.e., $RHS$
$\Rightarrow \cos \left( {2x} \right)$
Or It can be re-written as
$\Rightarrow \cos \left( {x + x} \right)$
Use a cosine sum identity: $\cos \left( {a + b} \right) = \cos \left( a \right)\cos \left( b \right) - \sin \left( a \right)\sin \left( b \right)$
Here, $a = x$ and $b = x$. Then on substituting we have
$\Rightarrow \,\,\cos \left( {x + x} \right) = \cos \left( x \right)\cos \left( x \right) - \sin \left( x \right)\sin \left( x \right)$
$\Rightarrow \,\,\cos \left( {2x} \right) = {\cos ^2}\left( x \right) - {\sin ^2}\left( x \right)$
$\Rightarrow \,\,{\cos ^2}\left( x \right) - {\sin ^2}\left( x \right)$--------(2)
Then, By equation (1) and (2)
$LHS = RHS$
Therefore, ${\cos ^4}\left( x \right) - {\sin ^4}\left( x \right) = \cos \left( {2x} \right)$
Hence, proved.

Note : The question involves the trigonometric functions and we have to prove the trigonometric function. When we simplify the trigonometric functions and which will be equal to the RHS then the function is proved. While simplifying the trigonometric functions we must know about the trigonometric ratios, trigonometric identities, formulas of sum and difference rule and double angles.