4 Most SAQ’s of Trigonometric Equations Chapter in Inter 1st Year Maths-1A (TS/AP)

4 Marks

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:

$$\theta = n\pi + (-1)^n \alpha, \ n \in \mathbb{Z}$$

$$\theta – \frac{\pi}{6} = n\pi + (-1)^n \frac{\pi}{4}, \ n \in \mathbb{Z}$$

$$\theta = n\pi + (-1)^n \frac{\pi}{4} + \frac{\pi}{6}, \ n \in \mathbb{Z}$$

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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 $$\alpha = \frac{\pi}{6}$$

$$x = 2n\pi \pm \alpha, \ n \in \mathbb{Z}$$

$$x – \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{6}$$

$$x = 2n\pi \pm \frac{\pi}{6} + \frac{\pi}{4}, \ n \in \mathbb{Z}$$

$$x = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{6}, \ n \in \mathbb{Z}$$

$$x = 2n\pi \pm \frac{\pi}{6} + \frac{\pi}{4}, \ n \in \mathbb{Z}$$

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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 $$\sin\theta = -\sqrt{3}$$

$$\sin\theta = \frac{\sqrt{3}}{2}$$

$$\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$$

$$\theta = n\pi + (-1)^n\alpha, n \in \mathbb{Z}$$

$$\theta = n\pi + (-1)^n\frac{\pi}{3}, n \in \mathbb{Z}$$

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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 $$\theta = n\pi + \tan^{-1} \frac{1}{2}$$ $$n \in \mathbb{Z}$$

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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 $$\theta = \frac{2n\pi \pm \alpha}{3}$$ $$n \in \mathbb{Z}$$

$$\Rightarrow 3\theta = 2n\pi \pm \frac{\pi}{3}$$

$$\Rightarrow \theta = \frac{2n\pi}{3} \pm \frac{\pi}{9}$$ $$n \in \mathbb{Z}$$

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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 $$\cos4\theta = -\frac{1}{2}$$

$$4\theta = 2n\pi \pm \frac{2\pi}{3}$$

$$\Rightarrow \theta = \frac{n\pi}{2} \pm \frac{\pi}{6}$$ $$n \in \mathbb{Z}$$

$$4\theta = 2n\pi \pm \alpha$$

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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:

$$\tan\theta_1 + \tan\theta_2 = -\frac{B}{A} = \frac{2b}{c + a}$$

$$\tan\theta_1 \cdot \tan\theta_2 = \frac{C}{A} = \frac{c – a}{c + a}$$

$$\tan(\theta_1 + \theta_2) = \frac{\tan\theta_1 + \tan\theta_2}{1 – \tan\theta_1\tan\theta_2}$$

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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:

$$\sin \alpha \cdot \sin \beta = \frac{c^2 – a^2}{a^2 + b^2}$$