Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(e) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1A Trigonometric Ratios up to Transformations Solutions Exercise 6(e)

I.

Question 1.
Prove that sin 50° – sin70° + sin 10° = 0
Solution:
LHS = sin 50° – sin 70° + sin 10°
= 2 cos($$\frac{50^{\circ}+70^{\circ}}{2}$$) . sin($$\frac{50^{\circ}-70^{\circ}}{2}$$) + sin 10°
= 2 cos 60° . sin (-10°) + sin 10°
= 2($$\frac{1}{2}$$) (-sin 10°) + sin 10°
= -sin 10° + sin 10°
= 0
= RHS
∴ sin 50° – sin 70° + sin 10° = 0

Question 2.
Prove that $$\frac{\sin 70^{\circ}-\cos 40^{\circ}}{\cos 50^{\circ}-\sin 20^{\circ}}=\frac{1}{\sqrt{3}}$$
Solution:

Question 3.
Prove that cos 55° +cos 65° + cos 175° = 0
Solution:
LHS = cos 55° + cos 65° + cos 175°
= cos 65° + cos 55° + cos (180° – 5°)
= 2 cos($$\frac{65^{\circ}+55^{\circ}}{2}$$) . cos($$\frac{65^{\circ}-55^{\circ}}{2}$$) – cos 5°
= 2 cos (60°) . cos (5°) – cos 5°
= 2($$\frac{1}{2}$$) cos 5° – cos 5°
= cos 5° – cos 5°
= 0
= RHS
∴ cos 55° + cos 65° + cos 175° = 0

Question 4.
Prove that 4(cos 66° + sin 84°) = √3 + √15
Solution:

Question 5.
Prove that cos 20° cos 40° – sin 5° sin 25° = $$\frac{\sqrt{3}+1}{4}$$
Solution:
cos 20° cos40° – sin 5° sin 25°
= $$\frac{1}{2}$$ [2 cos 20° cos 40° – 2 sin 5° sin 25°]
= $$\frac{1}{2}$$ [cos (20° + 40°) + cos (20° – 40°) – {cos (5° – 25°) – cos (5° + 25°)}]
= $$\frac{1}{2}$$ [cos 60° + cos 20° – cos 20° + cos 30°]
= $$\frac{1}{2}\left[\frac{1}{2}+\frac{\sqrt{3}}{2}\right]$$
= $$\frac{\sqrt{3}+1}{4}$$
= RHS
∴ cos 20° cos 40° – sin 5° sin 25° = $$\frac{\sqrt{3}+1}{4}$$

Question 6.
Prove that cos 48° . cos 12° = $$\frac{3+\sqrt{5}}{8}$$
Solution:
LHS = cos 48° . cos 12°
= $$\frac{1}{2}$$ (2 cos 48°. cos 12°)
= $$\frac{1}{2}$$ [cos (48° + 12°) + cos (48° – 12°)]
= $$\frac{1}{2}$$ [cos 60° + cos 36°]
= $$\frac{1}{2}\left[\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right]$$
= $$\frac{1}{2}\left[\frac{2+\sqrt{5}+1}{4}\right]$$
= $$\frac{3+\sqrt{5}}{8}$$
= RHS
∴ cos 48° . cos 12° = $$\frac{3+\sqrt{5}}{8}$$

II.

Question 1.
Prove that cos θ + cos[$$\frac{2 \pi}{3}$$ + θ] + cos[$$\frac{4 \pi}{3}$$ + θ] = 0
Solution:

Question 2.
Prove that sin2(α – π/4) + sin2(α + π/2) – sin2(α – π/2) = $$\frac{1}{2}$$
Solution:

Question 3.
If sin x + sin y = $$\frac{1}{4}$$ and cos x + cos y = $$\frac{1}{3}$$, then show that
(i) $$\tan \left(\frac{x+y}{2}\right)=\frac{3}{4}$$
(ii) cot (x + y) = $$\frac{7}{24}$$
Solution:

Question 4.
If neither [A – $$\frac{\pi}{12}$$] nor [A – $$\frac{5 \pi}{12}$$] is an integral multiple of π. Prove that cot(π/2 – A) + tan(π/12 + A) = $$\frac{4 \cos 2 A}{1-2 \sin 2 A}$$
Solution:

Question 5.
Prove that 4 cos 12° cos 48° cos 72° = cos 36°.
Solution:
LHS = 4 cos 12° cas 48° cos 72°
= 2 cos 12° {2 cos 72° cos 48°}
= 2 cos 12° {cos (72° + 48°) + cos (72° – 48°)}
= 2 cos 12° {cos (120°) + cos 24°)
= 2 cos 12° {$$\frac{1}{2}$$ + cos 24°}
= 2 cos 12° $$\left\{\frac{-1+2 \cos 24^{\circ}}{2}\right\}$$
= -cos 12° + 2 cos 24° cos 12°
= -cos 12° + {cos(24° + 12°) + cos (24° – 12°))
= -cos 12° + cos 36° + cos 12°
= cos 36°
= RHS
∴ 4 cos 12° cos 48° cos 72° = cos 36°

Question 6.
Prove that sin 10° + sin 20° + sin 40° + sin 50° = sin 70° + sin 80°
Solution:
LH.S. = sin 10° + sin 20° + sin 40° + sin 50°
= (sin 50°+ sin 10°) + (sin 40° + sin 20°)
= 2 sin($$\frac{50^{\circ}+10^{\circ}}{2}$$) . cos($$\frac{50^{\circ}-10^{\circ}}{2}$$) + 2 sin($$\frac{40^{\circ}+20^{\circ}}{2}$$) . cos($$\frac{40^{\circ}-20^{\circ}}{2}$$)
= 2 . sin (30°) . cos 20° + 2 sin 30° . cos 10°
= 2 sin 30° (cos 20° + cos 10°)
= 2($$\frac{1}{2}$$) [cos(90° – 70°) + cos(90° – 80°)]
= sin 70° + sin 80°
= RHS
∴ sin 10° + sin 20° + sin 40° + sin 50° = sin 70° + sin 80°

III.

Question 1.
If cos x + cos y = $$\frac{4}{5}$$ and cos x – cos y = $$\frac{2}{7}$$ find the value of $$14 \tan \left(\frac{x-y}{2}\right)+5 \cot \left(\frac{x+y}{2}\right)$$
Solution:

Question 2.
If none of the denominators is zero, prove that
$$\left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^{n}+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^{n}$$ = $$\begin{cases}2 \cot ^{n}\left(\frac{A-B}{2}\right), & \text { if } n \text { is even } \\ 0, & \text { if } n \text { is odd }\end{cases}$$
Solution:

Question 3.
If sin A = sin B and cos A = cos B, then prove that A = 2nπ + B for some integer n.
Solution:

Question 4.
If cos nα ≠ 0 and cos $$\frac{\alpha}{2}$$ ≠ 0. then show that $$\frac{\sin (n+1) \alpha-\sin (n-1) \alpha}{\cos (n+1) \alpha+2 \cos n \alpha+\cos (n-1) \alpha}$$ = tan $$\frac{\alpha}{2}$$
Solution:

Question 5.
If sec (θ + α) + sec (θ – α) = 2 sec θ and cos α ≠ 1, then show that cos θ = ±√2 cos $$\frac{\alpha}{2}$$.
Solution:

Question 6.
If none of x, y, z is an odd multiple of $$\frac{\pi}{2}$$ and if sin (y + z – x), sin (z + x – y), sin (x + y – z) are in A.P., then prove that tan x, tan y, tan z are also in A.P.
Solution:
sin (y + z – x), sin (z + x – y), sin (x + y – z) are in A.P.
⇒ sin (z + x – y) – sin (y + z – x) = sin (x + y – z) – sin (z + x – y)
⇒ 2 cos z sin (x – y) = 2 cos x sin (y – z)
⇒ cos z [sin x cos y – cos x sin y] = cos x [sin y cos z – cos y sin z]
Dividing with cos x cos y cos z, we get
$$\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}=\frac{\sin y}{\cos y}-\frac{\sin z}{\cos z}$$
⇒ tan x – tan y = tan y – tan z
⇒ tan x + tan z = 2 tan y
∴ tan x, tan y, tan z are in A.P.

Question 7.
If x, y, z are non zero real numbers and if x cos θ = y cos (θ + $$\frac{2 \pi}{3}$$) = z cos (θ + $$\frac{2 \pi}{3}$$) for some θ ∈ R, then show that xy + yz + zx = 0.
Solution:

Question 8.
If neither A nor A + B is an odd multiple of $$\frac{\pi}{2}$$ and if m sin B = n sin(2A + B), then prove that (m + n) tan A = (m – n) tan (A + B).
Solution:
Neither A nor (A + B) is an odd multiple of $$\frac{\pi}{2}$$
Given that m sin B = n sin (2A + B)

Question 9.
If tan (A + B) = λ tan (A – B), then show that (λ + 1) sin 2B = (λ – 1) sin 2A.
Solution: