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Question: How do you verify \({{\cos }^{4}}x-{{\sin }^{4}}x={{\cos }^{2}}x-{{\sin }^{2}}x?\)...

How do you verify cos⁡4x−sin⁡4x=cos⁡2x−sin⁡2x?{{\cos }^{4}}x-{{\sin }^{4}}x={{\cos }^{2}}x-{{\sin }^{2}}x?

Explanation

Solution

To verify the given question of trigonometric function cos⁡4x−sin⁡4x=cos⁡2x−sin⁡2x.{{\cos }^{4}}x-{{\sin }^{4}}x={{\cos }^{2}}x-{{\sin }^{2}}x. You have to solve this problem using basics of algebraic formula that is (a2−b2)=(a−b)(a+b)\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right) and also you have to used trigonometric identity which is cos⁡2x+sin⁡2x=1.{{\cos }^{2}}x+{{\sin }^{2}}x=1. For solving the problem, consider one of the sides of the equation.

Complete step-by-step answer:
Let us consider the left hand side of the equation of
cos⁡4x−sin⁡4x=cos⁡2x−sin⁡2x{{\cos }^{4}}x-{{\sin }^{4}}x={{\cos }^{2}}x-{{\sin }^{2}}x for the verification.
LHS =cos⁡4x−sin⁡4x={{\cos }^{4}}x-{{\sin }^{4}}x
Now, cos⁡4x{{\cos }^{4}}x and sin⁡4x{{\sin }^{4}}x can also be written as shown in below
i.e., cos⁡4x=(cos⁡2x)2{{\cos }^{4}}x={{\left( {{\cos }^{2}}x \right)}^{2}}and (sin⁡4x)=(sin⁡2x)2\left( {{\sin }^{4}}x \right)={{\left( {{\sin }^{2}}x \right)}^{2}}
Now, Putting the above values in the left hand side of the equation we have,
LHS =(cos⁡2x)2−(sin⁡2x)2={{\left( {{\cos }^{2}}x \right)}^{2}}-{{\left( {{\sin }^{2}}x \right)}^{2}}
Therefore, by using formula of (a2−b2)=(a−b)(a+b)\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right) above equation can be modified as
LHS =(cos⁡2x−sin⁡2x)(cos⁡2x+sin⁡2x)=\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)
Now, by applying the trigonometric identity that
(cos⁡2x+sin⁡2x)=1\left( {{\cos }^{2}}x+{{\sin }^{2}}x \right)=1 the above equation will be
LHS=(cos⁡2x−sin⁡2x)×1=\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)\times 1
LHS =(cos⁡2x−sin⁡2x)=\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) .. (consider as equation (i)(i))
The right hand side of the given equation in the problem is,
RHS =(cos⁡2x−sin⁡2x)=\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right) .. (consider as equation (ii)\left( ii \right))
On comparing the equation (i)(i) and equation (ii)(ii) we conclude that,
LHS=RHSLHS=RHS
cos⁡4x−sin⁡4x=cos⁡2x−sin⁡2x{{\cos }^{4}}x-{{\sin }^{4}}x={{\cos }^{2}}x-{{\sin }^{2}}x

Hence, it is verified that cos⁡4x−sin⁡4x=cos⁡2x−sin⁡2x{{\cos }^{4}}x-{{\sin }^{4}}x={{\cos }^{2}}x-{{\sin }^{2}}x

Additional Information:
Trigonometric functions are the function in mathematics. Which represent the ratios of the sides of the right angle triangle. There are six basic trigonometric functions in mathematics which are sine, cosine, tangent, cotangent, secant and cosecant.
Here, sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 this identity is get form Pythagoras theorem of right angled triangle with hypotenuse of having length 1.1.
From figure (1)(1) given one side of triangle is taken sin⁡x\sin x and another is cos⁡x\cos x having a hypotenuse length of triangle is 11

Then using Pythagoras theorem which states that the square of the hypotenuse side is equal to the sum of squares of the other two sides of the triangle now.
AC2=(AB)2+(BC)2A{{C}^{2}}={{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}
l=sin⁡2x+cos⁡2xl={{\sin }^{2}}x+{{\cos }^{2}}x
sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1

Note:
Points to remember while solving the solution of a given problem. Split the given trigonometric function to solve it further easily and then apply the arithmetic identity properly using suitable formula of (a2−b2)\left( {{a}^{2}}-{{b}^{2}} \right)which is equal to (a−b)(a+b)\left( a-b \right)\left( a+b \right)
sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1, this trigonometric function can be directly applied in the solution.