Question

Use the Chain Rule to show that if \(\theta \) is measured in degrees,

then

\(\frac{d}{{d\theta }}sin\left( \theta \right) = \frac{\pi }{{180}}cos\left( \theta \right)\)

(This gives one reason for the convention that radian measure is always used when dealing with trigonometric functions in calculus: the differentiation formulas would not be as simple if we used degree measure.)

Short answer

It is proved that \(\frac{d}{{d\theta }}\sin \left( \theta \right) = \frac{\pi }{{180}}\cos \left( \theta \right)\).

Step-by-step solution

Chain Rule of Derivative 

Let \(F\left( x \right) = f\left( {g\left( x \right)} \right)\) be a composition of function. The derivative of this function with respect to \(x\) is:

\(\begin{aligned}F'\left( x \right) &= \frac{d}{{dx}}f\left( {g\left( x \right)} \right)\\ &= f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\end{aligned}\)

So \(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\).

Proof of the statement

Here \(\theta \) is measured in degree so in radian it will be \({\theta ^ \circ } = \frac{\pi }{{180}}\theta \).

So \(\sin \left( {{\theta ^ \circ }} \right) = \sin \left( {\frac{\pi }{{180}}\theta } \right)\).

Now differentiating we get

\(\begin{aligned}\frac{d}{{d\theta }}\sin \left( {{\theta ^ \circ }} \right) &= \frac{d}{{d\theta }}\sin \left( {\frac{\pi }{{180}}\theta } \right)\\ &= \cos \left( {\frac{\pi }{{180}}\theta } \right) \cdot \frac{d}{{d\theta }}\left( {\frac{\pi }{{180}}\theta } \right)\\ &= \frac{\pi }{{180}}\cos \left( {\frac{\pi }{{180}}\theta } \right)\\ &= \frac{\pi }{{180}}\cos \left( {{\theta ^ \circ }} \right)\end{aligned}\)

Hence it is proved.