Question

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

Answer 1

Hint: To solve the question given above, use the trigonometric identities. Remember that while verifying trigonometric questions always make use of the identities and formulas. For this question use: \[\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b\] and \[\cos 2x = {\cos ^2}x - {\sin ^2}x\] .

 Formula used: In order to verify the above question, we will take help of the following formula:
\[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\].
\[{\sin ^2}x + {\cos ^2}x = 1\].
\[\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b\].
\[\cos 2x = {\cos ^2}x - {\sin ^2}x\]

Complete step-by-step answer:
We are given: \[{\cos ^4}x - {\sin ^4}x = \cos \left( {2x} \right)\]
Let us solve the L.H.S first,
\[{\cos ^4}x - {\sin ^4}x\]
Use the formula: \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\].
We get:
\[\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^{2x}}} \right)\]
Now, we know that \[\cos 2x = {\cos ^2}x - {\sin ^2}x\] and \[{\sin ^2}x + {\cos ^2}x = 1\]
So,
\[1 \times \left( {\cos 2x} \right) = \cos 2x = RHS\]
We get that \[LHS = RHS\].
Hence proved.
Additional information: we can also perform the above question in an alternative way:
We are given: \[{\cos ^4}x - {\sin ^4}x = \cos \left( {2x} \right)\].
Let us solve the L.H.S first,
\[{\cos ^4}x - {\sin ^4}x\]
Use the formula: \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\].
We get:
\[\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^{2x}}} \right)\]
Now, use the formula \[\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b\] and \[{\sin ^2}x + {\cos ^2}x = 1\].
We get:
\[
  1 \times \left( {{{\cos }^2}x - {{\sin }^2}x} \right) \\
   = \left( {{{\cos }^2}x - {{\sin }^2}x} \right) \\
   = \cos x\cos x - \sin x\sin x \\
 \]
This becomes:
\[
  \cos \left( {x + x} \right) \\
   = \cos \left( {2x} \right) \\
   = RHS \\
 \]
We get that \[LHS = RHS\].
Hence proved

Note: The trigonometric functions are real functions that link a right-angled triangle's angle to two-side length ratios. They are commonly used in all geodetic sciences, including navigation, solid mechanics, celestial mechanics, geodesy, and many others. They're one of the most basic periodic functions. The sine, cosine, and tangent are the most commonly used trigonometric functions in modern mathematics. The cosecant, secant, and cotangent are their reciprocals, which are less commonly used. Each of these six trigonometric functions has an inverse function (known as an inverse trigonometric function) and a hyperbolic function counterpart.
 Always remember that while verifying questions relating to trigonometry similar to the one given above, use the trigonometric identities and formulas. They make solving these questions easier and much more simpler.