Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,428 questions

17,804 answers

1,438 comments

47,198 users

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 in TRIGONOMETRY by johnkelly Apprentice

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 by richardson Scholar

 

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 by richardson Scholar

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 by richardson Scholar

Related questions

asked Nov 15, 2014 in TRIGONOMETRY by anonymous
asked Aug 14, 2014 in TRIGONOMETRY by anonymous
asked Aug 12, 2014 in TRIGONOMETRY by anonymous
asked Aug 11, 2014 in TRIGONOMETRY by anonymous
asked Aug 10, 2014 in TRIGONOMETRY by anonymous
asked Aug 9, 2014 in TRIGONOMETRY by anonymous
...