Find an approximate value of the sine of 61 degrees

Question

Use the fact that $\sin 61° = \sin (60°+1°)$ to give an approximate value of $\sin 61°$ in terms of $\pi$.

My textbook says that the answer is $(180\sqrt(3)+\pi)/360$, but I don't understand how it got there. I can expand the identity of $\sin (60°+1°)$ but I don't know where to go after that.

Answer 1

$$\sin61^{\circ}=\frac{\sqrt3}{2}\cos1^{\circ}+\frac12\sin1^{\circ}$$ But you are asking for it in terms of $\pi$.
One can also use series expansion for calculating $\sin(\frac\pi3+\frac\pi{180})$.
We use the $\sin(A+B)=\sin A\cos B+\sin B\cos A$ formula and then use: $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$$ to get an approximate form. (Substitute $\frac\pi{180}$ for $x$).

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Addition: Lets try something. Lets try to solve $\sin\theta=\theta$.
We observe that $0$ is a solution, but if we put $\theta=\frac\pi{180}\approx0.0174$,we get $\sin\theta\approx0.0174$. And since you want an approximation only, I think that solves the confusion. We are asked to evaluate $$\sin61^{\circ}=\frac{\sqrt3}{2}\cos1^{\circ}+\frac12\sin1^{\circ}\approx\frac{\sqrt3}{2}\cos0+\frac12\frac{\pi}{180}$$ ($\cos1^{\circ}$ is very close to $1$). So that gives us: $$\boxed{\frac{180\sqrt3+\pi}{360}}$$

Answer 2

Since you want an approximate value, you can use Taylor expansion $$\sin\left(\frac{\pi }{3}+x\right)=\frac{\sqrt{3}}{2}+\frac{x}{2}-\frac{\sqrt{3} x^2}{4}-\frac{x^3}{12}+O\left(x^4\right)$$ and make $x=\frac{\pi }{180}$. Depending on the number of terms you retain, you xill get more and more accurate results.

Using the above, you will get a value $0.8746197038$ to be compared to the exact value $0.8746197072$.