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Question: Find the approximate value of\(\sin 31^\circ \), given that\(\cos 30^\circ = 0.8660\)and\(1^\circ = ...

Find the approximate value ofsin⁡31∘\sin 31^\circ , given thatcos⁡30∘=0.8660\cos 30^\circ = 0.8660and1∘=0.0175c1^\circ = {0.0175^c}

Explanation

Solution

Take a = 30⁰ and h = 1⁰ = 0.0175 radians. Let f(x) =sin⁡x\sin x.
Use the approximation formula: f(a + h)≈\approxf(a) + hf’(a). Also, use the given information and the facts that cos x is the derivative of sin x and f(30⁰) =sin⁡30∘=0.5\sin 30^\circ = 0.5 to obtain the answer.

Complete step by step solution:
We are given the values of cos⁡30∘\cos 30^\circ and the value of 1∘1^\circ in radians.
We are asked to find the approximate value ofsin⁡31∘\sin 31^\circ .
Now, this is a problem related to derivatives. So, we will be using the approximation formula for derivatives here.
Since the question is about the sine function, but the cosine values, it can be understood that we need to use the relation between the two to find the answer.
We know that the derivative of the sine function is the cosine function.
Let’s use this fact to solve the problem at hand.
Consider the sine function and its derivative, the cosine function as follows:
Let f(x) =sin⁡x\sin x. This implies that f(31⁰) = sin⁡31∘\sin 31^\circ
Then the derivative of f(x) denoted by f’(x) iscos⁡x\cos x
Also, let a = 30⁰ and h = 1⁰ = 0.0175 radians
As a + h = 31⁰, we will be computing the value of f(a + h)
We will be using the following approximation formula:
f(a + h)≈\approxf(a) + hf’(a).
We need to plug in all the values in this formula.
f(31⁰)≈\approxf(30⁰) + 0.0175cf’(30⁰).
Now, we know that f(30⁰) =sin⁡30∘=0.5\sin 30^\circ = 0.5and f’(30⁰) = cos⁡30∘=0.8660\cos 30^\circ = 0.8660
Therefore, on substituting the values we get
sin⁡31∘≈\sin 31^\circ \approx 0.5 + 0.0175 × 0.8660
≈\approx0.5 + 0.0152
≈\approx0.5152
Hence the approximate value ofsin⁡31∘\sin 31^\circ is 0.5152.

Note: One should avoid using ‘=’ sign in the approximation formula. It has to be the wavy equal sign ‘≈\approx’ which indicates that the answer is not an exact value of the function but has been approximated. Also, to get an answer as close to the exact value, it is essential to include as many decimals as possible in your calculation.