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# What's the value of sin(2x) if tan x = 0.8 and sin>0?

### 7 Answers

- la consoleLv 73 years agoFavorite Answer
tan(x) = 0.8 → where: sin(x) > 0

You can see that tan(x) > 0, so the angle x belongs to the quadrant I or quadrant III.

Given that sin(x) > 0, you can deduce that the angle x is located in the quadrant I.

tan(x) = 0.8

sin(x)/cos(x) = 0.8

sin(x) = 0.8 * cos(x)

cos²(x) + sin²(x) = 1

cos²(x) + [0.8 * cos(x)]² = 1

cos²(x) + [0.8² * cos²(x)] = 1

cos²(x).[1 + 0.8²] = 1

cos²(x) * 1.64 = 1

cos²(x) = 1/1.64

cos²(x) = (± 1/√1.64)² → given that x is located in the quadrant I, you can say that: cos(x) > 0

cos(x) = 1/(√1.64) ← memorize this result as (1)

cos²(x) + sin²(x) = 1

sin²(x) = 1 - cos²(x)

sin²(x) = 1 - (1/1.64)

sin²(x) = 0.64/1.64

sin²(x) = [± √(0.64/1.64)]²

sin²(x) = [± 0.8/√(1.64)]² → given that x is located in the quadrant I, you can say that: sin(x) > 0

sin(x) = 0.8/(√1.64) ← memorize this result as (2)

Do you know this identity?

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = b = x

sin(x + x) = sin(x).cos(x) + cos(x).sin(x)

sin(2x) = 2.sin(x).cos(x) → recall (2): sin(x) = 0.8/(√1.64)

sin(2x) = 2.[0.8/(√1.64)].cos(x) → recall (1): cos(x) = 1/(√1.64)

sin(2x) = 2.[0.8/(√1.64)].[1/(√1.64)]

sin(2x) = (2 * 0.8) / (√1.64 * √1.64)

sin(2x) = 1.6/1.64

- PinkgreenLv 73 years ago
sin(2x)>0=> 2x is in the 1st or the 2nd quadrant.

tan x=0.8=>x=38.6598082...=>2x=77.3196165...

Thus sin(2x)=sin(77.3196165..)=0.97561 approximately.

- ComoLv 73 years ago
tan x = 4/5

sin x = 3 / √41

cos x = 5 / √41

sin 2x = 2 sin x cos x

sin 2x = 2 [ 3 / √41 ] [ 5 / √41 ]

sin 2x = 30 / 41

- TomVLv 73 years ago
sin(2x) = 2sin(x)cos(x)

If sin(x) > 0 and tan(x) > 0, then x is in Q1

tan(x) = sin(x)/cos(x) = 0.8

sin(x) = 0.8cos(x)

sin(2x) = 2(0.8cos²(x)) = 1.6cos²(x)

sin²(x) = 0.64cos²(x) = 1 - cos²(x)

1.64cos²(x) = 1

cos²(x) = 1/1.64

sin(2x) = 1.6/1.64 ≈ 0.97561

=========================

Corroboration:

[tan(x) = 0.8 → x = arctan(0.8) ≈ 38.660°

sin(2x) ≈ sin(2*38.660) ≈ sin(77.330) ≈ 0.97561]

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- RaymondLv 73 years ago
To keep it simple, do things one easy step at a time, and use the basic identitied.

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

therefore, cos^2(x) = 1 - sin^2(x)

and

tan(x) = sin(x) / cos(x)

tan(x) = 0.8

square both sides

tan^2(x) = 0.64

tan^2(x) = sin^2(x) / cos^2(x)

tan^2(x) = sin^2(x) / (1 - sin^2(x)) = 0.64

solve for sin^2(x)

sin^2(x) = 0.64(1 - sin^2(x))

sin^2(x) = 0.64 - 0.64*sin^2(x)

1.64 sin^2(x) = 0.64

sin^2(x) = 0.64 / 1.64 = 0.390244 (approx.)

square root both sides

sin(x) = 0.624695

set that aside for now

sin^2(x) = 0.390244

therefore

cos^2(x) = 1 - 0.390244 = 0.609756

cos(x) = 0.3718025

sin(2x) = 2sin(x)cos(x) = 0.464526

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Yes, there are shortcuts. But they require you to memorize more complicated things.

- alexLv 73 years ago
tan x = 0.8 and sin>0 ---> cosx = 1/√1.64 and sinx = 0.8/√1.64

sin(2x)=2sinx cosx=...