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6.1: Trigonometric Identities
https://math.libretexts.org/Courses/Coastline_College/Math_C180%3A_Calculus_I_(Nguyen)/06%3A_Appendices/6.01%3A_Trigonometric_Identities
$\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ \(\sin^2 ... $\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ $\sin^2 x = \dfrac{1- \cos 2x}{2}$ $\sin (-x) = -\sin x$ $\cos(-x) = \cos x$ $\tan (-x) = -\tan x$ $\sin\left(x \pm \frac{\pi}{2}\right) = \pm \cos x$ $\cos\left(x \pm \frac{\pi}{2}\right) = \mp \sin x$ $\sin(\pi - x) = \sin x$ $\cos(\pi - x) = -\cos x$ $\tan(\pi - x) = -\tan x$
6.1: Trigonometric Identities
https://math.libretexts.org/Courses/Coastline_College/Math_C180%3A_Calculus_I_(Everett)/06%3A_Appendices/6.01%3A_Trigonometric_Identities
$\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ \(\sin^2 ... $\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ $\sin^2 x = \dfrac{1- \cos 2x}{2}$ $\sin (-x) = -\sin x$ $\cos(-x) = \cos x$ $\tan (-x) = -\tan x$ $\sin\left(x \pm \frac{\pi}{2}\right) = \pm \cos x$ $\cos\left(x \pm \frac{\pi}{2}\right) = \mp \sin x$ $\sin(\pi - x) = \sin x$ $\cos(\pi - x) = -\cos x$ $\tan(\pi - x) = -\tan x$
6.4: Trigonometric Identities
https://math.libretexts.org/Courses/Penn_State_University_Greater_Allegheny/Math_140%3A_Calculus_1_(Gaydos)/06%3A_Appendices/6.04%3A_Trigonometric_Identities
$\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ \(\sin^2 ... $\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ $\sin^2 x = \dfrac{1- \cos 2x}{2}$ $\sin(α + β) = \sin(α) \cos(β) + \cos(α) \sin(β)$ $\sin(α - β) = \sin(α) \cos(β) - \cos(α) \sin(β)$ $\sin (-x) = -\sin x$ $\cos(-x) = \cos x$ $\sin\left(x \pm \frac{\pi}{2}\right) = \pm \cos x$ $\cos\left(x \pm \frac{\pi}{2}\right) = \mp \sin x$
6.1: Trigonometric Identities
https://math.libretexts.org/Courses/Coastline_College/Math_C180%3A_Calculus_I_(Tran)/06%3A_Appendices/6.01%3A_Trigonometric_Identities
$\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ \(\sin^2 ... $\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ $\sin^2 x = \dfrac{1- \cos 2x}{2}$ $\sin (-x) = -\sin x$ $\cos(-x) = \cos x$ $\tan (-x) = -\tan x$ $\sin\left(x \pm \frac{\pi}{2}\right) = \pm \cos x$ $\cos\left(x \pm \frac{\pi}{2}\right) = \mp \sin x$ $\sin(\pi - x) = \sin x$ $\cos(\pi - x) = -\cos x$ $\tan(\pi - x) = -\tan x$
8.1: Trigonometric Identities
https://math.libretexts.org/Courses/Coastline_College/Math_C185%3A_Calculus_II_(Tran)/08%3A_Appendices/8.01%3A_Trigonometric_Identities
$\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ \(\sin^2 ... $\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ $\sin^2 x = \dfrac{1- \cos 2x}{2}$ $\sin (-x) = -\sin x$ $\cos(-x) = \cos x$ $\tan (-x) = -\tan x$ $\sin\left(x \pm \frac{\pi}{2}\right) = \pm \cos x$ $\cos\left(x \pm \frac{\pi}{2}\right) = \mp \sin x$ $\sin(\pi - x) = \sin x$ $\cos(\pi - x) = -\cos x$ $\tan(\pi - x) = -\tan x$
21.1: A.1- Trigonometric Identities
https://math.libretexts.org/Under_Construction/Purgatory/Differential_Equations_and_Linear_Algebra_(Zook)/21%3A_Appendices/21.01%3A_A.1-_Trigonometric_Identities
$\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ \(\sin^2 ... $\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ $\sin^2 x = \dfrac{1- \cos 2x}{2}$ $\sin(α + β) = \sin(α) \cos(β) + \cos(α) \sin(β)$ $\sin(α - β) = \sin(α) \cos(β) - \cos(α) \sin(β)$ $\sin (-x) = -\sin x$ $\cos(-x) = \cos x$ $\sin\left(x \pm \frac{\pi}{2}\right) = \pm \cos x$ $\cos\left(x \pm \frac{\pi}{2}\right) = \mp \sin x$
Trigonometric Identities
https://math.libretexts.org/Courses/Monroe_Community_College/MTH_210_Calculus_I_(Seeburger)/Appendices/a0-Trigonometric_Identities
$\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ \(\sin^2 ... $\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ $\sin^2 x = \dfrac{1- \cos 2x}{2}$ $\sin(α + β) = \sin(α) \cos(β) + \cos(α) \sin(β)$ $\sin(α - β) = \sin(α) \cos(β) - \cos(α) \sin(β)$ $\sin (-x) = -\sin x$ $\cos(-x) = \cos x$ $\sin\left(x \pm \frac{\pi}{2}\right) = \pm \cos x$ $\cos\left(x \pm \frac{\pi}{2}\right) = \mp \sin x$
Section 16.0: Trigonometric Identities
https://math.libretexts.org/Courses/Al_Akhawayn_University/MTH2301_Multivariable_Calculus/16%3A_Appendices/16.00%3A_Trigonometric_Identities
$\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ \(\sin^2 ... $\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ $\sin^2 x = \dfrac{1- \cos 2x}{2}$ $\sin (-x) = -\sin x$ $\cos(-x) = \cos x$ $\tan (-x) = -\tan x$ $\sin\left(x \pm \frac{\pi}{2}\right) = \pm \cos x$ $\cos\left(x \pm \frac{\pi}{2}\right) = \mp \sin x$ $\sin(\pi - x) = \sin x$ $\cos(\pi - x) = -\cos x$ $\tan(\pi - x) = -\tan x$
16.1: Trigonometric Identities
https://math.libretexts.org/Courses/Reedley_College/Differential_Equations_and_Linear_Algebra_(Zook)/16%3A_Appendices/16.01%3A_Trigonometric_Identities
$\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ \(\sin^2 ... $\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ $\sin^2 x = \dfrac{1- \cos 2x}{2}$ $\sin (-x) = -\sin x$ $\cos(-x) = \cos x$ $\tan (-x) = -\tan x$ $\sin\left(x \pm \frac{\pi}{2}\right) = \pm \cos x$ $\cos\left(x \pm \frac{\pi}{2}\right) = \mp \sin x$ $\sin(\pi - x) = \sin x$ $\cos(\pi - x) = -\cos x$ $\tan(\pi - x) = -\tan x$
8.1: Trigonometric Identities
https://math.libretexts.org/Courses/Coastline_College/Math_C185%3A_Calculus_II_(Everett)/08%3A_Appendices/8.01%3A_Trigonometric_Identities
$\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ \(\sin^2 ... $\cos^2 x + \sin^2 x = 1$ $\sec^2 x - \tan^2 x = 1$ $\sin 2x = 2 \sin x \cos x$ $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x = 2 \cos^2 x - 1$ $\cos^2 x = \dfrac{1+ \cos 2x}{2}$ $\sin^2 x = \dfrac{1- \cos 2x}{2}$ $\sin (-x) = -\sin x$ $\cos(-x) = \cos x$ $\tan (-x) = -\tan x$ $\sin\left(x \pm \frac{\pi}{2}\right) = \pm \cos x$ $\cos\left(x \pm \frac{\pi}{2}\right) = \mp \sin x$ $\sin(\pi - x) = \sin x$ $\cos(\pi - x) = -\cos x$ $\tan(\pi - x) = -\tan x$

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