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

+1 vote
1). cosx-sinxcosx=0

2). 2sinxcosx= -1/2

3). cos^4 theta- sin^4 theta= cos2 theta
asked Feb 11, 2013

3 Answers

+2 votes

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.

answered Feb 11, 2013

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.

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.

+2 votes

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.

answered Feb 11, 2013

sin(2x) = - 1/2.

The function sin (x) has a period of , 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, πx ≤ 3π/2.

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

In fourth Quadrant, 3π/2x ≤ 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 = nπ + 7π/12 and x = n₂π + 11π/12, where n₁, n₂∈ Z.

+2 votes

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

answered Feb 11, 2013