Prove that: 2cos 8A +1 = (2cos 2A -1)(2cos 2A +1)(2cos 4A -1)

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,

= 2cos 8A +1

= 2cos (2*4A) +1

= 2(2cos² 4A -1) +1

= 4cos² 4A -2 +1

= 4cos² 4A -1

= (2cos 4A +1)(2cos 4A -1)

= {2cos (2*2A)+1} (2cos 4A -1)

= {2(2cos² 2A-1) +1} (2cos 4A -1)

= (4cos² 2A -2 +1)(2cos 4A -1)

= (4cos² 2A -1)(2cos 4A -1)

= (2cos 2A -1)(2cos 2A +1)(2cos 4A -1)

= RHS

Explanation to this solution:

Step 1: Write the LHS same from the question.

Step 2: Write 8A as 2*4A so that we can get the expression in the form of cos(2A) +1.

Step 3: We need all the terms in the RHS in form of cos. So, we use the formula of cos2A = 2cos²A -1.

Step 4: We perform the simple multiplication.

Step 5: -2+1 = -1

Step 6: As we had the expression in the form of a²-b², we write them as (a+b)(a-b).

Step 7: (2cos 4A -1) matches our RHS so, we leave it as it is. Now, we further perform operations on  (2cos 4A +1) which we can write as 2cos (2*2A)+1.

Step 8: cos2A = 2cos²A -1.

Step 9: Perform basic multiplication.

Step 10: Perform addition. 

Step 11: As we had a²-b², we can write as (a-b)(a+b).

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: 2cos 8A +1 = (2cos 2A -1)(2cos 2A +1)(2cos 4A -1). | Trigonometric Identities | SciPiPupil

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