Trigonometry Half Angle Formulas

Introduction to Trigonometry Half Angle Formulas:

In this topic trigonometry half angle formulas, trigonometry half angle formulas for trigonometry functions are derived from the sum of angles formulas. Sine, Cosine and Tangent are the trigonometry functions involved in trigonometry half angle formulas. Trigonometry half angle formulas are as follows, Having problem with Angle Sum Identity keep reading my upcoming posts, i will try to help you.

Sin`(theta/2)=+-sqrt((1-costheta)/2)`,

Cos`(theta/2)=+-sqrt((1+costheta)/2)`,

Tan`(theta/2)=+-(1-costheta)/sintheta`.

Let us see about trigonometry half angle formulas

Proving Trigonometry Sine Half Angle Formulas:
To Prove: Sin`(theta/2)=+-sqrt((1-costheta)/2)`.

Proof: We know that, cos2A = 1 − 2sin2A.

=> cos2A − 1 = −2sin2A,

=> 2sin2A = 1 − cos2A,

=> sin2A = `(1-cos2A)/2`,

Taking square root on both sides, we get,

=> sinA = `+-sqrt((1-cos2A)/2)`,

Now plug A = `theta/2` in the above equation, we get,

=> Sin`(theta/2)=+-sqrt((1-costheta)/2)`,

Hence proved that, Sin`(theta/2)=+-sqrt((1-costheta)/2)`,

Proving Trigonometry Cosine Half Angle Formulas:

To Prove: Cos`(theta/2)=+-sqrt((1+costheta)/2)`.

Proof: We know that, cos2A = 2cos2A − 1,

=> cos2A + 1 = 2cos2A,

=> cos2A = `(1+cos2A)/2` ,

Taking square root on both sides, we get,

=> cosA = `+-sqrt((1+cos2A)/2)`,

Now plug A = `theta/2` in the above equation, we get,

=> Cos`(theta/2)=+-sqrt((1+costheta)/2)`.

Hence proved that, Cos`(theta/2)=+-sqrt((1+costheta)/2)`. I have recently faced lot of problem while learning Changing Radians to Degrees, But thank to online resources of math which helped me to learn myself easily on net.

Proving Trigonometry Tangent Half Angle Formulas:

To Prove: Tan`(theta/2)=+-(1-costheta)/sintheta`.

Proof: We know that,`tantheta=sintheta/costheta`,

=>`tan(theta/2)=sin(theta/2)/cos(theta/2)`,

=>`tan(theta/2)=+-(sqrt((1-costheta)/2))/(sqrt((1+costheta)/2))`,

=>`tan(theta/2)=+-sqrt((1-costheta)/(1+costheta))`,

Taking conjugate, we get,

=>`tan(theta/2)=+-sqrt((1-costheta)/(1+costheta)xx(1-costheta)/(1-costheta))`,

=>`tan(theta/2)=+-sqrt((1-costheta)^2/(1-cos^2theta))`,

=>`tan(theta/2)=+-sqrt((1-costheta)^2/(sin^2theta))`,

=>`tan(theta/2)=+-(1-costheta)/sintheta`.

Hence proved that, `tan(theta/2)=+-(1-costheta)/sintheta`.