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Maths - Any Ideas

0 votes
Hello,

There are some questions that I am not sure how to tackle:

1) sin3A = 3sinA - 4sin^3A

2) cos3A = 4cos^3A - 3cosA

3) cotθ = (sin2θ) / (1 - cos2θ)

4) cos^4θ - sin^4θ = cos2θ

Thank you in advance.
asked Aug 11, 2014 in TRIGONOMETRY by anonymous

4 Answers

0 votes

(1). sin3A = 3sinA - 4sin^3A.

Left hand side identity = sin 3A

                                   = sin(2A + A)

Apply Sum and Difference Formula : sin(u+ v) = sin(u) cos(v) + cos(u) sin(v).

                                   = sin(2A) cos(A) + cos(2A) sin(A)

Apply Double angle Formulas : sin(2u) = 2 sin(u) cos(u) and cos(2u) = 1 - 2sin2(u).

                                   = [2 sin(A) cos(A)] cos(A) + [1 - 2sin2(A)] sin(A)

                                   = 2 sin(A) cos2(A) + sin(A) - 2sin3(A)

                                   = 2 sin(A) [1 - sin2(A)] + sin(A) - 2sin3(A)

                                   = 2 sin(A) - 2sin3(A) + sin(A) - 2sin3(A)

                                   = 3sin(A) - 2sin3(A) - 2sin3(A)

                                   = 3sin(A) - 4sin3(A).

                                   = Right hand side identity.

answered Aug 11, 2014 by casacop Expert
0 votes

(2). cos3A = 4cos^3A - 3cosA.

Left hand side identity = cos 3A

                                   = cos(2A + A)

Apply Sum and Difference Formula : cos(u+ v) = cos(u) cos(v) - sin(u) sin(v).

                                   = cos(2A) cos(A) - sin(2A) sin(A)

Apply Double angle Formulas : sin(2u) = 2 sin(u) cos(u) and cos(2u) = 2 cos2(u) - 1.

                                   = [2 cos2(A) - 1] cos(A) - 2 sin(A) cosA sin(A)

                                   = 2 cos3(A) - cos(A) - 2 sin2(A) cosA

                                   = 2 cos3(A) - cos(A) - 2[1- cos2(A)] cosA

                                   = 2 cos3(A) - cos(A) - 2 cosA + 2 cos3(A)

                                   = 4 cos3(A) - 3 cosA

                                   = Right hand side identity.

answered Aug 11, 2014 by casacop Expert
edited Aug 11, 2014 by casacop
0 votes

(3). cotθ = (sin2θ) / (1 - cos2θ)

Left hand side identity = cot(θ)

                                   = cos(θ)/sin(θ)

                                   = [ cos(θ)/sin(θ) ] * [ 2sin(θ) / 2sin(θ) ]

                                   = 2 cos(θ)sin(θ) / 2sin2(θ)

Apply Double angle Formulas : 2 sin(u) cos(u) = sin(2u) and 2sin2(u) = 1 - cos(2u).

                                   = sin(2θ) / 1 - cos(2θ)

                                   = Right hand side identity.

answered Aug 11, 2014 by casacop Expert
0 votes

(4). cos^4θ - sin^4θ = cos2θ

Left hand side identity = cos4(θ) - sin4(θ)

Apply Difference of Two Squares : a2- b2 = (a + b)(a - b).


                                   = [cos2(θ) + sin2(θ)] * [cos2(θ) - sin2(θ)]

Apply Pythagorean identity : sin2(u) + cos2(u) = 1.


                                   = [cos2(θ) - sin2(θ)]

Apply Double angle Formulas : cos(2u) = cos2(u) - sin2(u).


                                   = cos(2θ).

                                   = Right hand side identity.

answered Aug 11, 2014 by casacop Expert

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