Some Typical Examples

Some Typical Examples

(a) Given that (1 + cos α) (1 + cos β) (1 + cos γ) = (1 - cos α) (1 - cos β) (1 - cos γ) show that one of the values of each member of this equality is sin α sin β sin γ.

Solution :

Multiplying both sides of the given equality by (1 + cos α) (1 + cos β) (1 + cos γ) we
get (1 + cos α)² (1 + cos β)² (1 + cos γ)² = sin²α sin²β sin²γ. So that one of the values
of (1 + cos α) (1 + cos β) is (1 + cos γ) is sin α sin β sin γ which is also one of the values of (1 - cos α) (1 - cos β) (1 - cos γ)

(b) If a² + b² + 2ab cos θ = 1, c² + d2 + 2cd cos θ = 1 and ac + bd + (ad + bc) cos θ = 0 prove that a² + c² = cosec2 θ.

Solution:

From the given relations, we have

(b + a cos θ)² = 1 - a² sin² θ and (1)

(d + c cos θ)² = 1 - c² sin² θ (2)

Multiplying (1) and (2) we get

(b + a cos θ)² (d + c cos θ)² = (1 – a² sin² θ) (1 − c² sin²θ) (3)

also (b + a cos θ) (d + c cos θ) = bd + (ad + bc) cos θ + ac cos² θ

= bd + ac + (ad + bc) cos θ - ac sin² θ = -ac sin² θ [given in condition]

Hence from (3) it follows that

(1 – a² sin² θ) (1 – c² sin² θ) = a²c² sin⁴ θ

⇒ 1 – (a² + c²) sin² θ + a²c² sin4 θ = a²c² sin⁴ θ

⇒ (a² + c²) sin² θ = 1 ⇒ a² + c² = cosec² θ

(c) Find the values of a for which the equation sin4x + cos4x = a has real solution.

Solution:

We have sin⁴x + cos⁴x = a

⇒ (sin²x + cos²x) −2sin²xcos²x = a

⇒ 1 − 1/2sin²2x = a

⇒ sin²2x = 2(1 −a)

since 0 ≤ sin²2x ≤ 1

so 0 ≤ 2(1 − a) ≤ 1 or 1/2 ≤ a ≤ 1

Next (sin²x + cos²x)² - 2 sin²x cos²x = a

⇒ 1/2 sin²x = 1-a.

Or 1 – a ≤ 1/2 or 1/2 ≤ a implying that 1/2 ≤ a ≤ 1.

(d) Prove that 4 sin 27° = (5 + √5 )½ - (3 - √5 )½

Solution:

sin 54° = cos 36° =

cos 54° =

⇒ 1 – 2 sin² 27° =

⇒ 2 sin² 27 =

⇒ 16 sin² 27° = = {(5 + √5)½ - (3 - √5)½}²

⇒ 4 sin 27° = (5 + √5)½ - (3 - √5)½

(e) If A,B,C are the angles of a triangle such that angle A is obtuse, prove that tan B tan C < 1.

Solution:

Since A > π/2  we have 0 < B + C < π/2

⇒ tan (B+C) > 0 ⇒

⇒ tan B tan C < 1 as tan B and tan C > 0