given some information about theta, find sin2 theta, cos2 theta, and tan2 theta. A.) cos theta=-1/4 and pi/2 is less than theta less than pi.

Question

B.) sec theta = 5/2 and sin theta is greater than 0. C.) sin theta=3/5 and tan theta is less than 0

Answer 1

The category is double angle identity, so I assume your question was originally written like this:

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Find sin(2θ), cos(2θ), and tan(2θ) when...

A) cos(θ) = -1/4 and π/2 < θ < π

B) sec(θ) = 5/2 and sin(θ) > 0

C) sin(θ) =3/5 and tan(θ) < 0.

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Double angle identities:

sin(2θ) = 2sin(θ)cos(θ)

cos(2θ) = cos²(θ) - sin²(θ)

tan(2θ) : I'd rather use the fact that tangent is sine over cosine.

Useful: sin²(θ) + cos²(θ) = 1

Now I'll start.

A) cos(θ) = -1/4 and π/2 < θ < π

From sin²(θ) + cos²(θ) = 1, sin²(θ) = 15/16, so sin(θ) = ±√15/4.

Because π/2 < θ < π, sin(θ) must be positive. Therefore sin(θ) = √15/4.

sin(2θ) = 2sin(θ)cos(θ) = 2*(√15/4)*(-1/4) = -√15/8

cos(2θ) = cos²(θ) - sin²(θ) = 1/16 - 15/16 = -7/8

tan(2θ) = sin(2θ) / cos(2θ) = √15/7

B) sec(θ) = 5/2 and sin(θ) > 0

sec(θ) = 5/2 means cos(θ) = 2/5.

From sin²(θ) + cos²(θ) = 1, sin²(θ) = 21/25, so sin(θ) = ±√21/5. Well, sin(θ) > 0, so it is √21/5.

sin(2θ) = 2sin(θ)cos(θ) = 2*(√√21/5)*(2/5) = 4√21/25

cos(2θ) = cos²(θ) - sin²(θ) = 4/25 - 21/25 = -17/25

tan(2θ) = sin(2θ) / cos(2θ) = -4√21/17

C) sin(θ) =3/5 and tan(θ) < 0.

From sin²(θ) + cos²(θ) = 1, cos²(θ) = 16/25, so cos(θ) = ±4/5.

If sin(θ) > 0 and tan(θ) < 0, then cos(θ) < 0. Therefore cos(θ) = -4/5.

sin(2θ) = 2sin(θ)cos(θ) = 2*(3/5)*(-4/5) = -24/25

cos(2θ) = cos²(θ) - sin²(θ) = 16/25 - 9/25 = 7/25

tan(2θ) = sin(2θ) / cos(2θ) = -24/7

Hope this helps!