Sin 1

The value of sin 1 is 0.8414709848, in radian. In trigonometry, the complete trigonometric functions and formulas are based on three primary ratios, i.e., sine, cosine, and tangent in trigonometry. These trigonometric ratios help us in finding angles and lengths of sides, in a triangle. The basic angles, which are commonly used for solving trigonometric problems are 0, 30, 45, 60, 90 degrees. These angles are also expressed in the form of radians, such as π/2, π/3, π/4, π/6, π and so on.

Let us find here how to calculate the value of sin 1.

What is the Value of Sin 1?

The value of sine 1 in radian is 0.8414709848.

We know, π/3 = 1.047198≈1

Sin (π/3) = √3/2 and sin π = 0

Now using these data, we can write;

sin1=sin[π/3−(π/3−1)]

⟹sin1=sin(π/3)cos(π/3-1)−cosπ/3sin(π/3−1)

The angle π/3−1=0.047198 is a very small angle.

We know that, for small angles θ,

Sinθ ≈ θ and cos⁡θ ≈ 1

hence,

Sin1 ≈ (√3/2×1)−[1/2×(π/3−1)]

Therefore,

⟹sin1 ≈ 0.842427

Also, find:

How to find Sine 1 value?

The sine of an angle, say x, can take either radian or degrees, as its argument. The rule is radian measurement.

Now, π radian = 180 degree

so, 1 rad = 180/π degree

1 rad = 57.2957795131 degree

In terms of degree, we know,

sin 0° = 0, sin 90° = 1

In radians,

sin 0 = 0 and sin (π/2)=1

Now, π = 3.14159265359, π/2=1.5707963268

Thus,

sin (1.5707963268)= 1, when the angle is in radian

So,

sin (1) = 0.8414709848 [when angle is in radian]

sin (57.2957795131) = 0.8414709848 [when angle is in degree]

Taylor’s Series to Find Sine 1

As per the Taylor series, we know;

\(f(x)=f(a) \frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{(3)}(a)}{3 !}(x-a)^{3}+\dots\)

Hence, if we put sine function in this equation, we can get the approx value of sin 1, such as:

x=1

\(Sin (1)=1-\frac{1}{3 !}+\frac{1}{5 !}-\frac{1}{7 !}+\cdots \\ Sin (1)= 1-\frac{1}{6}+\frac{1}{120}-\frac{1}{5040}+\frac{1}{362880}-\frac{1}{39916800}+\cdots\)

By solving the above, we can get;

Sin 1 ≈ 0.82

What is the value of Inverse of sin 1?

The inverse sin of 1, i.e., sin-1 (1) is a very unique value for the inverse sine function. Sin-1(x) will give us the angle whose sine is x. Hence, sin-1 (1) is equal to the angle whose sine is 1.

We know,

Sin 90 = 1

Therefore,

sin-1(1) = 90 ( in degrees)

sin-1(1) = Π/2 (in radian)

Since the inverse sin-1 (1) is 90° or Π/2. ‘1’ denotes the maximum value of the sine function. Hence, for every 90 degrees it will happen, such as at Π/2, 3Π/2, and so on.

By this we can conclude that;

sin-1(1) = Π/2+2Πk (for any integer k)

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