Conic SectionsTrigonometric Identities Reference
Sum Identities
\(\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\cdot\sin\beta\)
\(\cos(\alpha + \beta) = \cos\alpha\cdot\cos\beta - \sin\alpha\cdot\sin\beta\)
\(\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\cdot\tan\beta}\)
\(\tan(\alpha + \beta) = \dfrac{\sin(2\alpha) + \sin(2\beta)}{\cos(2\alpha) + \cos(2\beta)}\)
\(\sin(2\alpha) = 2\sin\alpha\cdot\cos\alpha\)
\(\cos(2\alpha) = \cos^{2}\alpha - \sin^{2}\alpha\)
\(\cos(2\alpha) = 2\cos^{2}\alpha - 1 = \) \( 1 - 2\sin^{2}\alpha\)
\(\tan(2\alpha) = \dfrac{2\tan\alpha}{1 - \tan^{2}\alpha}\)
\( a\sin\theta + b\cos\theta = \sqrt{a^2+b^2}\sin\left(\theta + \arctan\frac{b}{a}\right)\)
Squares (Pythagorean) Identities
\( \sin^{2}\alpha + \cos^{2}\alpha = 1 \)
\( \sec^{2}\alpha = 1 + \tan^{2}\alpha \)
\( \csc^{2}\alpha = 1 + \cot^{2}\alpha \)
Half-Angle Identities
\( \cos\frac{\alpha}{2} = \sqrt{\frac{1+\cos\alpha}{2}} \)
\( \sin\frac{\alpha}{2} = \sqrt{\frac{1-\cos\alpha}{2}} \)
\( \tan\frac{\alpha}{2} = \sqrt{\frac{1-\cos\alpha}{1-\cos\alpha}} \)
\( \tan\frac{\alpha}{2} = \frac{\sin\alpha}{1+\cos\alpha} \)
\( \tan\frac{\alpha}{2} = \frac{1-\cos\alpha}{\sin\alpha} \)
Triple-Angle Identities
\( \sin(3\theta) = 3\sin\theta - 4\sin^{3}\theta \)
\( \cos(3\theta) = 4\cos^{3}\theta - 3\cos\theta \)
\( \tan(3\theta) = \frac{3\tan\theta - \tan^{3}\theta}{1-3\tan^{2}\theta} \)
For more multiple-angle identities of sine and cosine, see Cosine Identities.
Sum to Product Identities
\( \cos m + \cos n = 2\cos\left(\frac{m+n}{2}\right)\cdot \cos\left(\frac{m-n}{2}\right) \)
\( \sin m + \sin n = 2\sin\left(\frac{m+n}{2}\right)\cdot \cos\left(\frac{m-n}{2}\right) \)
Product to Sum Identities
\( \sin\alpha\cdot\cos\beta = \frac{1}{2}\left[ \sin(\alpha + \beta) + \sin(\alpha - \beta) \right] \)
\( \cos\alpha\cdot\sin\beta = \frac{1}{2}\left[ \sin(\alpha + \beta) - \sin(\beta - \alpha) \right] \)
\( \sin\alpha\cdot\sin\beta = \frac{1}{2}\left[ \cos(\alpha - \beta) - \cos(\alpha + \beta) \right] \)
\( \cos\alpha\cdot\cos\beta = \frac{1}{2}\left[ \cos(\alpha + \beta) + \cos(\alpha - \beta) \right] \)
Inverse Function Identities
\( \arcsin x \pm \arcsin y = \arcsin\left( x\sqrt{1-y^2} \pm y\sqrt{1-x^2} \right) \)
\( \arccos x \pm \arcsin y = \arccos\left( xy \mp \sqrt{(1-x^2)(1-y^2)} \right) \)
\( \arctan x \pm \arctan y = \arctan\left( \frac{x \pm y}{1 \mp xy}\right) \)
Composite Identities
\( \sin(\arccos x) = \sqrt{1-x^2} \)
\( \cos(\arcsin x) = \sqrt{1-x^2} \)
\( \sin(\arctan x) = \frac{x}{\sqrt{1+x^2}} \)
\( \cos(\arctan x) = \frac{1}{\sqrt{1+x^2}} \)
\( \tan(\arcsin x) = \frac{x}{\sqrt{1-x^2}} \)
\( \tan(\arccos x) = \frac{\sqrt{1-x^2}}{x} \)
Interesting Values
\(\cos 20^{\circ}\cdot \cos 40^{\circ} \cdot \cos 80^{\circ} = \) \( \cos\frac{\pi}{9}\cdot\cos\frac{2\pi}{9}\cdot\cos\frac{4\pi}{9} = \) \( -\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{4\pi}{7} = \frac{1}{8} \)
\(\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{3\pi}{7} = \frac{1}{8} \)
\(\sin 20^{\circ}\cdot \sin 40^{\circ} \cdot \sin 80^{\circ} = \) \( \sin\frac{\pi}{9}\cdot\sin\frac{2\pi}{9}\cdot\sin\frac{4\pi}{9} = \frac{\sqrt{3}}{8} \)
\( \sin\frac{\pi}{7}\cdot\sin\frac{2\pi}{7}\cdot\sin\frac{4\pi}{7} = \) \(\sin\frac{\pi}{7}\cdot\sin\frac{2\pi}{7}\cdot\sin\frac{3\pi}{7} = \frac{\sqrt{7}}{8} \)
\( \tan 50^{\circ}\cdot\tan 60^{\circ}\cdot\tan 70^{\circ} = \tan 80^{\circ}\)
\( \tan 40^{\circ}\cdot\tan 30^{\circ}\cdot\tan 20^{\circ} = \tan 10^{\circ}\)
Trigonometric Excursions
Problem: If \(\sin x + \cos x = n \), what is \(\sin^{2} x + \cos^{2} x \), \(\sin^{3} x + \cos^{3} x \), and \(\sin^{4} x + \cos^{4} x \) in terms of n?
Solution: We know that for any x, \(\sin^{2} x + \cos^{2} x = 1 \). We’ll square both sides to find the quantity \(\sin x\cos x\) however, because this quantity is useful for higher powers.
(i) \( n^2 = (\sin x + \cos x)^2 \)
(ii) \( n^2 = \sin^2 x + 2\sin x\cos x + \cos^2 x \)
(iii) \( n^2 = 1 + 2\sin x\cos x \)
(iv) \( \sin x\cos x = \frac{1}{2}(n^2 - 1) \)
Next, we will cube the sum of sine and cosine.
(i) \( n^3 = (\sin x + \cos x)^3 \)
(ii) \( n^3 = \sin^3 x + 3\sin^2 x\cos x + 3\sin x\cos^2 x + \cos^3 x \)
(iii) \( n^3 = \sin^3 x + \cos^3 x + 3\sin x\cos x(\sin x + \cos x) \)
(iv) \( \sin^3 x + \cos^3 x = n^3 - 3\cdot\frac{1}{2}(n^2-1)(n) \) (Substituting the value we found above.)
(v) \( \sin^3 x + \cos^3 x = n^3 - \frac{3}{2}n^3 + \frac{3}{2}n \)
(vi) \( \sin^3 x + \cos^3 x = -\frac{1}{2}n^3 + \frac{3}{2}n \)
Now for the fourth power.
(i) \( n^4 = (\sin x + \cos x)^4 \)
(ii) \( n^4 = \sin^4 x + 4\sin^3 x\cos x + 6\sin^2 x\cos^2 x + 4\sin x\cos^3 x + \cos^4 x \)
(iii) \( n^4 = \sin^4 x + \cos^4 x - 2\sin x\cos x (2\sin^2 x + 3\sin x\cos x + 2\cos^2 x) \)
(iv) \( \sin^4 x + \cos^4 x = n^4 - (n^2-1)\left[2 + 3\cdot\frac{1}{2}(n^2-1)\right] \)
(v) \( \sin^4 x + \cos^4 x = -\frac{1}{2}n^4 + n^2+\frac{1}{2} \)