Test: Basic Trigonometric Formula

We know, 
cosA cosB + sinA sinB = cos(A-B)

cos 68° cos 8° + sin 68° sin 8° = Cos (68-8) = Cos60°
=1/2

In triangle ABC,
► Sum of all three angles = 180⁰
► A + B + C = 180

cosec A (sin B cos C + cos B sin C) = cosec A × sin(B+C)
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1

We know that:
Sin θ = Opposite / Hypotenuse

∴ SinA = 1/√10
CosA= 3/√10
similarly, SinB = 1/√5
CosB= 2/√5

⇒ Multiply:
Cos(A+B)= CosA x CosB - SinA x SinB

Substituting the value in above equation we get:

= 3/√10 x 2/√5 - 1/√10 x 1/√5
= 6/√50 - 1/√50
= 6-1/5√2. ........(√50=5√2)
= 1/ √2

we know that, sin 45 = 1/ √ 2 therefore
sinθ / cosθ = 45

3 × tan (x – 15) = tan (x + 15)
⇒ tan(x + 15) / tan(x – 15) = 3/1

► {tan (x + 15) + tan (x – 15)} / {tan (x + 15) – tan (x – 15)} = (3 + 1) / (3 – 1)

► {tan (x + 15) + tan (x – 15)} / {tan (x + 15) – tan (x – 15)} = 2

sin(x + 15 + x – 15) / sin(x + 15 – x + 15) = 2
sin 2x / sin 30 = 2
sin 2x / (1/2) = 2
2 × sin 2x = 2
Sin 2x = 1

∴ Sin 2x = Sin 90
2x = 90
x = 45

tan 20 × tan 40 × tan 80
= tan 40 × tan 80 × tan 20
= [{sin 40 × sin 80} / {cos 40 × cos 80}] × (sin 20/cos 20)
= [{2 * sin 40 × sin 80} / {2 × cos 40 × cos 80}] × (sin 20/cos 20)
= [{cos 40 – cos 120} / {cos 120 + cos 40}] × (sin 20/cos 20)
= [{cos 40 – cos (90 + 30)} / {cos (90 + 30) + cos 40}] × (sin 20/cos 20)
= [{cos 40 + sin30} / {-sin30 + cos 40}] × (sin 20/cos 20)
= [{(2 × cos 40 + 1)/2} / {(-1 + cos 40)/2}] × (sin 20/cos 20)
= [{2 × cos 40 + 1} / {-1 + cos 40}] × (sin 20/cos 20)
= [{2 × cos 40 × sin 20 + sin 20} / {-cos 20 + cos 40 × cos 20}]
= (sin 60 – sin 20 + sin 20) / (-cos 20 + cos 60 + cos 20)
= sin 60/cos 60
= tan 60
So, tan 20 × tan 40 × tan 80 = tan 60

sinA + cosA = √2(sinA/√2+cosA/√2)
=√2(sinAcos(π/4)+cosAsin(π/4))
=√2sin(A+π/4)
by using sin(A+B) formula

(a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²
⇒ c² + (a × sin x – b × cos x)² = a² + b²
⇒ (a × sin x – b × cos x)² = a² + b² – c²

L.H.S. = sin(60+A)cos(30−B)+cos(60+A)sin(30−B)        
= sin[(60+A)+(30−B)]            (Using, sin(A+B)sinAcosB+cosAsinB)            
= sin(90+A−B)            
= sin(90+(A−B))            
= cos(A−B)            (Using, sin(90+θ)=cosθ)       = R.H.S.Hence Proved.

cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)

2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)

2 sin(B/2) cos((A-C)/2) = 4 sin² (B/2) 
cos((A-C)/2) = 2 sin (B/2)
cos((A-C)/2) = 2 cos((A+C)/2)
cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)

2 sin(A/2) sin(C/2) = sin(B/2)
⇒ 2 {√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac

2(s – b) = b
a + b + c – 2b = b
a + c – b = b
a + c = 2b

sin(n+1)Asin(n+2)A + cos(n+1)Acos(n+2)A = cos (n+1)Acos(n+2)A + sin(n+1)Asin(n+2)A = cos{A(n+2-n-1)} = cos (A.1) = cos A

Cos(π/4-x)cos (π/4-y) - sin (π/4-x) sin(π/4-y)

= CosA*Cos B - Sin A*Sin B
= Cos (A+B)
= cos(π/4-x+π/4-y)
= cos(π/2-x-y)
= cos{π/2 - (x+y)}
= sin(x+y)

3A= A+ 2A
⇒ tan 3A = tan (A + 2A)
⇒ tan 3A = (tan A + tan 2A) / (1 – tan A . tan 2A)
⇒ tan A + tan 2A = tan 3A – tan 3A x tan 2A . tan A
⇒ tan 3 A – tan 2A – tan A = tan 3A . tan 2A . tan A