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Find sin^2a-cos^2a, where cosa+Sina=1/2, 0<a<2π​

User BCsongor
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1 Answer

3 votes

Answer:

We know that:

sin^2a + cos^2a = 1 (1)

Also, we have:

cos a + sin a = 1/2

Squaring both sides of the above equation, we get:

cos^2a + 2cos a sin a + sin^2a = 1/4

Using equation (1), we can simplify this to:

1 + 2cos a sin a = 1/4

Subtracting 1 from both sides, we get:

2cos a sin a = -3/4

Squaring both sides, we get:

4cos^2a sin^2a = 9/16

Using the identity:

sin^2a = 1 - cos^2a

We can rewrite the above equation as:

4cos^2a (1 - cos^2a) = 9/16

Expanding and rearranging, we get:

4cos^4a - 4cos^2a + 9/16 = 0

Multiplying both sides by 16, we get:

64cos^4a - 64cos^2a + 9 = 0

Letting x = cos^2a, we can rewrite this as a quadratic equation:

64x^2 - 64x + 9 = 0

Solving for x using the quadratic formula, we get:

x = (64 ± √16384)/128

x = 1/2 or x = 9/64

Since 0 < a < 2π, we know that cos a is positive, so we can take the positive square root:

cos a = √(1/2) = 1/√2

Substituting this into the equation sin a + cos a = 1/2, we get:

sin a = 1/2 - cos a = 1/2 - 1/√2 = (√2 - 1)/2

Therefore:

sin^2a - cos^2a = ((√2 - 1)/2)^2 - (1/√2)^2

= (3 - 2√2)/4

Therefore, sin^2a - cos^2a = (3 - 2√2)/4.

User RickK
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