Half Angle formulas are trigonometric identities to find the function values of half an angle measure.The half angle formulas are derived from double angle formulas, substituting $\frac{\theta }{2}$ for θ.

Half angle formulas are used in
1.Evaluating functions of half angles
2.Solving trigonometric identities
3.Solving trigonometric equations

In this page you can find half angle formulas for sin, cosine and tangent functions, their proofs and sample problems based on these formulas.

The half angle formula for sin is
sin $\frac{\theta }{2}$ = ± $\sqrt{\frac{1-cos\theta }{2}}$

The sign of sin $\frac{\theta }{2}$ is determined by the quadrant in which $\frac{\theta }{2}$ lies.

Let us verify the formula for θ = 30º, using the table values for cos 30º and checking with calculator returned value for sin 15º.

Let θ = 30º ⇒ $\frac{\theta }{2}$ = 15º.

sin 15º = $\sqrt{\frac{1-cos30}{2}}$ We consider the positive square root as 15º falls in the first quadrant.

= $\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$ cos 30 = $\frac{\sqrt{3}}{2}$

= $\sqrt{\frac{2-\sqrt{3}}{4}}$ Simplified

≈ 0.2588
This agrees with the calculated return value for sin 15º. Thus the formula is verified.
The half angle formula for cos is
cos $\frac{\theta }{2}$ = ± $\sqrt{\frac{1+cos\theta }{2}}$ 
The sign of cos $\frac{\theta }{2}$ is determined by the quadrant in which the terminal side of angle $\frac{\theta }{2}$ lies.

let us verify the formula for θ = 60º, using the table values of cos 60º and cos 30º

Half Angle Formula Cos

Let θ = 60º       ⇒       $\frac{\theta }{2}$  = 30º

cos 30º = $\sqrt{\frac{1+cos60}{2}}$       Positive square root is considered as 30º falls in the first quadrant.

             = $\sqrt{\frac{1+\frac{1}{2}}{2}}$           cos 60 = $\frac{1}{2}$

             = $\sqrt{\frac{2+1}{4}}$

             = $\frac{\sqrt{3}}{2}$

This agrees with the function value given for cos 60 in the table.
The half angle formula for the tangent is
tan $\frac{\theta }{2}$ = ±$\sqrt{\frac{1-cos\theta }{1+cos\theta }}$

The sign of tan $\frac{\theta }{2}$ is determined by the quadrant in which the terminal side of angle $\frac{\theta }{2}$ falls.

Let us verify the formula for θ = 120º evaluating the value of cos 120º and checking with the table value for tan 60º.
Let θ 120º ⇒ $\frac{\theta }{2}$ = 60º.

cos 120 = cos (180 -60) = - cos 60º = -$\frac{1}{2}$.

tan 60º = $\sqrt{\frac{1-cos120}{1+cos120}}$ Positive square root is considered as 60º falls in the first quadrant.

= $\sqrt{\frac{1-(-\frac{1}{2})}{1+(-\frac{1}{2})}}$

= $\sqrt{\frac{2+1}{2-1}}$ = $\sqrt{3}$

Which is indeed the tan function value given for 60º in the table. Thus the identity is verified.
Using the double angle formula for cosine
cos 2θ = 2cos2 θ - 1 = 1 - 2sin2 θ, the power reducing formulas for sine and cosine are got as follows:

sin2 θ = $\frac{1-cos2\theta }{2}$ and cos2 θ = $\frac{1+cos2\theta }{2}$


And using the above two, the power reducing formula for tan can be written as follows:

tan2 θ = $\frac{1-cos2\theta }{1+cos2\theta }$

The half angle formulas are derived from power reducing formulas by taking square root and
letting θ = $\frac{\theta }{2}$.


When θ is changed to $\frac{\theta }{2}$, 2θ will be written as θ.

Now the power reducing formulas can be written as,

sin2 $\frac{\theta }{2}$ = $\frac{1-cos\theta }{2}$

cos2 $\frac{\theta }{2}$ = $\frac{1+cos\theta }{2}$

tan2 $\frac{\theta }{2}$ = $\frac{1-cos\theta }{1+cos\theta }$

Taking square roots, we get the half angle formulas as

sin $\frac{\theta }{2}$ = ± $\sqrt{\frac{1-cos\theta }{2}}$

cos $\frac{\theta }{2}$ = ± $\sqrt{\frac{1+cos\theta }{2}}$ and

tan $\frac{\theta }{2}$ = ± $\sqrt{\frac{1-cos\theta }{1+cos\theta }}$

While using the half angle formulas as stated earlier, the sign of the function value is determined by the quadrant in which

$\frac{\theta }{2}$ falls.

Solved Examples

Question 1: Evaluating half angle functions:
    cos A = $\frac{1}{8}$ and 300º < A < 360º,  evaluate sin $\frac{A}{2}$, cos $\frac{A}{2}$ and tan $\frac{A}{2}$ using half angle formulas.
Solution:
 
300º < A < 360º      ⇒   150º < $\frac{A}{2}$ < 180º

   ∠ $\frac{A}{2}$ falls in the second quadrant. Hence sin $\frac{A}{2}$ is positive and cos $\frac{A}{2}$ and tan $\frac{A}{2}$ are negative.

 sin $\frac{A}{2}$ = $\sqrt{\frac{1-cosA}{2}}$

                             = $\sqrt{\frac{1-\frac{1}{8}}{2}}$ = $\sqrt{\frac{7}{16}}$

                             = $\frac{\sqrt{7}}{4}$

 cos $\frac{A}{2}$ = -$\sqrt{\frac{1+cosA}{2}}$

                              = -$\sqrt{\frac{1+\frac{1}{8}}{2}}$ = -$\sqrt{\frac{9}{16}}$

                              = -$\frac{3}{4}$.

 tan $\frac{A}{2}$ = -$\sqrt{\frac{1-cosA}{1+cosA}}$

                            = -$\sqrt{\frac{1-\frac{1}{8}}{1+\frac{1}{8}}}$

                            = -$\frac{\sqrt{7}}{3}$
 

Question 2: Use half angle formulas to find the exact values of sin $\frac{3\pi }{8}$, cos $\frac{3\pi }{8}$ and tan $\frac{3\pi }{8}$.
Solution:
 
Let θ = $\frac{3\pi }{4}$     then $\frac{\theta }{2}$ = $\frac{3\pi }{8}$

    cos $\frac{3\pi }{4}$ = $\frac{\sqrt{2}}{2}$

    All the three functions are positive as 0 < $\frac{3\pi }{8}$ < $\frac{\pi }{2}$

    sin $\frac{\theta }{2}$$\sqrt{\frac{1-cos\theta }{2}}$

    sin $\frac{3\pi }{8}$ = $\sqrt{\frac{1-cos\frac{3\pi }{4}}{2}}$

                                   = $\sqrt{\frac{1-(-\frac{\sqrt{2}}{2})}{2}}$

    sin $\frac{3\pi }{8}$ = $\sqrt{\frac{2+\sqrt{2}}{4}}$

  cos $\frac{\theta }{2}$ = $\sqrt{\frac{1+cos\theta }{2}}$

  cos $\frac{3\pi }{8}$ = $\sqrt{\frac{1+cos\frac{3\pi }{4}}{2}}$

                                  = $\sqrt{\frac{1+(-\frac{\sqrt{2}}{2})}{2}}$

 cos $\frac{3\pi }{8}$ = $\sqrt{\frac{2-\sqrt{2}}{4}}$

 tan $\frac{3\pi }{8}$ = $\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}$

                                 = $\sqrt{\frac{(2+\sqrt{2})(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})}}$                                          Rationalizing the denominator

                                 = $\frac{2+\sqrt{2}}{\sqrt{2}}$

                                 = $\frac{2+2\sqrt{2}}{2}$                                                            Rationalizing the denominator

                                 = $\sqrt{2}+1$                                                          Factored and simplified.
 

Question 3: Verify the identity using the half angle formulas for sin and cos sin $\frac{x}{2}$ cos $\frac{x}{2}$ = $\frac{sinx}{2}$
Solution:
 
Using half angle formulas sin $\frac{x}{2}$ = $\sqrt{\frac{1-cosx}{2}}$     and cos $\frac{x}{2}$ = $\sqrt{\frac{1+cosx}{2}}$

    sin $\frac{x}{2}$ cos $\frac{x}{2}$  = $\sqrt{\frac{1-cosx}{2}}$ x $\sqrt{\frac{1+cosx}{2}}$

                                                          = $\frac{\sqrt{(1-cosx)(1+cosx)}}{2}$

                                                          = $\frac{\sqrt{1-cos^{2}x}}{2}$                               Multiplied and simplified

                                                          = $\frac{\sqrt{1-cos^{2}x}}{2}$                               Pythagorean identity

                                                          = $\frac{sinx}{2}$
 

Question 4: Prove the following identity tan $\frac{x}{2}$ = csc x - cot x
Solution:
 
Tan $\frac{x}{2}$ = $\sqrt{\frac{1-cosx}{1+cosx}}$         cos x ≠ -1

                                                                           = $\sqrt{\frac{(1-cosx)(1-cosx)}{(1+cosx)(1-cosx)}}$   Multiplying and dividing by the same quantity.

                                                                           = $\sqrt{\frac{(1-cosx)^{2}}{1-cos^{2}x}}$

                                                                           = $\sqrt{\frac{(1-cosx)^{2}}{sin^{2}x}}$     Pythagorean identity

                                                                           = $\frac{1-cosx}{sinx}$

                                                                           = $\frac{1}{sinx}$ - $\frac{cosx}{sinx}$       Separated and divided

                                                                           = csc x - cotx             Reciprocal and Quotient identities.