Unit 7: Trigonometry

\(\begin{aligned} & 1. sin2A = 2sinAcosA, \frac{2tanA}{1 + tan^{2}A} \\ & 2. cos2A = cos^{2}A – sin^{2}A, 2cos^{2}A – 1, 1 – 2sin^{2}A \\ & 3. tan2A = \frac{2tanA}{1 – tan^{2}A} \\ & 4. sin3A = 3sinA – 4sin^{3}A \\ & 5. cos3A = 4cos^{3}A – 3cosA \\ & 6. tan3A = \frac{3tanA – tan^{3}A}{1-3tan^{2}A} \end{aligned}\)

\(\begin{aligned} & 1. sinA = 2sin\frac{A}{2}cos\frac{A}{2} \\ & 2. cosA = cos^{2}\frac{A}{2} – sin^{2}\frac{A}{2}, 2cos^{2}\frac{A}{2} – 1, 1 – 2sin^{2}\frac{A}{2} \\ & 3. tanA = \frac{2tan\frac{A}{2}}{1 – tan^{2}\frac{A}{2}} \\ & 4. sinA = 3sin\frac{A}{3} – 4sin^{3}\frac{A}{3} \\ & 5. cosA = 4cos^{3}\frac{A}{3} – 3cos\frac{A}{3} \\ & 6. tanA = \frac{3tan\frac{A}{3} – tan^{3}\frac{A}{3}}{1-3tan^{2}\frac{A}{3}} \end{aligned}\)

\(\begin{aligned} & 1. 2sinA.cosB = sin(A+B) + sin(A-B) \\ & 2. 2cosA.sinB = sin(A+B) - sin(A-B) \\ & 3. 2cosA.cosB = cos(A+B) + cos(A-B) \\ & 4. 2sinA.sinB = cos(A-B) - sin(A+B) \\ \end{aligned}\)

\(\begin{aligned} & 1. sinC + sinD = 2sin(\frac{C+D}{2})cos(\frac{C-D}{2}) \\ & 2. sinC - sinD = 2cos(\frac{C+D}{2})sin(\frac{C-D}{2}) \\ & 3. cosC + cosD = 2cos(\frac{C+D}{2})cos(\frac{C-D}{2}) \\ & 4. cosC - cosD = 2sin(\frac{C+D}{2})sin(\frac{C-D}{2}) \\ \end{aligned}\)

\(\begin{aligned} & Solve. (0 \leq \theta \leq 360^\circ) \\ & \text{Q. } cos\theta + cos3\theta = 2cos2\theta \\ & \text{Solution: } \\ & \text{or, } cos\theta + 4cos^{3}\theta – 3cos\theta = 2(2cos^{2}\theta - 1) \\ & \text{or, } 4 cos^{3}\theta – 2 cos\theta = 4 cos^{2}\theta - 2 \\ & \text{or, } 4 cos^{3}\theta – 2 cos\theta - 4cos^{2}\theta + 2 = 0 \\ & \text{or, } 4 cos^{3}\theta - 4cos^{2}\theta – 2 cos\theta + 2 = 0 \\ & \text{or, } 4 cos^{2}\theta (cos\theta - 1) - 2 (cos\theta – 1) = 0 \\ & \text{or, } (cos\theta - 1) (4 cos^{2}\theta - 2)= 0 \\ & \text{Either } cos\theta – 1 = 0 \\ & \text{or, } cos\theta = 1 \\ & \text{or, } cos\theta = cos0^\circ, cos360^\circ \\ & \therefore \theta = 0^\circ, 360^\circ \\ \\ & \text{OR, } 4 cos^{2}\theta – 2 = 0 \\ & \text{or, } 4 cos^{2}\theta = 2 \\ & \text{or, } cos^{2}\theta = \frac{1}{2} \\ & \text{or, } cos\theta = \frac{1}{\sqrt{2}} \\ & \text{or, } cos\theta = cos45^\circ, cos(360^\circ – 45^\circ) \\ & \text{or, } cos\theta = cos45^\circ, cos(315^\circ) \\ & \therefore \theta = 45^\circ , 315^\circ \\ & \text{Hence, } \theta = 0^\circ , 45^\circ , 315^\circ , 360^\circ \text{ (Answer)} \end{aligned}\)

Q. Prove That: $\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)\left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right) = \frac{1}{8}$ $LHS = \left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)\left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$
Use $\cos(\pi-\theta) = -\cos\theta$:
$\cos\frac{5\pi}{8} = \cos\left(\pi-\frac{3\pi}{8}\right) = -\cos\frac{3\pi}{8}$
$\cos\frac{7\pi}{8} = \cos\left(\pi-\frac{\pi}{8}\right) = -\cos\frac{\pi}{8}$
So: $LHS = \left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)\left(1-\cos\frac{3\pi}{8}\right)\left(1-\cos\frac{\pi}{8}\right)$ $= \left(1-\cos^{2}\frac{\pi}{8}\right)\left(1-\cos^{2}\frac{3\pi}{8}\right)$ $= \sin^{2}\frac{\pi}{8}\cdot \sin^{2}\frac{3\pi}{8} \quad [\because 1-\cos^2\theta = \sin^2\theta]$ Now $\sin^2\theta = \frac{1-\cos 2\theta}{2}$: $\sin^{2}\frac{\pi}{8} = \frac{1-\cos\frac{\pi}{4}}{2} = \frac{1-\frac{1}{\sqrt2}}{2} = \frac{2-\sqrt2}{4}$ $\sin^{2}\frac{3\pi}{8} = \frac{1-\cos\frac{3\pi}{4}}{2} = \frac{1-(-\frac{1}{\sqrt2})}{2} = \frac{2+\sqrt2}{4}$ Hence: $LHS = \frac{2-\sqrt2}{4}\cdot\frac{2+\sqrt2}{4} = \frac{4-2}{16} = \frac{2}{16} = \frac18 = RHS$ Verified numerically $LHS \approx 0.125 = 1/8$. Proved.

\( \begin{aligned} \text{LHS} &=\frac{\cos A-\sin A+1}{\cos A+\sin A+1}\\ &=\frac{(\cos A-\sin A+1)(1+\sin A)}{(\cos A+\sin A+1)(1+\sin A)}\\ &=\frac{\cos A+\cos A\sin A-\sin A-\sin^2A+1+\sin A}{(\cos A+\sin A+1)(1+\sin A)}\\ &=\frac{\cos A(1+\sin A+\cos A)}{(\cos A+\sin A+1)(1+\sin A)}\\ &=\frac{\cos A}{1+\sin A}\\ &=\frac{\cos A(1-\sin A)}{1-\sin^2A}\\ &=\frac{1-\sin A}{\cos A}\\ &=\text{RHS}. \end{aligned} \)

\( \begin{aligned} \text{LHS} &=\frac{\cos A-\sin A+1}{\cos A+\sin A-1}\\ &=\frac{(\cos A-\sin A+1)(1+\cos A)}{(\cos A+\sin A-1)(1+\cos A)}\\ &=\frac{(1+\cos A)(1+\cos A-\sin A)}{(\cos A-1+\sin A)(1+\cos A)}\\ &=\frac{(1+\cos A)(1+\cos A-\sin A)}{(\cos^2A-1)+\sin A(1+\cos A)}\\ &=\frac{(1+\cos A)(1+\cos A-\sin A)}{-\sin^2A+\sin A(1+\cos A)}\\ &=\frac{(1+\cos A)(1+\cos A-\sin A)}{\sin A(1+\cos A-\sin A)}\\ &=\frac{1+\cos A}{\sin A}\\ &=\text{RHS}. \end{aligned} \)