Prove that: $cos^6A-sin^6A = cos2A \left (1- \frac{1}{4}sin^22A \right)$

This is a class 10 Question From Values of Trigonometric Ratios chapter of Unit Trigonometry. All the steps for the solutions are mentioned in the description below. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts.

Solution:

Taking LHS,

= $cos^6A-sin^6A$

= $(cos^2A)^3 -(sin^2A)^3$

= $(cos^2A-sin^2A)\{(cos^2A)^2+cos^2Asin^2A +(sin^2A)^2\}$

= $cot2A\{(cos^2A)^2+(sin^2A)^2+cos^2Asin^2A)\}$

= cot2A{(cos²A+sin²A)²-2cos²Asin²A+cos²Asin²A}

= $cot2A\{1-cos^2Asin^2A\}$

= $cot2A\{1- \frac{4cos^2Asin^2A}{4}\}$

= $cot2A\{1- \frac{(2sinAcosA)^2}{4}\}$

= $cot2A\{1- \frac{1}{4}sin^2 2A\}$

Explanation to this solution:

Step 1: Write the LHS same from the question.

Step 2: We need to use the formula of a³-b³ in this solution. We know this by seeing the RHS of the question.

Step 3: Expand the formula of (a³-b³) using the factorisation formula (a-b)(a²+ab+b²).

Step 4: On the factor form we have (cos²A-sin²A) which is equal to cos2A (multiple angle formula). And, then we write (cos²A)²+(sin²A)² together.

Step 5: Its turn to convert (cos²A)²+(sin²A)² into (cos²A+sin²A)². For this, we use (a²+b²) = (a+b)²-2ab.

Step 6: On adding cos²Asin²A and (-2cos²Asin²A), we get -cos²Asin²A.

Step 7: Now, we divide and multiply -cos²Asin²A by 4. This is because on our RHS of the question, we have 4 as the denominator of last term and sin²2A.  

Step 8: As we got 4cos²Asin²A, we can write this as (2sinAcosA)² 

Step 9: (sin2A)² = sin²2A. Then, we prove that the LHS is equal to the RHS.


Here is the Facebook link to the solution of this question in image. 

Related Notes:

Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles

Question: Prove that: cos⁶A-sin⁶A = cos2A (1- (1/4)sin² 2A). | Trigonometric Identities | SciPiPupil

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