(2cosA+1) (2cosA-1)=2cos2A+1 prove that

In this case, we'll solve the right hand side (2cos2A + 1) of the equation with the aim of getting the left hand side of the equation (2cosA + 1)(2cosA - 1)

Solving the right hand side: 2cos2A + 1

i. We know that cos2A = cos(A+A) = cosAcosA - sinAsinA

Therefore;

cos2A = cos²A - sin²A

ii. We also know that: cos²A + sin²A = 1

Therefore;

sin²A = 1 - cos²A

iii. Now re-write the right hand side by substituting the value of cos2A as follows;

2cos2A + 1 = 2(cos²A - sin²A) + 1

iv. Expand the result in (iii)

2cos2A + 1 = 2cos²A - 2sin²A + 1

v. Now substitute the value of sin²A in (ii) into the result in (iv)

2cos2A + 1 = 2cos²A - 2(1 - cos²A) + 1

vi. Solve the result in (v)

2cos2A + 1 = 2cos²A - 2 + 2cos²A + 1

2cos2A + 1 = 4cos²A - 2 + 1

2cos2A + 1 = 4cos²A - 1

2cos2A + 1 = (2cosA)² - 1²

vii. Remember that the difference of the square of two numbers is the product of the sum and difference of the two numbers. i.e

a² - b² = (a+b)(a-b)

This means that if we put a = 2cosA and b = 1, the result from (vi) can be re-written as;

2cos2A + 1 = (2cosA)² - 1²

2cos2A + 1 = (2cosA + 1)(2cosA - 1)

Since, we have been able to arrive at the left hand side of the given equation, then we can conclude that;

(2cosA + 1)(2cosA - 1) = 2cos2A + 1

VirαtKσhli VirαtKσhli · Answer 2 · 2021-07-13T13:38:00+02:00

Answer:

[tex]\boxed{\sf LHS = RHS }[/tex]

Step-by-step explanation:

We need to prove that ,

[tex]\sf\implies (2 cosA +1)(2cosA-1) = 2cos2A+1[/tex]

We can start by taking RHS and will try to obtain the LHS . The RHS is ,

[tex]\sf\implies RHS= 2cos2A + 1 [/tex]

We know that , cos2A = 2cos²A - 1 ,

[tex]\sf\implies RHS= 2(2cos^2-1)-1 [/tex]

Simplify the bracket ,

[tex]\sf\implies RHS= 4cos^2A - 2 +1 [/tex]

Add the constants ,

[tex]\sf\implies RHS= 4cos^2-1 [/tex]

Write each term in form of square of a number ,

[tex]\sf\implies RHS= (2cos^2A)^2-1^2 [/tex]

Using (a+b)(a-b) = a² - b² , we have ,

[tex]\sf\implies RHS= (2cosA+1)(2cosA-1) [/tex]

This equals to LHS , therefore ,

[tex]\sf\implies \boxed{\pink{\textsf{\textbf{ RHS= LHS }}}} [/tex]

Hence Proved !