15. Trigonometrical Identities Solutions¶

1. $\because A + B + C = \pi \therefore A + B = \pi - C$

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

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

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

$\Rightarrow \sin^A + \sin^2B + \cos^2C - 1 = 2\sin A\sin B\cos C$

$\Rightarrow \sin^2A + \sin^2B - \cos^2C = 2\sin A\sin B\cos C$

2. $A + B + C = 180^\circ \Rightarrow \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ \Rightarrow \frac{A}{2} + \frac{B}{2} = 90^\circ - \frac{C}{2}$

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

$\Rightarrow \cos\frac{A}{2}\cos\frac{B}{2} - \sin\frac{A}{2}\sin\frac{B}{2} = \sin\frac{C}{2}$

$\Rightarrow \sin\frac{C}{2} + \sin\frac{A}{2}\sin\frac{B}{2} = \cos\frac{A}{2}\cos\frac{B}{2}$

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

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

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

3. Let $A + B = C \Rightarrow \cos(A + B) = \cos C$

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

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

$\Rightarrow \sin^2A + \sin^2B + 2\sin A\sin B\cos C = \sin^2C$

$\Rightarrow \sin^2A + \sin^2B + 2\sin A\sin B\cos(A + B) = \sin^2(A + B)$

4. Given, $A + B + C = 180^\circ \Rightarrow A + B = 180^\circ - C \Rightarrow \cos(A + B) = -\cos C$

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

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

$\Rightarrow \cos^2A + \cos^2B + \cos^2C + 2\cos A\cos B\cos C = 1$

5. We have just proved that $\cos^2A + \cos^2B + \cos^2C + 2\cos A\cos B\cos C = 1$

$\Rightarrow 3 - \sin^2A - \sin^2B - \sin^2C + 2\cos A\cos B\cos C = 1$

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

6. Given, $A + B + C = 180^\circ \Rightarrow A + B = 180^\circ - C \Rightarrow \cos(A + B) = -\cos C$

$\Rightarrow \cos A\cos B = \sin A\sin B - \cos C \Rightarrow \cos^2A\cos^2B = \sin^2A\sin^2B + \cos^2C - 2\sin A\sin B\cos C$

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

$\Rightarrow \cos^2A + \cos^2B - \cos^2C = 1 - 2\sin A\sin B\cos C$

7. Given, $A + B + C = 180^\circ \Rightarrow \frac{A + B + C}{2} = 90^\circ$

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

$\Rightarrow \cos\frac{A}{2}\cos\frac{B}{2} - \sin\frac{C}{2} = \sin\frac{A}{2}\sin\frac{B}{2}$

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

$\Rightarrow \cos^2\frac{A}{2} + \cos^2\frac{B}{2} - \cos^2\frac{C}{2} = 2\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}$

8. Given, $A + B + C = 180^\circ \Rightarrow \frac{A + B + C}{2} = 90^\circ$

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

$\Rightarrow \cos\frac{A}{2}\cos\frac{B}{2} = \sin\frac{A}{2}\sin\frac{B}{2} + \sin\frac{C}{2}$

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

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

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

9. Given, $A + B + C = \frac{\pi}{2} \Rightarrow A + B = \frac{\pi}{2} - C \Rightarrow \cos(A + B) = \sin C$

$\Rightarrow \cos A\cos B = \sin A\sin B + \sin C$

$\Rightarrow \cos^2A\cos^2B = \sin^2A\sin^2B + \sin^2C + 2\sin A\sin B\sin C$

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

$\Rightarrow \sin^2A + \sin^2B + \sin^2C = 1 - 2\sin A\sin B\sin C$

10. We have just proven that $\sin^2A + \sin^2B + \sin^2C = 1 - 2\sin A\sin B\sin C$ in previous problem.

$\Rightarrow 1 - \cos^2A + 1 - \cos^2B + 1 - \cos^2C = 1 - 2\sin A\sin B\sin C$

$\Rightarrow \cos^2A + \cos^2B + \cos^2C = 2 + 2\sin A\sin B\sin C$

11. Givem $A + B + C = 2\pi \Rightarrow A + B = 2\pi - C \Rightarrow \cos(A + B) = \cos C$

$\Rightarrow \cos A\cos B - \cos C = \sin A\sin B$

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

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

12. Given $A + B = C \Rightarrow \cos(A + B) = \cos C$

$\Rightarrow \cos A\cos B - \cos C = \sin A\sin B$

$\Rightarrow \cos^2A\cos^2B + \cos^2C - 2\cos A\cos B\cos C = \sin^2A\sin^2B$

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

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

13. Given $A + B = \frac{\pi}{3} \Rightarrow \cos(A + B) = \cos\frac{\pi}{3} = \frac{1}{2}$

$\Rightarrow \cos A\cos B - \frac{1}{2} = \sin A\sin B$

$\Rightarrow \cos^2A\cos^2B - \cos A\cos B + \frac{1}{4} = \sin^2A\sin^2B = (1 - \cos^2A)(1 - \cos^2B)$

$\Rightarrow \cos^2A + \cos^2B - \cos A\cos B = \frac{3}{4}$

14. From problem 12 we have $A + B = C$ and $\cos^2A + \cos^2B + \cos^2C - 2\cos A\cos B\cos C = 1$

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

$\Rightarrow \cos^2B + \cos^2(A + B) - 2\cos A\cos B\cos(A + B) = 1 - \cos^2A = \sin^2A$ which is independent of $B$

15. Given $A + B + C = \pi$ and $A + B = 2C \Rightarrow C = \frac{\pi}{3} \Rightarrow A + B = \pi - \frac{pi}{3}$

$\cos(A + B) = -\cos\frac{\pi}{3}\Rightarrow \cos A\cos B = \sin A\sin B - \frac{1}{2}$

$\Rightarrow \cos^2A\cos^2B = \sin^2A\sin^2B - \sin A\sin B + \frac{1}{4}$

$\Rightarrow (1 - \sin^2A)(1 - \sin^2B) = \sin^2A\sin^2B - \sin A\sin B + \frac{1}{4}$

$\Rightarrow 4(\sin^2A + \sin^2B - \sin A\sin B) = 3$

16. Given $A + B + C = 2\pi \Rightarrow \cos(B + C) = \cos(2\pi - A) = \cos A$

$\Rightarrow \cos B\cos C - \cos A = \sin B\sin C$

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

$\Rightarrow \cos^2B + \cos^2C - \sin^2A - 2\cos A\cos B\cos C = 0$

17. Given $A + B + C = 0 \Rightarrow \cos(A + B) = \cos C$

$\Rightarrow \cos A\cos B - \cos C = \sin A\sin B$

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

$\Rightarrow \cos^2A + \cos^2B + \cos^2C = 1 + 2\cos A\cos B\cos C$

18. Putting $A = B - C, B = C - A$ and $C = A - B$ in 17 we can obtain the desired result.

19. Given $A + B + C = \pi,$ we have to prove that $\sin A\cos B\cos C + \sin B\cos C\cos A + \sin C\cos A\cos B= \sin A\sin B\sin C$

Dividing both sides by $\sin A\sin B\sin C,$ we get

$\cot B\cot C + \cot C\cot A + \cot A\cot B = 1$

$A + B = \pi - C\Rightarrow \cot(A + B) = -\cot C$

$\Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = -\cot C$

$\Rightarrow \cot B\cot C + \cot C\cot A + \cot A\cot B = 1$

20. Given, $A + B + C = \pi \Rightarrow A + B = \pi - C$

$\Rightarrow \tan(A + B) = \tan(\pi - C) = -\tan C$

$\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C$

$\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C$

21. Given $A + B + C = \pi \Rightarrow \frac{A + B}{2} = \frac{\pi - C}{2}$

$\Rightarrow \tan\frac{A + B}{2} = \tan\frac{\pi - C}{2}$

$\Rightarrow \frac{\tan\frac{A}{2} + \tan\frac{B}{2}}{1 - \tan\frac{A}{2}\tan\frac{B}{2}} = \cot\frac{C}{2} = \frac{1}{\tan\frac{C}{2}}$

$\Rightarrow \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1$

22. Let $B + C - A = \alpha, C + A - B = \beta, A + B - C = \gamma$

$\alpha + \beta + \gamma = A + B + C = \pi$

We have just proven that if $A + B + C = \pi$ then $\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C$

Thus, substituting we get, $\Rightarrow \tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma$

$\Rightarrow \tan(B + C - A) + \tan(C + A - B) + \tan(A + B - C) = \tan(B + C - A)\tan(C + A - B)\tan(A + B - C)$

23. Given $A + B + C = \pi\Rightarrow A + B = \pi - C \Rightarrow \cot(A + B) = \cot(\pi - C)$

$\Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = -\cot C$

$\Rightarrow \cot B\cot C + \cot C\cot A + \cot A\cot B = 1$

24. From previosu problem if $A + B + C = \pi$ then $\Rightarrow \cot B\cot C + \cot C\cot A + \cot A\cot B = 1$

Given $\cot A + \cot B + \cot C = \sqrt{3}$

$\Rightarrow \cot^2A + \cot^2B + \cot^2C + 2(\cot A\cot B + \cot B\cot C + \cot C\cot A) = 3$

$\cot^2A + \cot^2B + \cot^2C = 1$

$2\cot^2A + 2\cot^2B + 2\cot^2C - 2 = 0$

$2\cot^2A + 2\cot^2B + 2\cot^2C - 2(\cot A\cot B + \cot B\cot C + \cot C\cot A) = 0$

$(\cot A - \cot B)^2 + (\cot B - \cot C)^2 + (\cot C - \cot A)^2 = 0$

This is possible only if $\cot A - \cot B = 0$ i.e. $\cot A = \cot B,$ $\cot B - \cot C = 0$ i.e. $\cot B = \cot C$ and $\cot C - \cot A = 0$ i.e. $\cot C = \cot A$

$\therefore \cot A = \cot B = \cot C \Rightarrow A = B = C$

25. $\because A + B + C + D = 2\pi \Rightarrow A + B = 2\pi - C - D$

$\Rightarrow \tan(A + B) = -\tan(C + D)$

$\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\frac{\tan C + \tan D}{1 - \tan C\tan D}$

$\Rightarrow (\tan A + \tan B)(1 - \tan C\tan D) = -(1 - \tan A\tan B)(\tan C + \tan D)$

$\Rightarrow \tan A + \tan B + \tan C + \tan D = \tan A\tan B\tan C + \tan A\tan C\tan D + \tan A\tan B\tan D + \tan B\tan C\tan D$

Dividing both sides by $\tan A\tan B\tan C\tan D,$ we get

$\frac{\tan A + \tan B + \tan C + \tan D}{\tan A\tan B\tan C\tan D} = \frac{1}{\tan A} + \frac{1}{\tan B} + \frac{1}{\tan C} + \frac{1}{\tan D}$

$\Rightarrow \frac{\tan A + \tan B + \tan C + \tan D}{\cot A + \cot B + \cot C + \cot D} = \tan A\tan B\tan C\tan D$

26. Given $A + B + C = \frac{\pi}{2}\Rightarrow A + B = \frac{\pi}{2} - C$

$\Rightarrow \cot(A + B) = \cot\left(\frac{\pi}{2} - C\right)$

$\Rightarrow \frac{\cot A\cot B - 1}{\cot A + \cot B} = \tan C = \frac{1}{\cot C}$

$\Rightarrow \cot A + \cot B + \cot C = \cot A\cot B\cot C$

27. We have just proven in 26 that $\Rightarrow \cot A + \cot B + \cot C = \cot A\cot B\cot C$

Dividing both sides by $\cot A\cot B\cot C,$ we get

$\tan A\tan B + \tan B\tan C + \tan C\tan A = 1$

28. Given $A + B + C = \pi \Rightarrow 3(A + B + C) = 3\pi \Rightarrow 3A + 3B = 3\pi - 3C$

$\Rightarrow \tan(3A + 3B) = \tan(3\pi - 3C) = -\tan3C$

$\Rightarrow \frac{\tan 3A + \tan 3B}{1 - \tan3A\tan3B} = -\tan3C$

$\Rightarrow \tan 3A + \tan 3B + \tan 3C = \tan 3A\tan 3B\tan 3C$

29. Given $A + B + C = \pi \Rightarrow \frac{A + B}{2} = \frac{\pi - C}{2}$

$\Rightarrow \cot\frac{A + B}{2} = \cot\frac{\pi - C}{2}$

$\Rightarrow \frac{\cot\frac{A}{2}\cot\frac{B}{2} - 1}{\cot\frac{A}{2} + \cot\frac{B}{2}} = \tan\frac{C}{2} = \frac{1}{\cot\frac{C}{2}}$

$\Rightarrow \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$

30. We have to prove that $\frac{\cot A + \cot B}{\tan A + \tan B} + \frac{\cot B + \cot C}{\tan B + \tan C} + \frac{\cot C + \cot A}{\tan C + \tan A} = 1$

Putting $\tan A = \frac{1}{\cot A}, \tan B = \frac{1}{\cot B}, \tan C = \frac{1}{\cot C},$ we get

$\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

We have already proven above in problem 19.

31. Let $A - B = \alpha, B - C = \beta, C - A = \gamma,$ then

$\alpha + \beta + \gamma = 0$

$\Rightarrow \tan(\alpha + \beta) = -\tan\gamma$

$\Rightarrow \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = -\tan\gamma$

$\tan\alpha + \tan\beta + \tan\gamma = \tan\alpha\tan\beta\tan\gamma$

Substituting back the values, we get

$\tan(A - B) + \tan(B - C) + \tan(C - A) = \tan(A - B)\tan(B - C)\tan(C - A)$

32. We have already proven in problem 19 that if $A + B + C = 0,$ then

$\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

Let $A = x + y - z, B = z + x - y, C = y + z - x,$ then

$A + B + C = x + y + z = 0$

$\Rightarrow \cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

Substituting back the values, we get

$\cot(x + y - z)\cot(z + x - y) + \cot(x + y - z)\cot(y + z - x) + \cot(y + z - x)\cot(z + x - y) = 1$

33. Given $A + B + C= n\pi \Rightarrow \tan(A + B) = \tan(n\pi - C) = -\tan C$

$\Rightarrow \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C$

$\Rightarrow \tan A + \tan B + \tan C = \tan A\tan B\tan C$

34. L.H.S $= (\sin2A + \sin2B) + \sin2C = 2\sin(A + B)\cos(A - B) + \sin2C$

$= 2\sin(\pi - C)\cos(A - B) + \sin2C = 2\sin C\cos(A - B) + 2\sin C\cos C$

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

$= 4\sin A\sin B\sin C$

35. L.H.S. $= (\cos A + \cos B) + \cos C - 1 = 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} + \cos C - 1$

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

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

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

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

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

$= 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$

36. We have proven in 34 and 35 that $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$ and $\cos A + \cos B + \cos C - 1 = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$ respectively. Thus,

$\frac{\sin 2A + \sin 2B + \sin 2C}{\cos A + \cos B + \cos C - 1} = \frac{4\sin A\sin B\sin C}{4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}$

$= \frac{4.2\sin\frac{A}{2}\cos\frac{A}{2}.2\sin\frac{B}{2}\cos\frac{B}{2}.2\sin\frac{C}{2}\cos\frac{C}{2}}{4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}$

$= 8\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

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

$= 2\cos\frac{A + B}{4}\cos\frac{A - B}{4} + \sin\frac{\pi - C}{2}$

$= 2\cos\frac{\pi - C}{4}\cos\frac{A - B}{4} + 2\sin\frac{\pi - C}{4}\cos\frac{\pi - C}{4}$

$= 2\cos\frac{\pi - C}{4}\left[\cos\frac{A - B}{4} + \cos\left(\frac{\pi}{2} - \frac{\pi - C}{4}\right)\right]$

$= 2\cos\frac{\pi - C}{4}2\cos\frac{\pi + A + C - B}{8}\cos\frac{\pi + C - A + B}{8}$

$= 4\cos\frac{\pi - A}{4}\cos\frac{\pi - B}{4}\cos\frac{\pi - C}{4}$

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

$= 2\sin\frac{A + B}{4}\cos\frac{A - B}{4} + \cos\frac{\pi - C}{2}$

$= 2\sin\frac{\pi - C}{4}\cos\frac{A - B}{4} + 1 - 2\sin^2\frac{\pi - C}{4}$

$=1 + 2\sin\frac{\pi - C}{4}\left[\cos\frac{A - B}{4} - \sin\frac{\pi - C}{4}\right]$

$= 1 + 2\sin\frac{\pi - C}{4}\left[\cos\frac{A - B}{4} - \cos\frac{\pi + C}{4}\right]$

$= 1 + 2\sin\frac{\pi - C}{4}.2\sin\frac{\pi + A + C - B}{8}\sin\frac{\pi + C - A + B}{8}$

$= 1 + 4\sin \frac{B + C}{4}\sin \frac{C + A}{4}\sin \frac{A + B}{4}$

39. L.H.S. $= \frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} - \frac{1 - \cos C}{2}$

$= \frac{1}{2} - \frac{1}{2}[\cos A + \cos B - \cos C]$

$\cos A + \cos B - \cos C = 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} - \cos C$

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

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

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

$= -1 + 2\sin\frac{C}{2}.2\cos\frac{A}{2}\cos\frac{B}{2}$

$= -1 + 4\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}$

Thus, L.H.S. $= 1 - 2\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}$

40. L.H.S. $= 1 + \cos56^\circ + (\cos58^\circ - \cos66^\circ)$

$= 2\cos^228^\circ + 2\sin62^\circ\sin4^\circ$

$=2\cos^228^\circ + 2\cos28^\circ\sin4^\circ$

$= 2\cos28^\circ[\sin4^\circ + \cos28^\circ]$

$= 4\cos28^\circ\cos29^\circ\sin33^\circ$

41. Given $A + B + C = \pi,$ we have to prove that $\cos 2A + \cos 2B - \cos 2C = 1 - 4\sin A\sin B\cos C$

L.H.S. $=\cos 2A + \cos 2B - \cos 2C = \cos 2A + \cos 2B - \cos[2\pi - 2(A + B)]$

$= 2\cos(A + B)\cos(A - B) - \cos2(A + B) = 2\cos(A + B)\cos(A - B) - 2\cos^2(A + B) + 1$

$= 1 + 2\cos(A + B)[\cos(A - B) - \cos(A + B)]$

$= 1 - 4\sin A\sin B\cos C[\because\cos(A + B) = \cos(\pi - C) = -\cos C]$

42. Given $A + B + C = \pi,$ we have to prove that $\sin 2A + \sin 2B - \sin 2C = 4\cos A\cos B\sin C$

L.H.S. $= \sin 2A + \sin 2B - \sin 2C = 2\sin(A + B)\cos(A - B) - 2\sin C\cos C$

$[\because \sin(A + B) = \sin(\pi - C) = \sin C, \cos C = \cos[\pi - (A + B)] = -\cos(A + B)]$

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

$= 4\cos A\cos B\sin C$

43. Given $A + B + C = \pi,$ we have to prove that $\sin A + \sin B + \sin C = 4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

L.H.S. $= \sin A + \sin B + \sin C = 2\sin\frac{A + B}{2}\cos\frac{A - B}{2} + 2\sin\frac{C}{2}\cos\frac{C}{2}$

$= 2\sin\frac{\pi - C}{2}\cos\frac{A - B}{2} + 2\sin\frac{C}{2}\cos\frac{C}{2}$

$= 2\cos\frac{C}{2}\cos\frac{A - B}{2} + 2\sin\frac{\pi - A - B}{2}\cos\frac{C}{2}$

$= 2\cos\frac{C}{2}[\cos\frac{A - B}{2} + \cos\frac{A + B}{2}]$

$= 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$

44. L.H.S. $= \cos A + \cos B - \cos C = 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} - 1 + 2\sin^2\frac{C}{2}$

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

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

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

$= 4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2} - 1$

45. $B + C - A = \pi - A - A = \pi - 2A, C + A - B = \pi - 2B, A + B - C = \pi - 2C$

$\Rightarrow$ L.H.S. $= \sin 2A + \sin 2B + \sin 2C$

We have proven in problem 34 that $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$

$\therefore \sin(B + C - A) + \sin(C + A - B) + \sin(A + B - C) = 4\sin A\sin B\sin C$