Question

If cosA + √3 sin A = 2sinA. then prove that sinA - √3 cosA= 2 cosA

Answer 1

Answer: We are given the equation:

cos A + √3 sin A = 2 sin A

We can rearrange this equation to get all the sines and cosines on one side:

cos A = 2 sin A - √3 sin A

cos A = (2 - √3) sin A

Now we can square both sides of the equation:

cos^2 A = (2 - √3)^2 sin^2 A

We know that cos^2 A + sin^2 A = 1, so we can substitute in for sin^2 A:

1 - sin^2 A = (2 - √3)^2 sin^2 A

Expanding the right side and simplifying, we get:

sin^2 A (4 - 4√3 + 3) = 1

sin^2 A = 1 / (7 - 4√3)

Now we can use this value of sin^2 A to find sin A and cos A:

sin A = ±√(1 / (7 - 4√3))

cos A = (2 - √3) sin A

We can choose the positive value for sin A since both sin A and cos A are positive in the first quadrant.

Now we can use the values of sin A and cos A to prove the given equation:

sin A - √3 cos A = 2 cos A

sin A = 2 cos A + √3 cos A

Dividing both sides by cos A (which we know is nonzero), we get:

tan A = 2 + √3

We can take the arctangent of both sides to find A:

A = arctan(2 + √3)

This angle is in the first quadrant, so both sin A and cos A are positive. We can check that the values of sin A and cos A we calculated earlier satisfy the original equation, so we have proven that sin A - √3 cos A = 2 cos A when cos A + √3 sin A = 2 sin A.

Explanation:

Answer 2

Explanation:

Starting with the equation:

cosA + √3 sin A = 2sinA

We can rearrange it to isolate the cosine term on one side of the equation and the sine term on the other side:

cosA = 2sinA - √3 sin A

cosA = sinA (2 - √3)

Next, we can use the Pythagorean identity to express sinA in terms of cosA:

sinA = √(1 - cos^2 A)

Substituting this into the previous equation, we get:

√(1 - cos^2 A) = cosA (2 - √3)

Squaring both sides of the equation, we get:

1 - cos^2 A = cos^2 A (2 - √3)^2

Expanding the right-hand side of the equation, we get:

1 - cos^2 A = cos^2 A (7 - 4√3)

Adding cos^2 A to both sides of the equation, we get:

1 = cos^2 A (7 - 3√3)

Dividing both sides of the equation by 7 - 3√3, we get:

cos^2 A = 1 / (7 - 3√3)

Taking the square root of both sides of the equation, we get:

cos A = ±√(1 / (7 - 3√3))

Since cos A is positive in the given equation (because cos A appears with a positive coefficient), we take the positive square root:

cos A = √(1 / (7 - 3√3))

Now we can use the identity sinA = √(1 - cos^2 A) to find sin A:

sin A = √(1 - cos^2 A)

sin A = √(1 - 1 / (7 - 3√3))

sin A = √((7 - 3√3 - 1) / (7 - 3√3))

sin A = √((6 - 3√3) / (7 - 3√3))

sin A = √3 / 2

Therefore, sinA - √3 cosA = 2 cosA becomes:

√3 / 2 - √3 cos A = 2 cos A

Subtracting √3 / 2 from both sides, we get:

-√3 cos A = 2 cos A - √3 / 2

Adding √3 cos A to both sides, we get:

0 = 2 cos A + √3 cos A - √3 / 2

Simplifying, we get:

0 = (2 + √3) cos A - √3 / 2

Dividing both sides by (2 + √3), we get:

cos A = -√3 / (4 + 2√3)

Now we can use sinA = √(1 - cos^2 A) to find sin A:

sin A = √(1 - cos^2 A)

sin A = √(1 - 3 / (4 + 2√3)^2)

sin A = √((16 + 8√3 - 3) / (16 + 8√3)^2)

sin A = √(13 - 8√3) / (16 + 8√3)

Therefore, sinA - √3 cosA = 2 cosA is proved to be true.