trignometry help please?

Question

prove
1-sin2x
------------- = tan^2[pi/4 -x]
1+sin2x


2. if A & B be the two distinct two real numbers satisfying the equation acosx +bsinx = c then prove that
1. sin(A + B) = 2ab
---------
a^2 + B^2

2. tan (A+ B) = 2ab
-----------
a^2 - b^2



if A+B+C =pi then prove that


cosA + cos B + cosC
------------ ---------- ---------- = 2
sinBsinC sinCsinA sinAsinB
& also prove that

cos4A + cos4B + cos4C = -1+4cos2Acos2Bcos2C

Answer 1

1. RHS = tan^2 (π/4 - x)
= sin^2 (π/4 - x) / cos^2 (π/4 - x)
= [(1 - cos 2(π/4 - x)) / 2] / [(1 + cos 2(π/4 - x)) / 2]
= (1 - cos (π/2 - 2x)) / (1 + cos (π/2 - 2x))
= (1 - sin 2x) / (1 + sin 2x)
= LHS.

2. a cos x + b sin x = c
<=> a/√(a^2 + b^2) cos x + b/√(a^2 + b^2) sin x = c / √(a^2 + b^2)
Let φ be such that cos φ = a/√(a^2 + b^2) and sin φ = b/√(a^2 + b^2). Then we have
cos φ cos x + sin φ sin x = c/√(a^2 + b^2)
<=> cos (x - φ) = c / √(a^2 + b^2)
Let A be one solution. If A - φ = θ, then the other solution will be B - φ = -θ + 2kπ for any integer k.
So A + B = θ + φ - θ + φ + 2kπ
= 2φ + 2kπ
So sin (A + B) = sin (2φ + 2kπ) = sin (2φ)
= 2 sin φ cos φ
= 2 (b/√(a^2 + b^2)) (a/√(a^2 + b^2))
= 2ab / (a^2 + b^2).

Also, cos (A + B) = cos (2φ) = cos^2 φ - sin^2 φ
= (a/√(a^2 + b^2))^2 - (b/√(a^2 + b^2))^2
= (a^2 - b^2) / (a^2 + b^2)
so tan (A + B) = sin (A + B) / cos (A + B)
= (2ab / (a^2 + b^2)) / ((a^2 - b^2) / (a^2 + b^2))
= 2ab / (a^2 - b^2).

3.(i) 4 sin A sin B sin C
= 2 (2 sin A sin B) sin C
= 2 (cos (A-B) - cos (A+B)) sin C
= 2 cos (A-B) sin C - 2 cos (A+B) sin C
= (sin ((A-B) + C) - sin ((A-B) - C)) - (sin ((A+B) + C) - sin ((A+B) - C))
= sin (A-B+C) - sin (A-B-C) - sin (A+B+C) + sin (A+B-C)
Use A+B+C = π:
= sin (π-B - B) - sin (A - (π-A)) - sin π + sin (π-C - C)
= sin (π-2B) - sin (2A-π) + sin (π-2C)
= sin 2B + sin 2A + sin 2C
= 2 sin A cos A + 2 sin B cos B + 2 sin C cos C

So 4 sin A sin B sin C = 2 sin A cos A + 2 sin B cos B + 2 sin C cos C; divide through by 2 sin A sin B sin C to get
2 = cos A / (sin B cos B) + cos B / (sin A sin C) + cos C / (sin A sin B)
as required.

(ii) RHS = -1 + 4 cos 2A cos 2B cos 2C
= -1 + 2 (2 cos 2A cos 2B) cos 2C
= -1 + 2 (cos (2A+2B) + cos(2A-2B)) cos 2C
= -1 + 2 cos (2A+2B) cos 2C + 2 cos (2A-2B) cos 2C
= -1 + cos (2A+2B+2C) + cos (2A+2B-2C) + cos (2A-2B+2C) + cos (2A-2B-2C)
Use 2A+2B+2C = 2π:
= -1 + cos 2π + cos (2π-4C) + cos (2π-4B) + cos (4A-2π)
= -1 + 1 + cos 4C + cos 4B + cos 4A
= cos 4A + cos 4B + cos 4C
= LHS.

Answer 2

One problem per question please.

Prove the identity.

1) (1 - sin 2x) / (1 + sin 2x) = tan²(π/4 - x)

Right Hand Side = tan²(π/4 - x)

= {[(tan π/4) - (tan x)] / [1 + (tan π/4)(tan x)]}²

= {[1 - (tan x)] / [1 + (tan x)]}²

= {[(cos x) - (sin x)] / [(cos x) + (sin x)]}²

= {[cos²x - 2(sin x)(cos x) + sin²x]
/ [cos²x + 2(sin x)(cos x) + sin²x]

= {[1 - 2(sin x)(cos x)] / [1 + 2(sin x)(cos x)]

= (1 - sin 2x) / (1 + sin 2x) = Left Hand Side