Plotting the function $f(x)=\sin(x^\circ)$, it might look linear, but after checking it by recreating it as $g(x)=x\tan\left(\arctan\left(\frac{\sin(50^\circ)}{50}\right)\right)$, it is surely not.
What is the derivative of $\sin(x^\circ)$?
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What is $x^o$ here? – Cameron Williams Jan 15 '20 at 13:44
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@Cameron I don't quite understand. sin(x°) is a function – harieamjari Jan 15 '20 at 13:47
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2$\sin x = sin x^\circ\frac\pi{180^\circ}$, can you take it from here? – Maja Blumenstein Jan 15 '20 at 13:47
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What's the difficulty you are facing? Do you know what the derivative of $\sin{x}$ is normally? In the normal case, it's assumed that $x$ is in radians. You only need to convert the normal differentiation procedure to degrees and you're basically done. – Matti P. Jan 15 '20 at 13:49
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@Matt you mean its derivative is cos(x°)? I'm new to math BTW. – harieamjari Jan 15 '20 at 13:54
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Oh it's degrees. Calculus is not done in degrees because the expressions get unwieldy. Calculus with trig functions is almost universally done in radians. – Cameron Williams Jan 15 '20 at 13:57
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1Well, almost. $$ \sin{x^{\circ}} = \sin{\left( x \cdot \frac{\pi}{180} \right)} \quad \Rightarrow \frac{d}{dx}\sin{\left( x \cdot \frac{\pi}{180} \right)} = \frac{\pi}{180} \cos{\left( x \cdot \frac{\pi}{180} \right)} $$ – Matti P. Jan 15 '20 at 13:57
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From a complete circle, you can see the following- $$\frac {x^o}{360}=\frac {x}{2{\pi}}$$ $$x={{\pi}x^o\over {180}}$$ You can just replace $x^o$ and convert to radians as all trigonometric derivations are done in radians. Hence you can differentiate as usual. $$\frac {d}{dx}\sin{x^{\circ}} = \frac{d}{dx}\sin{\left( \frac{x\pi}{180} \right)} = \frac{\pi}{180} \cos{\left(\frac{x\pi}{180} \right)}=\frac{\pi}{180} \cos x^{\circ}$$
Sam
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I can now compute its taylor series so that I wouldn't be using scientific calculator anymore (they are expensive), but just an ordinary calculator. Thanks. – harieamjari Jan 15 '20 at 14:29