Sec 3 A Math: Basic Trigonometric Identities, practice questions & worked solutions

(b)

$$\begin{aligned} \text{L.H.S: } \cot A + \tan A &= \frac{1}{\tan A} + \tan A\\ &= \frac{\cos A}{\sin A} + \frac{\sin A}{\cos A}\\ &= \frac{\cos^2 A + \sin^2 A}{\sin A \cos A}\\ &= \frac{1}{\sin A \cos A}\\ &= \sec A \cosec A = \text{RHS} \end{aligned}$$

OR

$$\begin{aligned} \text{RHS: } \sec A \cosec A &= \frac{1}{\cos A} \times \frac{1}{\sin A}\\ &= \frac{1}{\cos A \sin A}\\ &= \frac{\sin^2 A + \cos^2 A}{\cos A \sin A}\\ &= \frac{\sin^2 A}{\cos A \sin A} + \frac{\cos^2 A}{\cos A \sin A}\\ &= \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}\\ &= \tan A + \cot A = \text{LHS} \end{aligned}$$

(a)

$$\begin{aligned} \frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x} &= \frac{\tan^2 x+(1+\sec x)^2}{\tan x(1+\sec x)}\\ &= \frac{\tan^2 x+1+2\sec x+\sec^2 x}{\tan x(1+\sec x)}\\ &= \frac{\sec^2 x+2\sec x+\sec^2 x}{\tan x(1+\sec x)}\\ &= \frac{2\sec^2 x+2\sec x}{\tan x(1+\sec x)}\\ &= \frac{2\sec x(\sec x+1)}{\tan x(1+\sec x)}\\ &= \frac{2}{\cos x}\div\frac{\sin x}{\cos x}\\ &= \frac{2}{\sin x}\\ &= 2\cosec x \end{aligned}$$

(c)

$$\begin{aligned} 2\cosec x &= k\\ \frac{2}{\sin x} &= k\\ \sin x &= \frac{2}{k} \end{aligned}$$ Since $-1\leq\sin x\leq 1$, $\sin x=\dfrac{2}{k}$ has no solutions when $\dfrac{2}{k}<-1$ or $\dfrac{2}{k}>1$, therefore $2\cosec x=k$ has no solutions when $-2<k<2$.

(a)

$$\begin{aligned} 2 - \sin^2\theta &= 2 - (1 - \cos^2\theta)\\ &= 1 + \cos^2\theta\\ &= 1 + \left(\frac{5}{13}\right)^2\\ &= \frac{194}{169} \end{aligned}$$

(b)

$$\begin{aligned} 2 - \sin^2\theta &= \frac{194}{169}\\ \sin^2\theta &= \frac{144}{169}\\ \tan^2(90^\circ - \theta) &= \frac{1}{\tan^2\theta}\\ &= \frac{\cos^2\theta}{\sin^2\theta}\\ &= \frac{\left(\dfrac{5}{13}\right)^2}{\dfrac{144}{169}}\\ &= \frac{25}{144} \end{aligned}$$

(a)

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

(b)

$$\begin{aligned} 2\cosec x &= 5\\ \cosec x &= \frac{5}{2}\\ \sin x &= \frac{2}{5}\\ \text{Basic angle} &= 0.41151\\ x &= 0.412,\ 2.73\ (\text{3 s.f}) \end{aligned}$$

(a)

$$\begin{aligned} \text{LHS} &= \frac{(\sin A - \cos A)(1 + \sin A\cos A)}{\cos^3 A}\\ &= \frac{\sin A - \cos A + \sin^2 A\cos A - \sin A\cos^2 A}{\cos^3 A}\\ &= \frac{\sin A - \cos A + (1 - \cos^2 A)\cos A - \sin A(1 - \sin^2 A)}{\cos^3 A}\\ &= \frac{\sin A - \cos A + \cos A - \cos^3 A - \sin A + \sin^3 A}{\cos^3 A}\\ &= \frac{\sin^3 A - \cos^3 A}{\cos^3 A}\\ &= \frac{\sin^3 A}{\cos^3 A} - \frac{\cos^3 A}{\cos^3 A}\\ &= \tan^3 A - 1 \end{aligned}$$

OR

$$\begin{aligned} \text{LHS} &= \frac{(\sin A - \cos A)(1 + \sin A\cos A)}{\cos^3 A}\\ &= \frac{(\sin A - \cos A)(\sin^2 A + \cos^2 A + \sin A\cos A)}{\cos^3 A}\\ &= \frac{\sin^3 A + \sin A\cos^2 A + \sin^2 A\cos A - \sin^2 A\cos A - \cos^3 A - \sin A\cos^2 A}{\cos^3 A}\\ &= \frac{\sin^3 A - \cos^3 A}{\cos^3 A}\\ &= \frac{\sin^3 A}{\cos^3 A} - \frac{\cos^3 A}{\cos^3 A}\\ &= \tan^3 A - 1 \end{aligned}$$

(b)

$$\begin{aligned} (\sin A - \cos A)(1 + \sin A\cos A) &= -2\cos^3 A\\ \frac{(\sin A - \cos A)(1 + \sin A\cos A)}{\cos^3 A} &= -2\\ \tan^3 A - 1 &= -2\\ \tan^3 A &= -1\\ \tan A &= -1 \end{aligned}$$ Basic angle $= \dfrac{\pi}{4}$

Quad: Q2, Q4 $$A = \frac{3\pi}{4},\ -\frac{\pi}{4}$$

(a)

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

(b)

$$\begin{aligned} \frac{1-\tan x}{1+\tan x} &= \frac{2}{3}\tan x\\ 3(1-\tan x) &= 2\tan x+2\tan^2 x\\ 2\tan^2 x+5\tan x-3 &= 0\\ (2\tan x-1)(\tan x+3) &= 0\\ \tan x=\frac{1}{2} \;&\text{or}\; -3\\ x=0.464 \;&\text{or}\; 3.61\\ x=1.89 \;&\text{or}\; 5.03 \end{aligned}$$

(a)

$$\begin{aligned} \text{LHS} &= \sin^3 x\sec^2 x + \sin x\\ &= \sin^3 x\left(\frac{1}{\cos^2 x}\right) + \sin x\\ &= \tan^2 x\sin x + \sin x\\ &= \sin x\left(\tan^2 x + 1\right)\\ &= \sin x\sec^2 x\\ &= \sin x\left(\frac{1}{\cos^2 x}\right)\\ &= \tan x\sec x\\ &= \text{RHS} \end{aligned}$$

(b)

$$\begin{aligned} \tan x\sec x &= 5\\ \frac{\sin x}{\cos x}\left(\frac{1}{\cos x}\right) &= 5\\ \frac{\sin x}{\cos^2 x} &= 5\\ \sin x &= 5\left(1 - \sin^2 x\right)\\ 5\sin^2 x + \sin x - 5 &= 0\\ \sin x &= \frac{-1 \pm \sqrt{(1)^2 - 4(5)(-5)}}{2(5)}\\ \sin x = 0.90498 \ &\text{or}\ -1.10498 \ (\text{NA}) \end{aligned}$$ Basic angle $= 64.8215^\circ$ $$x = 64.8^\circ,\ 115.2^\circ$$

(a)

$$\begin{aligned} \text{LHS} &= \frac{1}{\cosec x - \cot x} - \frac{1}{\sin x}\\[2pt] &= \frac{1}{\dfrac{1}{\sin x} - \dfrac{\cos x}{\sin x}} - \frac{1}{\sin x}\\[2pt] &= \frac{\sin x}{1 - \cos x} - \frac{1}{\sin x}\\[2pt] &= \frac{\sin x(1 + \cos x)}{1 - \cos^2 x} - \frac{1}{\sin x}\\[2pt] &= \frac{\sin x(1 + \cos x)}{\sin^2 x} - \frac{1}{\sin x}\\[2pt] &= \frac{1 + \cos x}{\sin x} - \frac{1}{\sin x}\\[2pt] &= \cot x\\[2pt] &= \text{RHS} \end{aligned}$$

(b)

$$\begin{aligned} \cot x &= \cot^2 x - 2\\ \cot^2 x - \cot x - 2 &= 0\\ (\cot x - 2)(\cot x + 1) &= 0\\ \cot x = -1 \ &\text{or} \ \cot x = 2\\ \tan x = -1 \ &\text{or} \ \tan x = 0.5 \end{aligned}$$ $$x = \frac{3\pi}{4},\ \frac{7\pi}{4} \text{or} \text{Basic angle} = 0.46364$$ $$x = 0.464,\ 3.61 \text{ radians (3 s.f.)}$$ $$x = 0.464,\ 2.36,\ 3.61,\ 5.50 \text{ radians (3 s.f.)}$$