University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.3 - Trigonometric Functions - Exercises - Page 28: 38


$$\cos(A+2\pi)=\cos A$$ $$\sin(A+2\pi)=\sin A$$ The results show that adding an angle of $2\pi$, which is the period of sine and cosine, to angle $A$ would not change its value of sine and cosine.

Work Step by Step

* Recall the addition formulas: $$\cos(A+B)=\cos A\cos B-\sin A\sin B$$ $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ If we take $B=2\pi$, then we have: 1) The addition formula for cosine: $$\cos(A+2\pi)=\cos A\cos(2\pi)-\sin A\sin(2\pi)$$ $$\cos(A+2\pi)=\cos A\times1-\sin A\times0$$ $$\cos(A+2\pi)=\cos A$$ 2) The addition formula for sine: $$\sin(A+2\pi)=\sin A\cos(2\pi)+\cos A\sin(2\pi)$$ $$\sin(A+2\pi)=\sin A\times1+\cos A\times0$$ $$\sin(A+2\pi)=\sin A$$ As we can see, if we add $2\pi$ to an angle, the value of its sine and cosine is unchanged. This is because $2\pi$ is exactly one period of cosine and sine. In other words, in the unit circle, adding an angle of $2\pi$ to an angle $A$ would lead you to the exact same spot, so the value of sine and cosine is still the same.
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