Trigonometric Ratios Upto Transformations (LAQs)

Maths-1A | 6. Trigonometric Ratios Upto Transformations – LAQs:
Welcome to LAQs in Chapter 6: Trigonometric Ratios Upto Transformations. This page contains the most Important FAQs for Long 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.

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LAQ-1 : If A + B + C = 180°, then show that sin2A + sin2B + sin2C = 4sinAsinBsinC

L.H.S

$$\sin^2A + \sin^2B + \sin^2C$$

$$= 2\sin\left(\frac{2A + 2B}{2}\right)\cos\left(\frac{2A – 2B}{2}\right) + \sin 2C$$

$$\text{[Using the identity } \sin C + \sin D = 2\sin\left(\frac{C + D}{2}\right) \cos\left(\frac{C – D}{2}\right) \text{]}$$

$$= 2\sin(A+B)\cos(A-B)+2\sin C\cos C$$

$$= 2\sin(180^\circ – C) \cos(A-B) + 2\sin C\cos C$$

$$\text{Since } (A+B)+C = 180^\circ$$

$$= 2\sin C \cos(A-B)+2\sin C\cos C$$

$$= 2\sin C[\cos(A-B)+\cos C]$$

$$= 2\sin C[\cos(A-B)+\cos(180^\circ-(A+B))]$$

$$= 2\sin C[\cos(A-B)-\cos(A+B)]$$

$$\text{Using the identity } \cos(A-B)-\cos(A+B) = -2\sin A\sin B$$

$$= 2\sin C (-2\sin A \sin B)$$

$$= 4\sin A \sin B \sin C = R.H.S$$

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LAQ-2 : If A,B,C are angles of a triangle, P.T cos2A + cos2B + cos2C = -1 -4cosA cosB cosC

L.H.S

$$\cos 2A + \cos 2B + \cos 2C$$

$$= 2\cos\left(\frac{2A + 2B}{2}\right)\cos\left(\frac{2A – 2B}{2}\right) + \cos 2C$$

$$\text{[Using the identity } \cos C + \cos D = 2\cos\left(\frac{C + D}{2}\right) \cos\left(\frac{C – D}{2}\right) \text{]}$$

$$= 2\cos(A+B)\cos(A-B)+(2\cos^2C-1)$$

$$= 2\cos(180^\circ-C)\cos(A-B)+2\cos^2C-1$$

$$= -2 \cos C \cos(A-B)+2\cos^2C-1$$

$$= -2 \cos C \left[\cos (A-B)-\cos C\right] – 1$$

$$= -2 \cos C\left[\cos (A-B)-\cos(180^\circ-(A+B))\right] – 1$$

$$= -2 \cos C \left[\cos (A-B)+\cos (A+B)\right] – 1$$

$$\text{Using the identity } \cos(A+B)+\cos(A-B)=2\cos A\cos B$$

$$= -2\cos C \left[2\cos A\cos B\right] – 1$$

$$= -4\cos A\cos B\cos C – 1 = R.H.S$$

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LAQ-3 : If A,B,C are angles of a triangle, then P.T sin²  A/2 + sin²  B/2 + sin²  C/2 = 1.2sin A/2 sin B/2 sin C/2

L.H.S

$$\sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2}$$

$$= \left(1 – \cos^2 \frac{A}{2}\right) + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2}$$

$$= 1 – \left(\cos^2 \frac{A}{2} – \sin^2 \frac{B}{2}\right) + \sin^2 \frac{C}{2}$$

$$= 1 – \left(\cos \frac{A+B}{2} \cdot \cos \frac{A-B}{2}\right) + \sin^2 \frac{C}{2}$$

$$\text{[Using the identity } \cos^2 A – \sin^2 B = \cos(A+B)\cos(A-B) \text{]}$$

$$= \left(1-\cos\left(180^\circ – \frac{C}{2}\right) \cdot \cos \frac{A-B}{2}\right) + \sin^2 \frac{C}{2}$$

$$= \left(1-\cos\left(90^\circ – \frac{C}{2}\right) \cdot \cos \frac{A-B}{2}\right) + \sin^2 \frac{C}{2}$$

$$= 1 – \sin \frac{C}{2} \cdot \cos \frac{A-B}{2} + \sin^2 \frac{C}{2}$$

$$= 1 – \sin \frac{C}{2}\left(\cos \frac{A-B}{2} – \sin \frac{C}{2}\right)$$

$$= 1 – \sin \frac{C}{2}\left(\cos \frac{A-B}{2} – \sin\left(180^\circ – \left(\frac{A+B}{2}\right)\right)\right)$$

$$= 1 – \sin \frac{C}{2}\left(\cos \frac{A-B}{2} – \sin\left(90^\circ – \frac{A+B}{2}\right)\right)$$

$$= 1 – \sin \frac{C}{2}\left(\cos \frac{A-B}{2} – \cos \frac{A+B}{2}\right)$$

$$\text{[Using the identity } \cos(A-B) – \cos(A+B) = -2\sin A \sin B \text{]}$$

$$= 1 – \sin \frac{C}{2} \cdot \left(-2\sin \frac{A}{2} \sin \frac{B}{2}\right)$$

$$= 1 – 2\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = R.H.S$$

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LAQ-4 : If A,B,C are angles of a triangle, then P.T sin²  A/2 + sin²  B/2 + sin²  C/2 = 1 – 2cos A/2 cos B/2 sin C/2

L.H.S

$$\sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} – \sin^2 \frac{C}{2}$$

$$= \sin^2 \frac{A}{2} + \left(1 – \cos^2 \frac{B}{2}\right) – \sin^2 \frac{C}{2}$$

$$= 1 + \left(\sin^2 \frac{A}{2} – \cos^2 \frac{B}{2}\right) – \sin^2 \frac{C}{2}$$

$$= 1 + \left(-\cos \frac{A+B}{2} \cdot \cos \frac{A-B}{2}\right) – \sin^2 \frac{C}{2}$$

$$\text{[Using the identity } \sin^2 A – \cos^2 B = -\cos(A+B)\cos(A-B) \text{]}$$

$$= \left(1 – \cos\left(180^\circ – \frac{C}{2}\right) \cdot \cos \frac{A-B}{2}\right) – \sin^2 \frac{C}{2}$$

$$= \left(1 – \cos\left(90^\circ – \frac{C}{2}\right) \cdot \cos \frac{A-B}{2}\right) – \sin^2 \frac{C}{2}$$

$$= 1 – \sin \frac{C}{2} \cdot \cos \frac{A-B}{2} – \sin^2 \frac{C}{2}$$

$$= 1 – \sin \frac{C}{2}\left(\cos \frac{A-B}{2} + \sin \frac{C}{2}\right)$$

$$= 1 – \sin \frac{C}{2}\left(\cos \frac{A-B}{2} + \sin\left(180^\circ – \left(\frac{A+B}{2}\right)\right)\right)$$

$$= 1 – \sin \frac{C}{2}\left(\cos \frac{A-B}{2} + \sin\left(90^\circ – \frac{A+B}{2}\right)\right)$$

$$= 1 – \sin \frac{C}{2}\left(\cos \frac{A-B}{2} + \cos \frac{A+B}{2}\right)$$

$$\text{[Using the identity } \cos(A+B) + \cos(A-B) = 2\cos A \cos B \text{]}$$

$$= 1 – \sin \frac{C}{2} \cdot \left(2\cos \frac{A}{2} \cos \frac{B}{2}\right)$$

$$= 1 – 2\cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} = R.H.S$$

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LAQ-5 : If A + B + C = π, then prove that cos²  A/2 + cos²  B/2 + cos²  C/2 = 2(1 + sin A/2 sin B/2 sin C/2)

Given $$A + B + C = \pi$$

$$\Rightarrow A + B + \frac{C}{2} = 90^\circ$$

$$\Rightarrow \frac{A + B}{2} = 90^\circ – \frac{C}{2}$$

L.H.S = $$\cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2} = \cos^2 \frac{A}{2} + \left(1 – \sin^2 \frac{B}{2}\right) + \cos^2 \frac{C}{2}$$

$$= 1 + \left(\cos^2 \frac{A}{2} – \sin^2 \frac{B}{2}\right) + \cos^2 \frac{C}{2}$$

$$= 1 + \left(\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\right) + \left(1 – \sin^2 \frac{C}{2}\right)$$

$$\text{[Using the identity } \cos^2 A – \sin^2 B = \cos(A+B)\cos(A-B) \text{]}$$

$$= 2 + \sin \frac{C}{2} \cos\left(\frac{A-B}{2}\right) – \sin^2 \frac{C}{2}$$

$$\text{Since } \cos\left(\frac{A+B}{2}\right) = \cos\left(90^\circ – \frac{C}{2}\right) = \sin \frac{C}{2}$$

$$= 2 + \sin \frac{C}{2}\left(\cos\left(\frac{A-B}{2}\right) – \sin \frac{C}{2}\right)$$

$$= 2 + \sin \frac{C}{2}\left(\cos\left(\frac{A-B}{2}\right) – \cos\left(\frac{A+B}{2}\right)\right)$$

$$= 2 + \sin \frac{C}{2}\left(2 \sin \frac{A}{2}\sin \frac{B}{2}\right)$$

$$= 2 + 2\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 2\left(1 + \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right) = R.H.S$$

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LAQ-6 : If A,B,C are angles in a triangle, then prove that cosA+cosB-cosC = -1+4cosA/2.cos B/2.sin C/2

Given A,B,C are angles of a triangle, then

$$A + B + C = 180^\circ$$

$$\Rightarrow A + B + \frac{C}{2} = 90^\circ$$

$$\Rightarrow \frac{A + B}{2} = 90^\circ – \frac{C}{2}$$

L.H.S = $$(\cos A + \cos B) – \cos C = 2\cos \left(\frac{A + B}{2}\right)\cos \left(\frac{A – B}{2}\right) – \cos C$$

$$= 2\cos\left(90^\circ – \frac{C}{2}\right)\cos\left(\frac{A – B}{2}\right) – \cos C = 2 \sin \frac{C}{2}\cos\left(\frac{A – B}{2}\right) – \left(1 – 2\sin^2 \frac{C}{2}\right)$$

$$= -1 + 2\sin \frac{C}{2}\cos\left(\frac{A – B}{2}\right) + 2\sin^2 \frac{C}{2} = -1 + 2\sin \frac{C}{2}\left(\cos\left(\frac{A – B}{2}\right) + \sin \frac{C}{2}\right)$$

$$= -1 + 2\sin \frac{C}{2}\left(\cos\left(\frac{A – B}{2}\right) + \sin\left(90^\circ – \left(\frac{A – B}{2}\right)\right)\right) = -1 + 2\sin \frac{C}{2}\left(\cos\left(\frac{A – B}{2}\right) + \cos\left(\frac{A + B}{2}\right)\right)$$

$$= -1 + 2\sin \frac{C}{2}\left(2\cos\left(\frac{A}{2}\right) \cos\left(\frac{B}{2}\right)\right)$$

$$\text{[Using the identity } \cos(A + B) + \cos(A – B) = 2\cos A \cos B \text{]}$$

$$= -1 + 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right) = R.H.S$$

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LAQ-7 : If A + B + C = 2S, then prove that sin(S-A) + sin(S-B) + sinC = 4cos((S-A)/2)cos((S-B)/2)sin C/2

L.H.S = $$\sin(S – A) + \sin(S – B) + \sin C$$

$$= 2\sin\left[\frac{(S-A)+(S-B)}{2}\right]\cos\left[\frac{(S-A)-(S-B)}{2}\right]+\sin C$$

$$= 2\sin\left[\frac{2S-(A+B)}{2}\right]\cos\left[\frac{B-A}{2}\right]+\sin C$$

$$= 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{B-A}{2}\right)+2\sin\left(\frac{C}{2}\right) \cos\left(\frac{C}{2}\right)$$

$$= 2\sin\left(\frac{C}{2}\right)\left[\cos\left(\frac{B-A}{2}\right)+\cos\left(\frac{C}{2}\right)\right]$$

$$= 2\sin\left(\frac{C}{2}\right)\left[2\cos\left(\frac{1}{2}[(B-A)+C]\right)\cos\left(\frac{1}{2}[(B-A)-C]\right)\right]$$

$$\text{Using the identity } \cos C + \cos D = 2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$$

$$= 2\sin\left(\frac{C}{2}\right)\left[2\cos\left(\frac{(B+C)-A}{2}\right)\cos\left(\frac{B-(A+C)}{2}\right)\right]$$

$$= 2\sin\left(\frac{C}{2}\right)\left[2\cos\left(\frac{2S-A}{2}\right)\cos\left(\frac{B-2S+B}{2}\right)\right]$$

$$= 2\sin\left(\frac{C}{2}\right)\left[2\cos\left(\frac{S-A}{2}\right)\cos\left(\frac{B-S}{2}\right)\right]$$

$$= 4\cos\left(\frac{S-A}{2}\right)\cos\left(\frac{S-B}{2}\right)\sin\left(\frac{C}{2}\right) = R.H.S$$

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LAQ-8 : If A + B + C = 2S, then prove that cos(S-A) + cos(S-B) + cosC = -1 + 4cos(S-A/2) cos((S-B)/2) cos C/2

L.H.S = $$\cos(S-A) + \cos(S-B) + \cos C$$

$$= 2\cos\left(\frac{(S-A) + (S-B)}{2}\right)\cos\left(\frac{(S-A) – (S-B)}{2}\right) + \cos C$$

$$= 2\cos\left(\frac{2S – (A + B)}{2}\right)\cos\left(\frac{B – A}{2}\right) + 2\cos^2\left(\frac{C}{2}\right) – 1$$

$$= -1 + 2\cos\left(\frac{C}{2}\right)\cos\left(\frac{B – A}{2}\right) + 2\cos^2\left(\frac{C}{2}\right)$$

$$= -1 + 2\cos\left(\frac{C}{2}\right)\left[\cos\left(\frac{B – A}{2}\right) + \cos\left(\frac{C}{2}\right)\right]$$

$$= -1 + 2\cos\left(\frac{C}{2}\right)\left[2\cos\left(\frac{1}{2}[(B – A) + C]\right)\cos\left(\frac{1}{2}[(B – A) – C]\right)\right]$$

$$\text{Using the identity } \cos C + \cos D = 2\cos\left(\frac{C + D}{2}\right)\cos\left(\frac{C – D}{2}\right)$$

$$= -1 + 2\cos\left(\frac{C}{2}\right)\left[2\cos\left(\frac{(B + C) – A}{2}\right)\cos\left(\frac{B – (A + C)}{2}\right)\right]$$

$$= -1 + 2\cos\left(\frac{C}{2}\right)\left[2\cos\left(\frac{2S – A}{2}\right)\cos\left(\frac{B – 2S + B}{2}\right)\right]$$

$$= -1 + 2\cos\left(\frac{C}{2}\right)\left[2\cos\left(\frac{S – A}{2}\right)\cos\left(\frac{B – S}{2}\right)\right]$$

$$= -1 + 4\cos\left(\frac{C}{2}\right)\cos\left(\frac{S – A}{2}\right)\cos\left(\frac{S – B}{2}\right)$$

$$= -1 + 4\cos\left(\frac{S – A}{2}\right)\cos\left(\frac{S – B}{2}\right)\cos\left(\frac{C}{2}\right) = R.H.S$$