Have a tough question .I expect an answer.If sinA+sin2A+sin3A = 1.then prove that .cos6A-4cos4A+8cos2A=4.the replier has his trigonometry almost perfect

Question

Answer

Let me ,make it simple ,bro has given good answer ,but square looks like '2' so I will re-write .sinA + sin²A + sin³A = 1=> sinA + sin³A = 1 - sin²A=> sinA ( 1+sin²A ) = cos​²A=> sinA ( 1 + 1 - cos²A ) = cos²A=> sinA ( 2 - cos²A ) = cos²A=> sinA = cos²A / ( 2 - cos²A)=> √ ( 1 - cos²A ) = cos²A / 2- cos²A=> 1 - cos²A = ( cos²A )² /  ( 2 - cos²A )²=> 1 - cos²A = cos⁴A / ( 4 + cos⁴A - 4 cos²A )=>  ( 1 - cos²A ) ( 4 + cos⁴A - 4 cos²A ) = cos⁴A=>  ( ​4 + cos²A - 4 cos²A -4cos²A - cos^6A + 4cos⁴A = cos⁴A=>  ( 4 -8cos²A - cos^6A + 4cos⁴A ) = cos⁴A - cos⁴A=>  ( 4 -8cos²A - cos^6A + 4 cos⁴A ) = 0=>  4 = 8cos²A + cos^6A - 4cos⁴A

Answer

SinA + sin2A + sin3A = 1=> sinA + sin3A = 1 - sin2A=> sinA ( 1+sin2A ) = cos​2A=> sinA ( 1 + 1 - cos2A ) = cos2A=> sinA ( 2 - cos2A ) = cos2A=> sinA = cos2A / ( 2 - cos2A)=> sqrt ( 1 - cos2A ) = cos2A / 2- cos2A=> 1 - cos2A = ( cos2A )2 /  ( 2 - cos2A )2=> 1 - cos2A = cos4A / ( 4 + cos4A - 4 cos2A )=>  ( 1 - cos2A ) ( 4 + cos4A - 4 cos2A ) = cos4A=>  ( ​4 + cos4A - 4 cos2A -4cos​2A - cos6A + 4cos4A = cos4A=>  ( 4 -8cos2A - cos6A + 4cos4A ) = cos4A - cos4A=>  ( 4 -8cos2A - cos6A + 4 cos4A ) = 0=>  4 = 8cos2A + cos6A - 4cos4A

Answer

Proof:

Given: sin(A) * sin(2A) * sin(3A) = 1

We need to prove: cos(6A) - 4cos(4A) + 8cos(2A) = 4

Identity:
To solve this problem, we will use the following trigonometric identity:

sin(2θ) = 2sin(θ)cos(θ)

Step 1: Expanding sin(3A)

To start, let's expand sin(3A) using the identity:

sin(3A) = sin(2A + A) = sin(2A)cos(A) + cos(2A)sin(A)

Step 2: Expanding sin(2A)

Now, let's expand sin(2A) using the same identity:

sin(2A) = sin(A + A) = sin(A)cos(A) + cos(A)sin(A)

Simplifying this expression, we get:

sin(2A) = 2sin(A)cos(A)

Substituting this into the previous equation for sin(3A), we get:

sin(3A) = 2sin(A)cos(A)cos(A) + cos(2A)sin(A)
= 2sin(A)cos^2(A) + cos(2A)sin(A)

Simplifying further, we have:

sin(3A) = sin(A)(2cos^2(A) + cos(2A))

Step 3: Substituting into the given equation

Now, let's substitute the expression for sin(3A) back into the given equation:

sin(A) * sin(2A) * sin(3A) = 1

sin(A) * sin(2A) * [sin(A)(2cos^2(A) + cos(2A))] = 1

Simplifying, we get:

2sin(A) * sin(A) * sin(2A) * cos^2(A) + sin(A) * sin(2A) * cos(2A) = 1

Step 4: Simplifying the equation

Using the identity sin^2(θ) = 1 - cos^2(θ), we can rewrite the equation as:

2sin^2(A) * sin(2A) * cos^2(A) + sin(A) * sin(2A) * cos(2A) = 1

Now, let's simplify each term individually:

2sin^2(A) * sin(2A) * cos^2(A) = 2sin(A) * cos^2(A) * sin(2A) = 2sin(A) * cos^2(A) * 2sin(A) * cos(A) = 4sin^2(A) * cos^3(A)

sin(A) * sin(2A) * cos(2A) = sin(A) * 2sin(A) * cos(A) * cos(2A) = 2sin^2(A) * cos(A) * cos(2A)

Substituting these values back into the equation, we have:

4sin^2(A) * cos^3(A) + 2sin^2(A

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