Trigonometric Equations (SAQs)
Maths-1A | 7. Trigonometric Equations – SAQs:
Welcome to SAQs in Chapter 7: Trigonometric Equations. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.
SAQ-1 : Solve √3 sinθ-cosθ=√2.
Given equation is:
$$\sqrt{3} \sin(\theta) – \cos(\theta) = \sqrt{2}$$
On dividing by $$\sqrt{(\sqrt{3})^2 + (-1)^2}$$
$$= \sqrt{3 + 1} = \sqrt{4} = 2$$ we get
$$\frac{\sqrt{3}}{2} \sin(\theta) – \frac{1}{2} \cos(\theta) = \frac{\sqrt{2}}{2}$$
$$\sin(\theta) \left(\frac{\sqrt{3}}{2}\right) – \cos(\theta) \left(\frac{1}{2}\right) = \frac{1}{\sqrt{2}}$$
$$\sin(\theta) \cos(30°) – \cos(\theta) \sin(30°) = \sin(45°)$$
$$\sin(\theta) \cos\left(\frac{\pi}{6}\right) – \cos(\theta) \sin\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{4}\right)$$
$$\sin\left(\theta – \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{4}\right)$$
$$\sin A \cos B – \cos A \sin B = \sin(A – B)$$
Here, P.V is $$\alpha = \frac{\pi}{4}$$
General solution is:
$$\theta = n\pi + (-1)^n \alpha, \ n \in \mathbb{Z}$$
$$\theta – \frac{\pi}{6} = n\pi + (-1)^n \frac{\pi}{4}, \ n \in \mathbb{Z}$$
$$\theta = n\pi + (-1)^n \frac{\pi}{4} + \frac{\pi}{6}, \ n \in \mathbb{Z}$$
SAQ-2 : Solve √2(sinx+cosx)=√3.
Given equation is: $$\sqrt{2}(\sin x + \cos x) = \sqrt{3}$$
$$\sqrt{2} \sin x + \sqrt{2} \cos x = \sqrt{3}$$
On dividing by $$\sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2$$ we get:
$$\frac{\sqrt{2}}{2} \sin x + \frac{\sqrt{2}}{2} \cos x = \frac{\sqrt{3}}{2}$$
$$\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{\sqrt{3}}{2}$$
$$\cos x \left(\frac{1}{\sqrt{2}}\right) + \sin x \left(\frac{1}{\sqrt{2}}\right) = \frac{\sqrt{3}}{2}$$
$$\cos x \cos \frac{\pi}{4} + \sin x \sin \frac{\pi}{4} = \cos \frac{\pi}{6}$$
$$\cos(x – \frac{\pi}{4}) = \cos \frac{\pi}{6}$$
$$\cos A \cos B + \sin A \sin B = \cos(A – B)$$
Here, P.V (Principal Value) is $$\alpha = \frac{\pi}{6}$$
$$x = 2n\pi \pm \alpha, \ n \in \mathbb{Z}$$
$$x – \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{6}$$
$$x = 2n\pi \pm \frac{\pi}{6} + \frac{\pi}{4}, \ n \in \mathbb{Z}$$
$$x = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{6}, \ n \in \mathbb{Z}$$
$$x = 2n\pi \pm \frac{\pi}{6} + \frac{\pi}{4}, \ n \in \mathbb{Z}$$
SAQ-3 : Solve 2cos2 θ – √3 sinθ + 1 = 0.
Given equation: $$2\cos^2\theta – \sqrt{3} \sin\theta + 1 = 0$$
$$2(1 – \sin^2\theta) – \sqrt{3} \sin\theta + 1 = 0$$
$$\sin^2\theta + \cos^2\theta = 1$$
$$2 – 2\sin^2\theta – \sqrt{3} \sin\theta + 1 = 0$$
$$2\sin^2\theta + \sqrt{3}\sin\theta – 3 = 0$$
$$2\sin^2\theta + 2\sqrt{3}\sin\theta – \sqrt{3}\sin\theta – (\sqrt{3})^2 = 0$$
$$2\sin\theta(\sin\theta + \sqrt{3}) – \sqrt{3}(\sin\theta + \sqrt{3}) = 0$$
$$(2\sin\theta – \sqrt{3})(\sin\theta + \sqrt{3}) = 0$$
$$2\sin\theta = \sqrt{3}$$ or $$\sin\theta = -\sqrt{3}$$
$$\sin\theta = \frac{\sqrt{3}}{2}$$
$$\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$$
$$\theta = n\pi + (-1)^n\alpha, n \in \mathbb{Z}$$
$$\theta = n\pi + (-1)^n\frac{\pi}{3}, n \in \mathbb{Z}$$
SAQ-4 : Solve 1 + sin2θ = 3sinθcosθ.
Given equation is $$1+ \sin^2\theta = 3\sin\theta\cos\theta$$ Dividing the given equation by $$\cos^2\theta$$
we get $$\frac{1}{\cos^2\theta} + \frac{\sin^2\theta}{\cos^2\theta} = \frac{3\sin\theta\cos\theta}{\cos^2\theta}$$
$$\Rightarrow \sec^2\theta + \tan^2\theta = 3\tan\theta$$
$$\Rightarrow (1 + \tan^2\theta) + \tan^2\theta = 3\tan\theta$$
$$\Rightarrow 2\tan^2\theta – 3\tan\theta + 1 = 0$$
$$\Rightarrow 2\tan^2\theta – 2\tan\theta – \tan\theta + 1 = 0$$
$$\Rightarrow 2\tan\theta(\tan\theta – 1) – (\tan\theta – 1) = 0$$
$$\Rightarrow (2\tan\theta – 1)(\tan\theta – 1) = 0$$
$$\Rightarrow \tan\theta = 1$$ or $$\tan\theta = \frac{1}{2}$$
Now, $$\tan\theta = 1 = \tan \frac{\pi}{4}$$
Here Principal Value (P.V) is $$\alpha = \frac{\pi}{4}$$
General solution is $$\theta = n\pi + \frac{\pi}{4}$$ $$n \in \mathbb{Z}$$
Also, $$\tan\theta = \frac{1}{2} \Rightarrow \theta = \tan^{-1} \frac{1}{2}$$
Principal Value (P.V) is $$\alpha = \tan^{-1} \frac{1}{2}$$
General solution is $$\theta = n\pi + \tan^{-1} \frac{1}{2}$$ $$n \in \mathbb{Z}$$
SAQ-5 : Solve cos2θ + cos8θ = cos5θ.
Given equation is $$\cos2\theta + \cos8\theta = \cos5\theta$$
$$\Rightarrow (\cos2\theta + \cos8\theta) – \cos5\theta = 0$$
Using the sum-to-product formula for cosine, $$\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$ we have:
$$\Rightarrow 2\cos\left(\frac{2\theta+8\theta}{2}\right)\cos\left(\frac{2\theta-8\theta}{2}\right)-\cos5\theta = 0$$
$$\Rightarrow 2\cos\left(\frac{10\theta}{2}\right)\cos\left(\frac{-6\theta}{2}\right)-\cos5\theta = 0$$
$$\Rightarrow 2\cos5\theta\cos(-3\theta)-\cos5\theta = 0$$
$$\Rightarrow 2\cos5\theta\cos3\theta-\cos5\theta = 0$$
$$\Rightarrow \cos5\theta(2\cos3\theta-1) = 0$$
$$\Rightarrow \cos5\theta = 0$$ or $$2\cos3\theta-1 = 0$$
i) If $$\cos5\theta = 0$$ then $$5\theta = \frac{(2n+1)\pi}{2}$$
$$\Rightarrow \theta = \frac{(2n+1)\pi}{10}$$ $$n \in \mathbb{Z}$$
ii) If $$2\cos3\theta-1 = 0$$ then
$$2\cos3\theta = 1$$
$$\Rightarrow \cos3\theta = \frac{1}{2} = \cos\frac{\pi}{3}$$
Here, Principal Value (P.V) is $$\alpha=\frac{\pi}{3}$$
General Solution (G.S) is $$\theta = \frac{2n\pi \pm \alpha}{3}$$ $$n \in \mathbb{Z}$$
$$\Rightarrow 3\theta = 2n\pi \pm \frac{\pi}{3}$$
$$\Rightarrow \theta = \frac{2n\pi}{3} \pm \frac{\pi}{9}$$ $$n \in \mathbb{Z}$$
SAQ-6 : Solve sinθ + sin5θ = sin3θ, 0 < θ < π.
Given equation is $$\sin\theta + \sin5\theta = \sin3\theta$$
$$\Rightarrow \sin\theta + (\sin5\theta-\sin3\theta) = 0$$
Using the sum-to-product formula for sine, $$\sin A – \sin B = 2\cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$$ we have:
$$\Rightarrow \sin\theta + 2\cos\left(\frac{5\theta + 3\theta}{2}\right) \sin\left(\frac{5\theta – 3\theta}{2}\right) = 0$$
$$\Rightarrow \sin\theta + 2\cos\left(\frac{8\theta}{2}\right)\sin\left(\frac{2\theta}{2}\right) = 0$$
$$\Rightarrow \sin\theta + 2\cos4\theta\sin\theta = 0$$
$$\Rightarrow \sin\theta(1 + 2\cos4\theta) = 0$$
$$\Rightarrow \sin\theta = 0$$ (or) $$1 + 2\cos4\theta = 0$$
$$\Rightarrow 2\cos4\theta = -1$$
$$\Rightarrow \cos4\theta = -\frac{1}{2}$$
i) If $$\sin\theta = 0$$ then $$\theta = n\pi$$ $$n \in \mathbb{Z}$$
ii) $$\cos4\theta = -\frac{1}{2}$$ which is equal to $$\cos\frac{2\pi}{3}$$ or $$\cos\frac{4\pi}{3}$$ (considering the cosine function’s symmetry and periodicity).
Here, Principal Value (P.V) is $$\alpha = \frac{2\pi}{3}$$
General Solution (G.S) for $$\cos4\theta = -\frac{1}{2}$$
$$4\theta = 2n\pi \pm \frac{2\pi}{3}$$
$$\Rightarrow \theta = \frac{n\pi}{2} \pm \frac{\pi}{6}$$ $$n \in \mathbb{Z}$$
$$4\theta = 2n\pi \pm \alpha$$
SAQ-7 : If acos2θ + bsinθ = c has θ1,θ2 as its solutions then show that i).tanθ1 + tanθ2 = 2b/(c+a). ii).tanθ1.tanθ2 = (c-a)/(c+a) and hence show that tan(θ1 + θ2) = b/a.
Given equation is $$a\cos2\theta + b\sin2\theta = c$$
The given equation can be transformed using the identities for cos2θ and sin2θ, which are $$\cos2\theta = \frac{1 – \tan^2\theta}{1 + \tan^2\theta}$$ and $$\sin2\theta = \frac{2\tan\theta}{1 + \tan^2\theta}$$ respectively. So, substituting these identities, we have:
$$a\left(\frac{1 – \tan^2\theta}{1 + \tan^2\theta}\right) + b\left(\frac{2\tan\theta}{1 + \tan^2\theta}\right) = c$$
$$a(1 – \tan^2\theta) + 2b\tan\theta = c(1 + \tan^2\theta)$$
$$a – a\tan^2\theta + 2b\tan\theta = c + c\tan^2\theta$$
$$(c + a)\tan^2\theta – 2b\tan\theta + (c – a) = 0$$
$$A\tan^2\theta + B\tan\theta + C = 0$$
Where $$A = c + a$$ $$B = -2b$$ and $$C = c – a$$
The sum and product of the roots (tanθ1 and tanθ2) of this quadratic equation are given by:
$$\tan\theta_1 + \tan\theta_2 = -\frac{B}{A} = \frac{2b}{c + a}$$
$$\tan\theta_1 \cdot \tan\theta_2 = \frac{C}{A} = \frac{c – a}{c + a}$$
$$\tan(\theta_1 + \theta_2) = \frac{\tan\theta_1 + \tan\theta_2}{1 – \tan\theta_1\tan\theta_2}$$
SAQ-8 : If α,β are the solutions of equation acosθ+bsinθ=c, where a,b,c ∈ R then show that i).Sinα + sinβ = 2bc/(a2+b2) ii).sinα.sinβ=(c2-a2)/(a2+b2).
Given Equation: $$a\cos \theta + b\sin \theta = c$$
$$a\cos \theta = c – b\sin \theta$$
Squaring Both Sides $$a^2\cos^2 \theta = (c – b\sin \theta)^2$$
$$a^2(1 – \sin^2 \theta) = c^2 + b^2\sin^2 \theta – 2bc \sin \theta$$
$$a^2\sin^2 \theta + b^2\sin^2 \theta – 2bc\sin \theta + c^2 – a^2 = 0$$
$$(a^2 + b^2)\sin^2 \theta – 2bc\sin \theta + (c^2 – a^2) = 0$$
Therefore, for our quadratic equation:
(i) Sum of Roots:
$$\sin \alpha + \sin \beta = \frac{-(-2bc)}{a^2 + b^2} = \frac{2bc}{a^2 + b^2}$$
(ii) Product of Roots:
$$\sin \alpha \cdot \sin \beta = \frac{c^2 – a^2}{a^2 + b^2}$$