We prove some results that are needed in the application of calculus to trigonometry.
Reference Image for Proof Theorem: If θ {\displaystyle \theta } ; is a positive angle, less than a right angle (expressed in radians), then 0 < sin ( θ ) < θ < tan ( θ ) {\displaystyle 0<\sin(\theta )<\theta <\tan(\theta )} .
Proof: Consider a circle, centre O {\displaystyle O} , radius r {\displaystyle r} , and choose two points A {\displaystyle A} and B {\displaystyle B} on the circumference such that ∠ A O B {\displaystyle \angle AOB} is less than a right angle. Draw a tangent to the circle at B {\displaystyle B} , and let O A → {\displaystyle {\overrightarrow {OA}}} produced intersect it at C {\displaystyle C} . Clearly
0 < area ( △ O A B ) < area ( ⌔ O A B ) < area ( △ O B C ) {\displaystyle 0<\operatorname {area} \left(\triangle OAB\right)<\operatorname {area} \left({\text{⌔}}OAB\right)<\operatorname {area} \left(\triangle OBC\right)} i.e.
0 < 1 2 r 2 sin ( θ ) < 1 2 r 2 θ < 1 2 r 2 tan ( θ ) {\displaystyle 0<{\frac {1}{2}}r^{2}\sin(\theta )<{\frac {1}{2}}r^{2}\theta <{\frac {1}{2}}r^{2}\tan(\theta )} and the result follows.
Corollary: If θ {\displaystyle \theta } is a negative angle, more than minus a right angle (expressed in radians), then 0 > sin ( θ ) > θ > tan ( θ ) {\displaystyle 0>\sin(\theta )>\theta >\tan(\theta )} . [This follows from sin ( − θ ) = − sin ( θ ) {\displaystyle \sin(-\theta )=-\sin(\theta )} and tan ( − θ ) = − tan ( θ ) {\displaystyle \tan(-\theta )=-\tan(\theta )} .]
Corollary: If θ {\displaystyle \theta } is a non-zero angle, less than a right angle but more than minus a right angle (expressed in radians), then 0 < | sin θ | < | θ | < | tan θ | {\displaystyle 0<\left\vert \sin \theta \right\vert <\left\vert \theta \right\vert <\left\vert \tan \theta \right\vert } .
Theorem: As θ → 0 , sin ( θ ) θ → 1 {\displaystyle \theta \rightarrow 0,{\frac {\sin(\theta )}{\theta }}\rightarrow 1} and tan ( θ ) θ → 1 {\displaystyle {\frac {\tan(\theta )}{\theta }}\rightarrow 1} .
Proof: Dividing the result of the previous theorem by sin ( θ ) {\displaystyle \sin(\theta )} and taking reciprocals,
1 > sin ( θ ) θ > cos ( θ ) {\displaystyle 1>{\frac {\sin(\theta )}{\theta }}>\cos(\theta )} .But cos ( θ ) {\displaystyle \cos(\theta )} tends to 1 {\displaystyle 1} as θ {\displaystyle \theta } tends to 0 {\displaystyle 0} , so the first part follows.
Dividing the result of the previous theorem by tan ( θ ) {\displaystyle \tan(\theta )} and taking reciprocals,
1 cos ( θ ) > tan ( θ ) θ > 1 {\displaystyle {\frac {1}{\cos(\theta )}}>{\frac {\tan(\theta )}{\theta }}>1} .Again, cos ( θ ) {\displaystyle \cos(\theta )} tends to 1 {\displaystyle 1} as θ {\displaystyle \theta } tends to 0 {\displaystyle 0} , so the second part follows.
Theorem: If θ {\displaystyle \theta } is as before, then cos ( θ ) > 1 − θ 2 2 {\displaystyle \cos(\theta )>1-{\frac {\theta ^{2}}{2}}} .
Proof:
cos ( θ ) = 1 − 2 sin 2 ( θ 2 ) {\displaystyle \cos(\theta )=1-2\sin ^{2}\left({\frac {\theta }{2}}\right)} sin ( θ 2 ) < θ 2 so {\displaystyle \sin \left({\frac {\theta }{2}}\right)<{\frac {\theta }{2}}{\text{ so}}} cos ( θ ) > 1 − 2 ( θ 2 ) 2 = 1 − θ 2 2 {\displaystyle \cos(\theta )>1-2\left({\frac {\theta }{2}}\right)^{2}=1-{\frac {\theta ^{2}}{2}}} .Theorem: If θ {\displaystyle \theta } is as before, then sin ( θ ) > θ − θ 3 4 {\displaystyle \sin(\theta )>\theta -{\frac {\theta ^{3}}{4}}} .
Proof:
sin ( θ ) = 2 sin ( θ 2 ) cos ( θ 2 ) = 2 tan ( θ 2 ) cos 2 ( θ 2 ) {\displaystyle \sin(\theta )=2\sin \left({\frac {\theta }{2}}\right)\cos \left({\frac {\theta }{2}}\right)=2\tan \left({\frac {\theta }{2}}\right)\cos ^{2}\left({\frac {\theta }{2}}\right)} .tan ( θ 2 ) > θ 2 so {\displaystyle \tan \left({\frac {\theta }{2}}\right)>{\frac {\theta }{2}}{\text{ so}}} sin ( θ ) > 2 ( θ 2 ) cos 2 ( θ 2 ) = θ ( 1 − sin 2 ( θ 2 ) ) {\displaystyle \sin(\theta )>2\left({\frac {\theta }{2}}\right)\cos ^{2}\left({\frac {\theta }{2}}\right)=\theta \left(1-\sin ^{2}\left({\frac {\theta }{2}}\right)\right)} .sin ( θ 2 ) < θ 2 so {\displaystyle \sin \left({\frac {\theta }{2}}\right)<{\frac {\theta }{2}}{\text{ so}}} 1 − sin 2 ( θ 2 ) > 1 − ( θ 2 ) 2 so {\displaystyle 1-\sin ^{2}\left({\frac {\theta }{2}}\right)>1-\left({\frac {\theta }{2}}\right)^{2}{\text{ so}}} sin ( θ ) < θ ( 1 − ( θ 2 ) 2 ) = θ − θ 3 4 {\displaystyle \sin(\theta )<\theta \left(1-\left({\frac {\theta }{2}}\right)^{2}\right)=\theta -{\frac {\theta ^{3}}{4}}} .Theorem: sin ( θ ) {\displaystyle \sin(\theta )} and cos ( θ ) {\displaystyle \cos(\theta )} are continuous functions.
Proof: For any h {\displaystyle h} ,
| sin ( θ + h ) − sin ( θ ) | = 2 | cos ( θ + h 2 ) | | sin ( h 2 ) | < h {\displaystyle |\sin(\theta +h)-\sin(\theta )|=2|\cos \left(\theta +{\frac {h}{2}}\right)||\sin \left({\frac {h}{2}}\right)|<h} ,since | cos ( θ ) | {\displaystyle \left\vert \cos(\theta )\right\vert } cannot exceed 1 {\displaystyle 1} and | sin ( θ ) | {\displaystyle \left\vert \sin(\theta )\right\vert } cannot exceed | x | {\displaystyle \left\vert x\right\vert } . Thus, as
h → 0 , sin ( θ + h ) → sin ( θ ) {\displaystyle h\rightarrow 0,\,\sin(\theta +h)\rightarrow \sin(\theta )} ,proving continuity. The proof for cos(θ) is similar, or it follows from
cos ( θ ) = sin ( π 2 − θ ) {\displaystyle \cos(\theta )=\sin \left({\frac {\pi }{2}}-\theta \right)} .