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Trigonometric Fundamental Identities

+2 votes
Simplify the expression:
A. sin x - sin3 x
B. sec x cos2 x + sec x sin2 x
C. cos x (cot x + tan x)
D. sin x csc x + sin2 x sec2 x
asked Feb 13, 2013 in TRIGONOMETRY by chrisgirl Apprentice

4 Answers

+1 vote

A). sin x - sin3x

Take out common term sin(x)

sin(x)[1 - sin2x]

Pythagorean Identities: sin2θ + cos2θ = 1⇒cos2θ = 1 - sin2θ

sin(x)cos2x.

answered Feb 13, 2013 by britally Apprentice
+1 vote

B). sec(x)cos2(x) + sec(x)sin2(x)

Take out common term sec(x)

sec(x)[cos2(x) + sin2(x)]

Pythagorean Identities: sin2θ + cos2θ = 1

sec(x)(1) = sec(x).

Therefore sec(x)cos2(x) + sec(x)sin2(x) = sec(x).

answered Feb 13, 2013 by britally Apprentice
+1 vote

C). cos(x)[cot(x) + tan(x)]

Quotient Identities: tanθ = sinθ/cosθ and cotθ = cosθ/sinθ

cos(x)[cos(x)/sin(x) + sin(x)/cos(x)]

Rewrite the expression with common denominator.

cos(x)[cos2(x)+ sin2(x)]/sin(x)cos(x)

Pythagorean Identities: sin2θ + cos2θ = 1

cos(x)[1]/sin(x)cos(x) = 1/sin(x)

Reciprocal identities: 1/sin(x) = coses(x)

Therefore cos(x)[cot(x) + tan(x)] = coses(x).

answered Feb 13, 2013 by britally Apprentice
+1 vote

D). sin(x)coses(x) + sin2(x)sec2(x)

Reciprocal identities: coses(x) = 1/sin(x) and cos(x) = 1/sec(x).

= sin(x)[1/sin(x)] + sin2(x)[1/cos2(x)]

Quotient Identities: tanθ = sinθ/cosθ

= 1 + tan2(x)

Pythagorean Identities:1+ tan2θ = sec2θ.

= sec2(x).

Therefore cos(x)[cot(x) + tan(x)] = sec2(x).

answered Feb 13, 2013 by britally Apprentice

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