trigonometry help!!!!!!!!!!!!!!!!

Question

1). cosx-sinxcosx=0

2). 2sinxcosx= -1/2

3). cos^4 theta- sin^4 theta= cos2 theta

Answer 1

1). cos(x) - sin(x)cos(x) = 0

Take out common term 'cos(x)'

cos(x)(1 - sin(x)) = 0

cos(x) = 0 or 1 - sin(x) = 0

Take cos(x) = 0

Trigonometry table in cos90 = 0

So, cos(x) = cos90 then x = 90

Take 1 - sin(x) = 0

Add 1 to each side.

sin(x) = 1

Trigonometry table in sin90 = 1

sin(x) = sin90 then x = 90.

Therefore x = 90.

Comments

  

cos(x) = 0 or 1 - sin(x) = 0.

• cos (x) = 0.

cos (x) = cos (π/2).

General solution : If cos θ = cos ∝, then the general solution of θ = 2nπ ± ∝, for n Є Z.

The general solution is x = 2nπ ± π/2 for n Є Z.

• 1 - sin(x) = 0.

sin(x) = 1

sin(x) = sin (π/2)

General solution : If sin θ = sin ∝, then the general solution of θ = 2nπ + ∝, for n Є Z.

The general solution is x = 2nπ + π/2 for n Є Z.

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cos(x) = 0 or 1 - sin(x) = 0.

• cos (x) = 0.

cos (x) = cos (π/2).

General solution : If cos θ = cos ∝, then the general solution of θ = 2nπ ± ∝, for n Є Z.

The general solution is x = 2nπ ± π/2 for n Є Z.

• 1 - sin(x) = 0.

sin(x) = 1

sin(x) = sin (π/2)

General solution : If sin θ = sin ∝, then the general solution of θ = nπ + (-1)ⁿ∝, for n Є Z.

The general solution is x = nπ + (-1)ⁿπ/2 for n Є Z.

Answer 2

2). 2sinxcosx= -1/2

Double Angle Formula: 2sinAcosA = sin2A

Trigonometry table in sin(210) = -1/2

sin2x = sin(210)

2x = 210

Divide each side by 2.

x = 105=7π/12.

Comments

sin(2x) = - 1/2.

The function sin (x) has a period of 2π, first find all solutions in the interval [0, 2π].

The function sin (x) is negative in third and fourth quadrant.

• sin(2x) = - 1/2.

In third Quadrant, π ≤ x ≤ 3π/2.

- 1/2 = sin (- π/6) = sin (π + π/6) = sin (7π/6).

In fourth Quadrant, 3π/2 ≤ x ≤ 2π.

- 1/2 = sin (- π/6) = sin (2π - π/6) = sin (11π/6).

Finally, add multiples of to each of these solutions to get the general form.

2x = 2n₁π + 7π/6 and 2x = 2n₂π + 11π/6, where n₁, n₂∈ Z.

So, the general solutions are x = n₁π + 7π/12 and x = n₂π + 11π/12, where n₁, n₂∈ Z.

 

Answer 3

3). cos⁴θ - sin⁴θ = cos2θ

Now take L.H.S is cos⁴θ - sin⁴θ

This expression it can be written as

(cos²θ)² - (sin²θ)²

MATHEMATICAL FORMULA: a² − b² = (a+b)(a − b)

(cos²θ + sin²θ)(cos²θ - sin²θ)

Pythagorean Identities: cos²θ + sin²θ = 1

1.(cos²θ - sin²θ)

Double Angle Formula: cos²A - sin²A = cos2A

So, cos2θ. That is R.H.S

Therefore cos⁴θ - sin⁴θ = cos2θ