Answers: Trigonometric identities (radians)
These are the answers to Questions: Trigonometric identities (radians).
Please attempt the questions before reading these answers!
Q1
1.1. \(\quad 2(6\sin^2(\theta))+3(4\cos^2(\theta)) = 12(\sin^2(\theta)+\cos^2(\theta)) = 12\)
1.2. \(\quad 10(7\sin^2(\theta))+14(5\cos^2(\theta)) = 70\)
1.3. \(\quad 5\left(\dfrac{6}{\csc^2(\theta)}\right)+15\left(\dfrac{2}{\sec^2(\theta)}\right) = 30\)
1.4. \(\quad (\cos^2(\theta)-\sin^2(\theta))^2 + 4\sin^2(\theta)\cos^2(\theta) = \cos^2(2\theta) + \sin^2(2\theta) = 1\)
1.5. \(\quad 2\sin(\pi/6)\cos(\pi/12)+2\cos(\pi/6)\sin(\pi/12) = 2\sin(\pi/6+\pi/12) = 2\sin(\pi/4) = \sqrt{2}\)
1.6. \(\quad 3\cos(\pi/4)\cos(\pi/12)-3\sin(\pi/4)\sin(\pi/12) = 3\cos(\pi/3) = \dfrac{3}{2}\)
1.7. \(\quad \sin(5\pi/6)+\sin(\pi/6) = 2\sin\left(\dfrac{180}{2}\right)\cos\left(\dfrac{120}{2}\right) = 2\sin(\pi/2)\cos(\pi/3) = 1\)
1.8. \(\quad \cos(5\pi/6)+\sin(\pi/6) = 2\cos(\pi/2)\cos(\pi/3) = 0\)
Q2
2.1. \(\quad \tan(\theta)\cos(-\theta) = \dfrac{\sin(\theta)}{\cos(\theta)}\cdot\cos(\theta) = \sin(\theta)\)
2.2 \(\quad \displaystyle\tan(-\theta)\csc(-\theta)\sec(-\theta) = \left(-\frac{\sin(\theta)}{\cos(\theta)}\right)\left(\frac{1}{-\sin(\theta)}\right)\left(\frac{1}{\cos(\theta)}\right) = \left(\frac{1}{\cos^2(\theta)}\right) = \sec^2(\theta)\)
2.3. \(\quad\tan^2(\theta)+\sin^2(\theta)+\cos^2(\theta) = \tan^2(\theta)+1 = \sec^2(\theta)\)
2.4. \(\quad \dfrac{2\sin(\theta)}{\cos(\theta)(1-\tan^2(\theta))} = \tan(2\theta)\)
2.5. \(\quad\displaystyle\frac{\sin(7\theta)+\sin(3\theta)}{\cos(7\theta)-\cos(3\theta)} = \frac{2\sin(5\theta)\cos(2\theta)}{-2\sin(5\theta)\sin(2\theta)} = -\cot(\theta)\)
2.6. \(\quad\displaystyle\frac{\sin(5\theta)-\sin(\theta)}{\cos(5\theta)+\cos(\theta)} = \tan(2\theta)\)
Q3
3.1. \(\cos(5\pi/6) = \dfrac{\sqrt{3}}{2}\)
3.2. Here \(\sin(3\pi/4) = \dfrac{1}{\sqrt{2}}\), and \(\sin(5\pi/4) = -\dfrac{1}{\sqrt{2}}\).
3.3. \(\cos(13\pi/18) = -0.766\) to three decimal places.
Q4
4.1. \(\sin(\pi/12) = \sin(\pi/4)\cos(\pi/6) - \cos(\pi/4)\sin(\pi/6) = \dfrac{\sqrt3}{2\sqrt2} - \dfrac{1}{2\sqrt2} = \dfrac{\sqrt3-1}{2\sqrt2}\)
4.2. \(\quad\cos(\pi/12) = \dfrac{\sqrt3+1}{2\sqrt2}\)
4.3. \(\quad \tan(\pi/12) =\dfrac{\sqrt3+1}{\sqrt3-1}\)
4.4. \(\quad \sin(5\pi/12) = \sin(\pi/4)\cos(\pi/6) + \cos(\pi/4)\sin(\pi/6) = \dfrac{\sqrt3+1}{2\sqrt2}\)
4.5. \(\quad\cos(5\pi/12) = \dfrac{\sqrt3-1}{2\sqrt2}\)
4.6. \(\quad\tan(5\pi/12) = \dfrac{\sqrt3+1}{\sqrt3-1}\)
Version history and licensing
v1.0: initial version created 08/23 by Dzhemma Ruseva as part of a University of St Andrews STEP project.
- v1.1: edited 05/24 by tdhc, and split into versions for both degrees and radians.