bikram57
Answered

Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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

Sagot :

stigawithfun

Step-by-step explanation:

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

We try to solve one side of the equation to get the other side of the equation.

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

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 !

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.