10 Most SAQ’s of Trigonometric Ratios Upto Transformations Chapter in Inter 1st Year Maths-1A (TS/AP)
4 Marks
SAQ-1 : Prove that tan A + cot A = 2cosec2A
L.H.S = $$\tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}$$
$$= \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A}$$
$$= \frac{2}{2 \sin A \cos A} = \frac{2}{\sin 2A}$$
$$= 2\csc 2A = R.H.S$$
SAQ-2 : Prove that cot A – tan A = 2cot 2A
L.H.S = $$\cot A – \tan A = \frac{\cos A}{\sin A} – \frac{\sin A}{\cos A}$$
$$= \frac{\cos^2 A – \sin^2 A}{\sin A \cos A} = \frac{\cos 2A}{\sin A \cos A}$$
$$= \frac{2\cos 2A}{2 \sin A \cos A} = \frac{2 \cos 2A}{\sin 2A}$$
$$= 2\cot 2A = R.H.S$$
SAQ-3 : Show that 1/(sin10°) – √3/(cos10°) = 4
L.H.S = $$\frac{1}{\sin 10^\circ} – \frac{\sqrt{3}}{\cos 10^\circ}$$
$$= \frac{\cos 10^\circ – \sqrt{3} \sin 10^\circ}{\sin 10^\circ \cos 10^\circ}$$
$$= \frac{2\left(\frac{1}{2} \cos 10^\circ – \frac{\sqrt{3}}{2} \sin 10^\circ\right)}{\sin 10^\circ \cos 10^\circ}$$
$$= \frac{2\left(\sin 30^\circ \cos 10^\circ – \cos 30^\circ \sin 10^\circ\right)}{\sin 10^\circ \cos 10^\circ}$$
$$= \frac{2\sin(30^\circ – 10^\circ)}{\sin 10^\circ \cos 10^\circ} = \frac{2\sin 20^\circ}{\sin 10^\circ \cos 10^\circ}$$
$$= 2 \cdot 2 \frac{\sin 10^\circ \cos 10^\circ}{\sin 10^\circ \cos 10^\circ}$$
$$= 4 = R.H.S$$
SAQ-4 : Show that √3 cos20°- sec20° = 4
L.H.S = $$\sqrt{3} \csc 20^\circ – \sec 20^\circ$$
$$= \frac{\sqrt{3}}{\sin 20^\circ} – \frac{1}{\cos 20^\circ}$$
$$= \frac{\sqrt{3} \cos 20^\circ – \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}$$
$$= \frac{2\left(\frac{\sqrt{3}}{2} \cos 20^\circ – \frac{1}{2} \sin 20^\circ\right)}{\sin 20^\circ \cos 20^\circ}$$
$$= \frac{2\left(\sin 60^\circ \cos 20^\circ – \cos 60^\circ \sin 20^\circ\right)}{\sin 20^\circ \cos 20^\circ}$$
$$= \frac{2 \sin(60^\circ – 20^\circ)}{\sin 20^\circ \cos 20^\circ} = \frac{2 \sin 40^\circ}{\sin 20^\circ \cos 20^\circ}$$
$$= 4 = R.H.S$$
SAQ-5 : If tanθ = b/a then prove that acos2θ + bsin2θ = a
L.H.S = $$a\cos 2\theta + b\sin 2\theta$$
Starting from the given, we use the double angle identities for cosine and sine:
$$= a \cos 2\theta + b(2\sin \theta \cos \theta)$$
Substituting bcosθ=asinθ into the equation:
$$= a\cos 2\theta + 2\sin \theta(b \cos \theta)$$
$$= a\cos 2\theta + 2\sin \theta(a \sin \theta)$$
Using the double angle identity for cosine, $$\cos 2\theta = 1 – 2\sin^2 \theta$$
$$= a(1 – 2\sin^2 \theta) + 2a\sin^2 \theta$$
Simplifying the expression:
$$= a – 2a\sin^2 \theta + 2a\sin^2 \theta$$
$$= a$$
SAQ-6 : Show that sin4 π/8 + sin4 3π/8 + sin4 5π/8 + sin4 7π/8 = 3/2
Given the trigonometric identities and the angles, we have:
$$\sin \frac{3\pi}{8} = \sin\left(4\pi – \frac{\pi}{8}\right) = \sin\left(\frac{\pi}{2} – \frac{\pi}{8}\right) = \cos \frac{\pi}{8}$$
$$\sin \frac{5\pi}{8} = \sin\left(4\pi + \frac{\pi}{8}\right) = \sin\left(\frac{\pi}{2} + \frac{\pi}{8}\right) = \cos \frac{\pi}{8}$$
$$\sin \frac{7\pi}{8} = \sin\left(8\pi – \frac{\pi}{8}\right) = \sin\left(\pi – \frac{\pi}{8}\right) = \sin \frac{\pi}{8}$$
L.H.S:
$$= \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8}$$
$$= \sin^4 \frac{\pi}{8} + \left(\cos \frac{\pi}{8}\right)^4 + \left(\cos \frac{\pi}{8}\right)^4 + \left(\sin \frac{\pi}{8}\right)^4$$
$$= \sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8}$$
$$= 2\left[\sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8}\right]$$
Using the identity $$[a^2 + b^2 = (a + b)^2 – 2ab]$$
$$= 2\left[\left(\sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8}\right)^2 – 2\sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}\right]$$
$$= 2\left[1 – 2\sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}\right]$$
$$= 2 – 4 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}$$
$$= 2 – \left[2\sin \frac{\pi}{8} \cos \frac{\pi}{8}\right]^2 = 2 – \left[\sin \frac{2\pi}{8}\right]^2$$
$$= 2 – \left[\sin \frac{\pi}{4}\right]^2 = 2 – \left(\frac{1}{\sqrt{2}}\right)^2$$
$$= 2 – \frac{1}{2} = \frac{3}{2} = R.H.S$$
SAQ-7 : Show that cos4 π/8 + cos4 3π/8 + cos4 5π/8 + cos4 7π/8 = 3/2
Given the trigonometric identities for angles:
$$\cos \frac{3\pi}{8} = \cos\left(4\pi – \frac{\pi}{8}\right) = \cos\left(\frac{\pi}{2} – \frac{\pi}{8}\right) = \sin \frac{\pi}{8}$$
$$\cos \frac{5\pi}{8} = \cos\left(4\pi + \frac{\pi}{8}\right) = \cos\left(\frac{\pi}{2} + \frac{\pi}{8}\right) = -\sin \frac{\pi}{8}$$
$$\cos \frac{7\pi}{8} = \cos\left(8\pi – \frac{\pi}{8}\right) = \cos\left(\pi – \frac{\pi}{8}\right) = -\cos \frac{\pi}{8}$$
L.H.S:
$$= \cos^4 \frac{\pi}{8} + \cos^4 \frac{3\pi}{8} + \cos^4 \frac{5\pi}{8} + \cos^4 \frac{7\pi}{8}$$
$$= \cos^4 \frac{\pi}{8} + \left(\sin \frac{\pi}{8}\right)^4 + \left(-\sin \frac{\pi}{8}\right)^4 + \left(-\cos \frac{\pi}{8}\right)^4$$
$$= \cos^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8} + \sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8}$$
$$= 2\left[\sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8}\right]$$
Using the identity for sum of squares and their relation to square of sums:
$$= 2\left[\left(\sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8}\right)^2 – 2\sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}\right]$$
Since $$\sin^2 \theta + \cos^2 \theta = 1$$
$$= 2\left[1 – 2\sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}\right]$$
$$= 2 – 4\sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}$$
Using the double-angle identity for sine:
$$= 2 – \left[2\sin \frac{\pi}{8} \cos \frac{\pi}{8}\right]^2 = 2 – \left[\sin \frac{2\pi}{8}\right]^2$$
$$= 2 – \left[\sin \frac{\pi}{4}\right]^2 = 2 – \left(\frac{1}{\sqrt{2}}\right)^2$$
$$= 2 – \frac{1}{2} = \frac{3}{2} = R.H.S$$
SAQ-8 : Show that (1+ cos π/8)(1+ cos 3π/8)(1+ cos 5π/8)(1+ cos 7π/8) = 1/8
Given the trigonometric properties:
$$\cos \frac{5\pi}{8} = \cos\left(8\pi – \frac{3\pi}{8}\right) = \cos\left(\pi – \frac{3\pi}{8}\right) = -\cos \frac{3\pi}{8}$$
$$\cos \frac{7\pi}{8} = \cos\left(8\pi – \frac{\pi}{8}\right) = \cos\left(\pi – \frac{\pi}{8}\right) = -\cos \frac{\pi}{8}$$
L.H.S:
$$= (1 + \cos \frac{\pi}{8})(1 + \cos \frac{3\pi}{8})(1 + \cos \frac{5\pi}{8})(1 + \cos \frac{7\pi}{8})$$
$$= (1 + \cos \frac{\pi}{8})(1 + \cos \frac{3\pi}{8})(1 – \cos \frac{3\pi}{8})(1 – \cos \frac{\pi}{8})$$
Using the identity $$\cos^2 \theta + \sin^2 \theta = 1$$ and $$\cos 2\theta = 2\cos^2 \theta – 1$$ we simplify:
$$= \left[(1 – \cos^2 \frac{\pi}{8})\right]\left[(1 – \cos^2 \frac{3\pi}{8})\right] = \sin^2 \frac{\pi}{8} \sin^2 \frac{3\pi}{8}$$
Given $$\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$$
$$= \sin^2 \frac{\pi}{8}\left[\cos^2 \frac{\pi}{8}\right]$$
Using the double-angle formula $$\sin 2\theta = 2\sin \theta \cos \theta$$
$$= \frac{1}{4}\left[4 \sin^2 \frac{\pi}{8} \cos^2 \frac{\pi}{8}\right] = \frac{1}{4}\left[2 \sin \frac{\pi}{8} \cos \frac{\pi}{8}\right]^2 = \frac{1}{4}\left[\sin \frac{2\pi}{8}\right]^2$$
$$= \frac{1}{4}\left[\sin \frac{\pi}{4}\right]^2 = \frac{1}{4}\left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$$
$$1/8 = R.H.S$$
SAQ-9 : Prove that (1+ cos π/10)(1+ cos 3π/10)(1+ cos 7π/10)(1+ cos 9π/10) = 1/16
Given:
$$\cos \frac{\pi}{10} = \cos 18^\circ$$
$$\cos \frac{3\pi}{10} = \cos 54^\circ$$
$$\cos \frac{7\pi}{10} = \cos 126^\circ = \cos(180^\circ – 54^\circ) = -\cos 54^\circ$$
$$\cos \frac{9\pi}{10} = \cos 162^\circ = \cos(180^\circ – 18^\circ) = -\cos 18^\circ$$
L.H.S :
$$= (1 + \cos 18^\circ)(1 + \cos 54^\circ)(1 – \cos 54^\circ)(1 – \cos 18^\circ)$$
This simplifies to:
$$= (1 – \cos^2 18^\circ)(1 – \cos^2 54^\circ)$$
Using the Pythagorean identity $$\sin^2 \theta + \cos^2 \theta = 1$$
$$= \sin^2 18^\circ \sin^2 54^\circ$$
$$= (\sqrt{5} – 1)/4)^2 ((\sqrt{5} + 1)/4)^2$$
$$1/16$$
$$R.H.S$$
SAQ-10 : Prove that cos A cos 2A cos 4A cos 8A = sin16A/16sinA
Given:
$$\text{L.H.S} = \cos A \cos 2A \cos 4A \cos 8A$$
$$\text{L.H.S} = \frac{2^4 \sin A}{2^4 \sin A} (\cos A)(\cos 2A)(\cos 4A)(\cos 8A)$$
$$= \frac{1}{16} \sin A (2 \sin A \cos A)(2 \cos 2A)(2 \cos 4A)(2 \cos 8A)$$
$$2 \sin A \cos A = \sin 2A$$
$$2 \sin 2A \cos 2A = \sin 4A$$
$$2 \sin 4A \cos 4A = \sin 8A$$
$$2 \sin 8A \cos 8A = \sin 16A$$
So, we have:
$$= \frac{1}{16} \sin A (2 \sin 2A \cos 2A)(2 \cos 4A)(2 \cos 8A)$$
$$= \frac{1}{16} \sin A (2 \sin 4A \cos 4A)(2 \cos 8A)$$
$$= \frac{1}{16} \sin A (2 \sin 8A \cos 8A)$$
$$= \frac{1}{16} \sin A (\sin 16A)$$
$$= \frac{\sin 16A}{16 \sin A} = \text{R.H.S}$$