In these trigonometry formulas, we will focus on all the identities used for solving the complex problems in trigonometry. Firstly, we will understand the functions used in trigonometries such as \sin, \cos,\tan,\csc, \sec and \cot and the basics of it. These functions are inter-related in the following way
\csc\theta = \frac{1}{\sin \theta}\\ \sec \theta = \frac{1}{\cos \theta}\\ \cot \theta = \frac{1}{\tan \theta}\\ \sin\theta = \frac{1}{\csc \theta}\\ \cos \theta = \frac{1}{\sec \theta}\\ \tan \theta = \frac{1}{\cot \theta}\\Above we have shown the basic formulas for this function using a right angled triangle in trigonometry table. Also, few more trigonometry formulas
I)In radians
\sin (\Pi/2 – \theta) = \cos \theta \:\&\: \cos (\Pi/2 – \theta) = \sin \theta \\ \sin (\Pi/2 + \theta) = \cos \theta \:\&\: \cos (\Pi/2 + \theta) = – \sin \theta \\ \sin (3\Pi/2 – \theta) = – \cos \theta \:\&\: \cos (3\Pi/2 – \theta) = – \sin \theta \\ \sin (3\Pi/2 + \theta) = – \cos \theta \:\&\: \cos (3\Pi/2 + \theta) = \sin \theta \\ \sin (\Pi – \theta) = \sin \theta \:\&\: \cos (\Pi – \theta) = – \cos \theta \\ \sin (\Pi + \theta) = – \sin \theta \:\&\: \cos (\Pi + \theta) = – \cos \theta \\ \sin (2\Pi – \theta) = – \sin \theta \:\&\: \cos (2\Pi – \theta) = \cos \theta \\ \sin (2\Pi + \theta) = \sin \theta \:\&\: \cos (2\Pi + \theta) = \cos \theta \\
II)Degree
\sin(90°−\theta) = \cos \theta\\ \cos(90°−\theta) = \sin \theta\\ \tan(90°−\theta) = \cot \theta\\ \cot(90°−\theta) = \tan \theta\\ \sec(90°−\theta) =\ csc \theta\\ \csc(90°−\theta) = \sec \theta\\
- Addition and subtraction formulas
\sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)\\ \cos(a+b) = \cos(a)\cos(b)–sin(a)\sin(b)\\ \tan(a+b) = \frac{(\tan a + \tan b)}{ (1−\tan a \times \tan b)}\\ \sin(a-b) = \sin(a)\cos(b)–\cos(a)\sin(b)\\ \cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)\\ \tan(a-b) = \frac{(\tan a–\tan b)} {(1+\tan a \times \tan b)}\\- Double and triple angle formula
\sin(2\theta) = 2\sin(\theta) \cos(\theta) = [\frac{2\tan \theta}{(1+\tan2 \theta)}]\\ \cos(2\theta) = \cos2(\theta)–\sin2(\theta) = \frac{(1-\tan2 \theta)}{(1+\tan2 \theta)}\\ \cos(2\theta) = 2\cos2(x\theta)−1 = 1–2\sin2(\theta)\\ \tan(2\theta) = \frac{2\tan(\theta)}{ 1−\tan2(\theta)}\\ \sec (2\theta) =\frac {\sec2 \theta}{(2-\sec2 \theta)}\\ \csc (2\theta) = \frac{(\sec \theta. \csc \theta)}{2}\\ \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]}\\ \sin \frac{\theta}{2}=\sqrt{±1−\cos\frac{\theta}{2}}\\ \cos \frac{\theta}{2}=\sqrt{±1+\cos\frac{\theta}{2}}\\ \tan\frac{\theta}{2}=\sqrt{\frac{1−\cos(\theta)}{1+\cos(\theta)}}\\ \tan\frac{\theta}{2}=\sqrt{\frac{1−\cos(\theta)}{1+\cos(\theta)}}=\sqrt{\frac{1−cos(\theta)}{sin(\theta)}} \sin \alpha .\cos \beta =\frac{\sin(\alpha +\beta)+\sin(\alpha -\beta)}{2}\\ \cos \alpha .\cos \beta =\frac{\cos(\alpha +\beta)+\cos(\alpha -\beta)}{2}\\ \sin \alpha .\sin \beta =\frac{\cos(\alpha -\beta)-\cos(\alpha +\beta)}{2}\\ \sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}\\ \sin\alpha−\sin\beta=2\cos\frac{\alpha+\beta}\sin\frac{\alpha-\beta}{2}\\ \cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}\cos\frac{\alpha-\beta}{2}\\ \cos\alpha−\cos\beta=−2\sin\frac{\alpha+\beta}\sin\frac{\alpha-\beta}{2}- Inverse trigonometry formula
\sin^{-1} (–\theta) = – \sin^{-1} \theta \\ \cos^{-1} (–\theta) = \Pi – \cos^{-1} \theta\\ \tan^{-1} (–\theta) = – \tan^{-1} \theta\\ \csc^{-1} (–\theta) = – \csc^{-1} \theta\\ \sec^{-1} (–\theta) = \Pi – \sec^{-1} \theta\\ \cot^{-1} (–\theta) = \Pi – \cot^{-1} \thetaDo you want to learn trigonometry equations and formulas with the best Experts? Contact the best maths expert and become a trigonometry expert today.
Need help with your Trigonometry? Feel free to get in touch with us.
Like this article? Don’t forget to share this blog with your colleagues & friends on Facebook and Twitter. Keep learning keep sharing.
Also, for daily updates subscribe to our weekly feed.