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Question: How do you verify \[{\cos ^4}x - {\sin ^4}x = 1 - 2{\sin ^2}x\]?...

How do you verify cos⁡4x−sin⁡4x=1−2sin⁡2x{\cos ^4}x - {\sin ^4}x = 1 - 2{\sin ^2}x?

Explanation

Solution

Trigonometric identities trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined.
In this question we have to prove LHS is equal to RHS, for doing this we use algebraic identity a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right), and simplify using the trigonometric identity, cos⁡2x−sin⁡2x=1−2sin⁡2x{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x, we will get the required RHS.

Complete step-by-step solution:
An identity is an equation that always holds true. A trigonometric identity is an identity that contains trigonometric functions and holds true for all right-angled triangles.
The six basic trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent. All these trigonometric ratios are defined using the sides of the right triangle, such as an adjacent side, opposite side, and hypotenuse side.
Given equation is,cos⁡4x−sin⁡4x{\cos ^4}x - {\sin ^4}x,
We have to prove that L.H.S term is equal to R.H.S,
Now using the algebraic identity a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right),
By equating LHS term to the identity we get,
Here a=cos⁡2x,b=sin⁡2xa = {\cos ^2}x,b = {\sin ^2}x,
Now rewriting the L.H.S given expression as,
\Rightarrow $$$${\left( {{{\cos }^2}x} \right)^2} - {\left( {{{\sin }^2}x} \right)^2},
Now applying the identity, we get,
\Rightarrow $$$${\left( {{{\cos }^2}x} \right)^2} - {\left( {{{\sin }^2}x} \right)^2} = \left( {{{\cos }^2}x + {{\sin }^2}x} \right)\left( {{{\cos }^2}x - {{\sin }^2}x} \right),
Now using trigonometric identitycos⁡2x+sin⁡2x=1{\cos ^2}x + {\sin ^2}x = 1, we get,
\Rightarrow $$$${\cos ^4}x - {\sin ^4}x = \left( 1 \right)\left( {{{\cos }^2}x - {{\sin }^2}x} \right),
Now again using the trigonometric identity, cos⁡2x−sin⁡2x=1−2sin⁡2x{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x, and substituting the value we get,
\Rightarrow $$$${\cos ^4}x - {\sin ^4}x = \left( 1 \right)\left( {1 - 2{{\sin }^2}x} \right),
Now simplifying we get,
cos⁡4x−sin⁡4x=1−2sin⁡2x{\cos ^4}x - {\sin ^4}x = 1 - 2{\sin ^2}x.
HENCE PROVED.

∴\therefore Using trigonometric identities we proved that cos⁡4x−sin⁡4x=1−2sin⁡2x{\cos ^4}x - {\sin ^4}x = 1 - 2{\sin ^2}x.

Note: An identity is an equation that always holds true. A trigonometric identity is an identity that contains trigonometric functions and holds true for all right-angled triangles. They are useful when solving questions with trigonometric functions and expressions. There are many trigonometric identities, here are some useful identities:
sin⁡2x=1−cos⁡2x{\sin ^2}x = 1 - {\cos ^2}x,
cos⁡2x+sin⁡2x=1{\cos ^2}x + {\sin ^2}x = 1,
sec⁡2x−tan⁡2x=1{\sec ^2}x - {\tan ^2}x = 1,
csc⁡2x=1+cot⁡2x{\csc ^2}x = 1 + {\cot ^2}x.
cos⁡2x−sin⁡2x=1−2sin⁡2x{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x,
cos⁡2x−sin⁡2x=2cos⁡2x−1{\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1.