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26. Given that tanθ almost equal to≈ 2.773 where 0<θ<π2, find the values of sinθ and cosθ. 27. Given that tanθ ≈ -1.559 where 3π2<θ<2π , find the values of sinθ and cosθ.

Sagot :

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Answer:

a. i. sinθ = +0.941

ii. cosθ = +0.339

b. i. sinθ = -0.842

ii.  cosθ = +0.540

Step-by-step explanation:

a. Given that tanθ almost equal to≈ 2.773 where 0<θ<π/2, find the values of sinθ and cosθ.

i. sinθ

Given that 1 + cot²θ = cosec²θ

cotθ = 1/tanθ

Since tanθ = 2.773, cotθ = 1/tanθ = 1/2.773 = 0.3606

So, 1 + cot²θ = cosec²θ

1 + (0.3606)² = cosec²θ

1 + 0.13 = cosec²θ

1.13 = cosec²θ

cosec²θ = 1.13

cosecθ = ±√1.13

cosecθ = ±1.063

1/sinθ = ±1.063

sinθ = ±1/1.063

sinθ = ±0.9407

sinθ ≅ ±0.941

Since we have 0<θ<π/2, sinθ is in the first quadrant, so we choose the positive value.

So, sinθ = +0.941 where 0<θ<π/2

ii. cosθ

Given that 1 + tan²θ = sec²θ

Since tanθ = 2.773,

So, 1 + tan²θ = sec²θ

1 + (2.773)² = sec²θ

1 + 7.6895 = sec²θ

8.6895 = sec²θ

sec²θ = 8.6895

secθ = ±√8.6895

secθ = ±2.9478

1/cosθ = ±2.9478

cosθ = ±1/2.9478

cosθ = ±0.3392

cosθ ≈ ±0.339

Since we have 0<θ<π/2, cosθ is in the first quadrant, so we choose the positive value.

So, cosθ = +0.339 where 0<θ<π/2

b. Given that tanθ ≈ -1.559 where 3π/2<θ<2π, find the values of sinθ and cosθ.

i. sinθ

Given that 1 + cot²θ = cosec²θ

cotθ = 1/tanθ

Since tanθ = -1.559, cotθ = 1/tanθ = 1/-1.559 = -0.6414

So, 1 + cot²θ = cosec²θ

1 + (-0.6414)² = cosec²θ

1 + 0.4114 = cosec²θ

1.4114 = cosec²θ

cosec²θ = 1.4114

cosecθ = ±√1.4114

cosecθ = ±1.1880

1/sinθ = ±1.1880

sinθ = ±1/1.1880

sinθ = ±0.8417

sinθ ≈ ±0.842

Since we have 3π/2<θ<2π, sinθ is in the fourth quadrant, so we choose the negative value.

So, sinθ = -0.842 where 3π/2<θ<2π

ii. cosθ

Given that 1 + tan²θ = sec²θ

Since tanθ = -1.559,

So, 1 + tan²θ = sec²θ

1 + (-1.559)² = sec²θ

1 + 2.4305 = sec²θ

3.4305 = sec²θ

sec²θ = 3.4305

secθ = ±√3.4305

secθ = ±1.8522

1/cosθ = ±1.8522

cosθ = ±1/1.8522

cosθ = ±0.5399

cosθ ≈ ±0.540

Since we have 3π/2<θ<2π, cosθ is in the fourth quadrant, so we choose the positive value.

So, cosθ = +0.540 where 3π/2<θ<2π

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