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Find 2 values of theta in degrees (0 deg <_ theta <_ 360 deg) for the angle sin theta= sqrt2/2

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Find 2 values of theta in degrees (0 deg <_ theta <_ 360 deg) for the angle sin theta= sqrt2/2

asked Apr 26, 2014 in TRIGONOMETRY by anonymous

2 Answers

–1 vote

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image in first and second quadrants.

 

image is the answer.

answered Apr 26, 2014 by joly Scholar
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The trigonometric equation is sin(θ) = √2/2, where 0o ≤ θ ≤ 360o.

sin(θ) = √2/2 = 1/√2.

The function sin(θ) has a period of 2π first find all solutions in the interval [0, 2π].

The function sin(θ) is positive in Quadrant I and Quadrant II.

In first quadrant, 0o ≤ θ ≤ 90o.

1/√2 = sin(θ)  -------> sin(θ) = sin(π/4) ---------> θ = π/4.

In second quadrant, 90o ≤ θ ≤ 180o.

1/√2 = sin(π/4) = sin(π - π/4) = sin(3π/4)  -------> sin(θ) = sin(3π/4) ---------> θ = 3π/4.

The values of θ are π/4 and 3π/4 in 0o ≤ θ ≤ 360o.

answered Apr 28, 2014 by steve Scholar

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