
How do you verify \[{\cos ^4}x - {\sin ^4}x = 1 - 2{\sin ^2}x\]?
Answer
544.2k+ views
Hint: 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 \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\], and simplify using the trigonometric identity, \[{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x\], 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,\[{\cos ^4}x - {\sin ^4}x\],
We have to prove that L.H.S term is equal to R.H.S,
Now using the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\],
By equating LHS term to the identity we get,
Here \[a = {\cos ^2}x,b = {\sin ^2}x\],
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 identity\[{\cos ^2}x + {\sin ^2}x = 1\], 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, \[{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x\], 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,
\[{\cos ^4}x - {\sin ^4}x = 1 - 2{\sin ^2}x\].
HENCE PROVED.
\[\therefore \]Using trigonometric identities we proved that \[{\cos ^4}x - {\sin ^4}x = 1 - 2{\sin ^2}x\].
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:
\[{\sin ^2}x = 1 - {\cos ^2}x\],
\[{\cos ^2}x + {\sin ^2}x = 1\],
\[{\sec ^2}x - {\tan ^2}x = 1\],
\[{\csc ^2}x = 1 + {\cot ^2}x\].
\[{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x\],
\[{\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1\].
In this question we have to prove LHS is equal to RHS, for doing this we use algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\], and simplify using the trigonometric identity, \[{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x\], 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,\[{\cos ^4}x - {\sin ^4}x\],
We have to prove that L.H.S term is equal to R.H.S,
Now using the algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\],
By equating LHS term to the identity we get,
Here \[a = {\cos ^2}x,b = {\sin ^2}x\],
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 identity\[{\cos ^2}x + {\sin ^2}x = 1\], 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, \[{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x\], 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,
\[{\cos ^4}x - {\sin ^4}x = 1 - 2{\sin ^2}x\].
HENCE PROVED.
\[\therefore \]Using trigonometric identities we proved that \[{\cos ^4}x - {\sin ^4}x = 1 - 2{\sin ^2}x\].
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:
\[{\sin ^2}x = 1 - {\cos ^2}x\],
\[{\cos ^2}x + {\sin ^2}x = 1\],
\[{\sec ^2}x - {\tan ^2}x = 1\],
\[{\csc ^2}x = 1 + {\cot ^2}x\].
\[{\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x\],
\[{\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1\].
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Which is the Longest Railway Platform in the world?

India Manned Space Mission Launch Target Month and Year 2025 Update

Which of the following pairs is correct?

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Plot a graph between potential difference V and current class 12 physics CBSE

