The $\cos(\sin 60^\circ)$

I stumbled across this question and I cannot figure out how to use the value of $\cos(\sin 60^\circ)$ which would be $\sin 0.5$ and $\cos 0.5$ seems to be a value that you can only calculate using a calculator or estimate at the very best.

• Isn't $\sin 60 = \sqrt3/2$ ? – Parth Thakkar Jul 7 '13 at 11:38
• And btw, what is the source of the question? – Parth Thakkar Jul 7 '13 at 11:40
• It seems that you are asking for $\cos(\sin(60))$. Yes you would need a calculator to compute that. – Tpofofn Jul 7 '13 at 11:40

$\cos(\sin(\pi/3))=\cos(\sqrt{3}/2)$.

(You can confirm the first step by observing the 30-60-90 special triangle.)

I believe this is far as you can go without the use of a calculator.

Using a calculator I get the final answer to be $0.6478593448524569104400717351567034556620295591904946...$ (according to Wolfram Alpha).

Notice that $\cos(60)=\cos(\pi/3)$ so to remove confusion we need to write some extra notation. These are related by $\cos_{deg}(\theta)=\cos_{rad}(\frac{\pi\theta }{180})$

In degrees, I used my calculator and obtain $\sin_{deg}(60)=0.8660254038$ and hence $$\cos_{deg}(\sin_{deg}(60))=\cos_{deg}(0.8660254038) = 0.9998857706.$$ Notice the convention of degrees or radians matters here. I assume that the cosine in question is the cosine in degrees function.

This is not to be confused with the usual radian-measure based cosine. For radian-based trig functions: $$\cos_{rad}(\sin_{rad}(\pi/3))=\cos_{rad}(0.88660254038) = 0.6478593448$$