If 2sin^2∆ =1+Cos∆, 0°<∆<90° find the value of∆

Answer 1

To find the value of ∆, we need to solve the given trigonometric equation:

2sin^2∆ = 1 + cos∆

Using the trigonometric identity sin^2∆ + cos^2∆ = 1, we can rewrite sin^2∆ as 1 - cos^2∆:

2(1 - cos^2∆) = 1 + cos∆

Expanding the left side of the equation, we get:

2 - 2cos^2∆ = 1 + cos∆

Rearranging the terms, we have:

2cos^2∆ + cos∆ - 1 = 0

This is a quadratic equation in terms of cos∆. To solve for cos∆, we can use the quadratic formula:

cos∆ = [-b ± √(b^2 - 4ac)] / 2a

In this case, a = 2, b = 1, and c = -1. Plugging these values into the formula, we get:

cos∆ = [-1 ± √(1 + 8)] / 4

cos∆ = [-1 ± √9] / 4

cos∆ = [-1 ± 3] / 4

This gives us two possible solutions for cos∆:

1) cos∆ = (3 - 1) / 4 = 2 / 4 = 0.5

2) cos∆ = (-3 - 1) / 4 = -4 / 4 = -1

Since 0° < ∆ < 90° (first quadrant), the value of cos∆ cannot be negative. Therefore, cos∆ = 0.5.

Now, we know that cos∆ = 0.5, and we can use the trigonometric identity sin^2∆ + cos^2∆ = 1 to find sin∆:

sin^2∆ + (0.5)^2 = 1
sin^2∆ + 0.25 = 1
sin^2∆ = 0.75
sin∆ = √0.75 = √(3/4) = √3 / 2

Since we are in the first quadrant, sin∆ is positive. Therefore, sin∆ = √3 / 2.

Finally, we can find the value of ∆ using sin∆ and cos∆:

tan∆ = sin∆ / cos∆ = (√3 / 2) / 0.5 = √3
∆ = arctan(√3)
∆ ≈ 60°

Therefore, the value of ∆ is approximately 60°.