Double Angle and Half Angle Formulas: Examples & Types |

Q: What are double-angle and half-angle formulas?

Double-angle and half-angle formulas are formulas used in finding the trigonometric values for angles that are doubled or halved.

Q: How are double angle and half angle formulas used?

Double-angle and half-angle formulas are used by applying directly the formulas when finding double/half-angle trigonometric identities.

Q: What is an example of double-angle and half-angle formulas?

An example of a double angle formula is sin2A = 2sinAcosA, while that of half-angle is sinA/2 = square root ((1-cosA)/2).

Q: How do you derive double angle formulas?

To derive double angle formulas you would need to apply the sum of trigonometric function formulas.

Q: What are the types of double angle formulas?

The types of double angle formulas are those of sin, cos, sec, cosec, tan and cot.

Jump to a key chapter

$\sin 2 θ \neq 2 \sin θ$

and

$\cos \frac{θ}{2} \neq \frac{\cos θ}{2}$ ?

In this article, you would understand what happens when trigonometric identities are either doubled or halved.

Double-angle formulas

Trigonometric functions can be doubled but not in the same way as normal numbers are doubled.

If you have the expression 3y and you are to double it, it is easy to multiply 3y by 2 to get 6y. Note that sin30° is 0.5 and doubling the angle gives 60°, but sin60° would not give you 1. As normal mathematical operations would multiply 0.5 by 2 to give 1, trigonometric identities would require their own formula to double their function.

Deriving the double-angle formula for Sine function

We intend on finding the formula for sin2θ. Note that

$\sin 2 θ = \sin (θ + θ)$

Recall that,

$\sin (A + B) = \sin A \cos B + \sin B \cos A$

Now take $A = B = θ,$ we get

$\sin 2 θ = \sin θ \cos θ + \sin θ \cos θ = 2 \sin θ \cos θ$

The double angle formula for the sine function is given by $\sin (2 θ) = 2 \sin θ \cos θ .$

Find $\sin 60 °$ using the double angle formula.

Solution:

We have

$60 ° = 2 (30 °)$

Thus

$\sin 60 ° = \sin (2 \times 30 °) = 2 \sin 30 ° \cos 30 °$

but,

$\sin 30 ° = \frac{1}{2}, \cos 30 ° = \frac{\sqrt{3}}{2}$

Then,

$\sin 60 ° = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} \sin 60 ° = \frac{\sqrt{3}}{2}$

Given that $90 ° < θ < 180 °$ , find $\sin 2 θ$ if

$\sin θ = \frac{4}{5}$

Solution:

We have sinθ in the given, but in order to apply our formula, we need to find cosθ.

Recall that,

$\cos^{2} θ + \sin^{2} θ = 1$

Thus,

$\cos^{2} θ = 1 - \sin^{2} θ = 1 - {(\frac{4}{5})}^{2} = 1 - \frac{16}{25} \cos^{2} θ = \frac{9}{25}$

We take the square root of both sides to get,

$\cos θ = \pm \frac{3}{5}$

Note that the range of the angle is between 90° and 180°, this means that θ is in the second quadrant. The cosine of angles in the second quadrant has negative values. Thus,

$\cos θ = - \frac{3}{5}$

Now we have to apply our double angle formula,

$\sin 2 θ = 2 \sin θ \cos θ = 2 \times \frac{4}{5} \times (- \frac{3}{5}) = - \frac{24}{25}$

Deriving the double-angle formula for Cosine function

We will now develop a formula for a double-angle for the cosine function. We derive three equal formulas.

We first note that

$\cos 2 θ = \cos (θ + θ)$

Now, recall that

$\cos (A + B) = \cos A \cos B - \sin A \sin B$

By taking $A = B = θ,$ we get

$\cos 2 θ = \cos (θ + θ) = \cos θ \cos θ - \sin θ \sin θ = \cos^{2} θ - \sin^{2} θ$

Thus, we derive the first formula for $\cos 2 θ$ ,

$\cos 2 θ = \cos^{2} θ - \sin^{2} θ$

We now recall the identity

\cos^{2} θ + \sin^{2} θ = 1

, thus we have

\cos^{2} θ = 1 - \sin^{2} θ

Now we replace this with the obtained formula for $\cos 2 θ,$ to get

$\cos 2 θ = 1 - \sin^{2} θ - \sin^{2} θ = 1 - 2 \sin^{2} θ$

Thus, the second formula for $\cos 2 θ$ is

$\cos 2 θ = 1 - 2 \sin^{2} θ$

In a similar way, we have $\sin^{2} θ = 1 - \cos^{2} θ$ .

Substituting the value of $\sin^{2} θ$ into the formula of $\cos 2 θ$ , we have

$\cos 2 θ = \cos^{2} θ - (1 - \cos^{2} θ) = \cos^{2} θ - 1 + \cos^{2} θ = 2 \cos^{2} θ - 1$

Thus, the third formula for $\cos 2 θ$ is

$\cos 2 θ = 2 \cos^{2} θ - 1$

The double angle formulas for the cosine function are given by,

$\cos 2 θ = \cos^{2} θ - \sin^{2} θ = 2 \cos^{2} θ - 1 = 1 - 2 \sin^{2} θ$

Given that $90 ° < θ < 180 °$ , find $\cos 2 θ$ if

$\sin θ = \frac{4}{5}$

Solution:

Method 1.

The direct way to find $\cos 2 θ$ is to use the formula $\cos 2 θ = 1 - 2 \sin^{2} θ$ , since we are given the value of $\sin θ$ .

So,

$\cos 2 θ = 1 - 2 \sin^{2} θ = 1 - 2 {(\frac{4}{5})}^{2} = 1 - 2 (\frac{16}{25}) = 1 - \frac{32}{25} = \frac{25 - 32}{25} = - \frac{7}{25}$

Method 2.

We can use either of the other formulas to find id="2967226" role="math" $\cos 2 θ$ , we will use id="2967228" role="math" $\cos 2 θ = 2 \cos^{2} θ - 1 .$ We need thus to find $\cos θ$ .

We recall that $\cos^{2} θ + \sin^{2} θ = 1$ , thus

$\cos^{2} θ = 1 - \sin^{2} θ = 1 - {(\frac{4}{5})}^{2} = 1 - \frac{16}{25} = \frac{9}{25}$

Taking the square root of both sides, we get

$\cos θ = \pm \frac{3}{5}$

Note that $90 ° < θ < 180 °$ , this means that $θ$ is in the second quadrant. The cosine of angles in the second quadrant have negative values. Thus,

$\cos θ = - \frac{3}{5}$

So, we can apply our formula

$\cos 2 θ = 2 \cos^{2} θ - 1 = 2 {(- \frac{3}{5})}^{2} - 1 = 2 \times \frac{9}{25} - 1 = \frac{18}{25} - 1 = - \frac{7}{25}$

For $180 ° < θ < 270 °$ , find $\cos 2 θ$ when

$\cos θ = - \frac{1}{3}$

Solution:

In solving this problem, it is faster to use the formula $\cos 2 θ = 2 \cos^{2} θ - 1$ .

Thus,

$\cos 2 θ = 2 \times {(\frac{- 1}{3})}^{2} - 1 = 2 \times \frac{1}{9} - 1 = \frac{2}{9} - 1 = - \frac{7}{9}$

Deriving the double-angle formula for Tangent function

We will develop a formula for a double-angle of the tangent function.

We recall that

$\tan θ = \frac{\sin θ}{\cos θ}$

and,

$\sin 2 θ = 2 \sin θ \cos θ$

$\cos 2 θ = \cos^{2} θ - \sin^{2} θ$

Thus,

$\tan 2 θ = \frac{\sin 2 θ}{\cos 2 θ}$

Substituting $\sin 2 θ$ and $\cos 2 θ$ by their expressions, we get

$\tan 2 θ = \frac{2 \sin θ \cos θ}{\cos^{2} θ - \sin^{2} θ}$

To simplify this further, we divide both the numerator and the denominator of the right-hand side of the equation by $\cos^{2} θ$ , to get

$\tan 2 θ = \frac{\frac{2 \sin θ \cos θ}{\cos^{2} θ}}{\frac{\cos^{2} θ - \sin^{2} θ}{\cos^{2} θ}} = \frac{\frac{2 \sin θ}{\cos θ}}{\frac{\cos^{2} θ}{\cos^{2} θ} - \frac{\sin^{2} θ}{\cos^{2} θ}} = \frac{2 \tan θ}{1 - \tan^{2} θ}$

The double angle formula for the tangent function is given by,

$\tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ}$

Given that $90 ° < θ < 180 °,$ find $\tan 2 θ$ if

$\sin θ = \frac{4}{5}$

Solution:

We need to find $\tan θ$ , this means that we firstly have to find $\cos θ$ .

We recall that $\cos^{2} θ + \sin^{2} θ = 1$ , thus $\cos^{2} θ = 1 - \sin^{2} θ$ . Replacing $\sin θ$ by its value, we get

$\cos^{2} θ = 1 - {(\frac{4}{5})}^{2} = 1 - \frac{16}{25} = \frac{9}{25}$

We take the square root of both sides, to get

$\cos θ = \pm \frac{3}{5}$

Note that $90 ° < θ < 180 °$ , this means that $θ$ is in the second quadrant. The cosine of angles in the second quadrant have negative values. Thus, $\cos θ = - \frac{3}{5}$ .

Therefore,

$\tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{4}{5}}{- \frac{3}{5}} = \frac{4}{5} \times (- \frac{5}{3}) - \frac{4}{3}$

Thus,

$\tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ} = \frac{2 \times (\frac{- 4}{3})}{1 - {(\frac{- 4}{3})}^{2}} = \frac{- \frac{8}{3}}{1 - \frac{16}{9}} = \frac{- \frac{8}{3}}{- \frac{7}{9}} = - \frac{8}{3} \times (\frac{- 9}{7}) = \frac{24}{7}$

Deriving the double-angle formulas for Secant, Cosecant and Cotangent functions

Secant, cosecant, cotangent functions are the reciprocals of cosine, sine and tangent respectively. In order to derive their double-angle formulas, you just need to find the multiplicative inverse of the corresponding double-angle formulas.

Double angle formula for Secant

We recall by the definition of the secant function that

$s e c θ = \frac{1}{\cos θ}$ so $s e c 2 θ = \frac{1}{\cos 2 θ}$

but from the double angle formula for cosine we have $\cos 2 θ = \cos^{2} θ - \sin^{2} θ$ , thus

$s e c 2 θ = \frac{1}{\cos^{2} θ - \sin^{2} θ}$

Now let us express $s e c 2 θ$ in terms of $s e c θ$ and $c s c θ$ .

In fact, $\cos θ = \frac{1}{s e c}$ and $c s c θ = \frac{1}{\sin θ}$ , thus we have

$s e c 2 θ = \frac{1}{{(\frac{1}{s e c θ})}^{2} - {(\frac{1}{c s c θ})}^{2}} = \frac{1}{\frac{1}{s e c^{2} θ} - \frac{1}{c s c^{2} θ}} = \frac{1}{\frac{c s c^{2} θ - s e c^{2} θ}{s e c^{2} θ c s c^{2} θ}} = \frac{s e c^{2} θ c s c^{2} θ}{c s c^{2} θ - s e c^{2} θ} .$

The double angle formula for the secant function is given by,

$s e c 2 θ = \frac{s e c^{2} θ c s c^{2} θ}{c s c^{2} θ - s e c^{2} θ}$

Double angle formula for Cosecant

We recall by the definition of the secant function that

$c s c θ = \frac{1}{\sin θ}$ so $c s c 2 θ = \frac{1}{\sin 2 θ}$

but from the double angle formula for sine function, we have $\sin 2 θ = 2 \sin θ \cos θ$ , thus

$c s c 2 θ = \frac{1}{2 \sin θ \cos θ} = \frac{1}{2} \times \frac{1}{\sin θ} \times \frac{1}{\cos θ} = \frac{1}{2} \times c s c θ \times s e c θ c s c 2 θ = \frac{1}{2} c s c θ s e c θ$

The double angle formula for the cosecant function is given by,

$c s c 2 θ = \frac{1}{2} c s c θ s e c θ$

Double angle formula for Cotangent

We recall by the definition of the secant function that

$c o t θ = \frac{1}{\tan θ}$ so $c o t 2 θ = \frac{1}{\tan 2 θ}$

We recall from the double angle formula for the tangent function that $\tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ}$ , we have

$c o t 2 θ = \frac{1}{\frac{2 \tan θ}{1 - \tan^{2} θ}} = \frac{1 - \tan^{2} θ}{2 \tan θ}$

The double angle formula for the cotangent function is given by,

$c o t 2 θ = \frac{1 - \tan^{2} θ}{2 \tan θ}$

Given that $90 ° < θ < 180 °,$ find $s e c 2 θ, c s c 2 θ$ and $c o t 2 θ$ given that

$\sin θ = \frac{4}{5}$

Solution:

We have the value of sinθ, but in order to apply these formulas, we need to find cosθ.

We recall that, $\cos^{2} θ + \sin^{2} θ = 1$

$\cos^{2} θ = 1 - \sin^{2} θ = 1 - {(\frac{4}{5})}^{2} = 1 - \frac{16}{25} = \frac{9}{25}$

Thus $\cos θ = \pm \frac{3}{5}$ , but since $90 ° < θ < 180 °,$ thus $\cos θ = - \frac{3}{5}$ .

$s e c θ = - \frac{5}{3}, c s c θ = \frac{5}{4}$

Hence, we have

$s e c 2 θ = \frac{s e c^{2} θ c s c^{2} θ}{c s c^{2} θ - s e c^{2} θ} = \frac{{(\frac{- 5}{3})}^{2} {(\frac{5}{4})}^{2}}{{(\frac{5}{4})}^{2} - {(\frac{- 5}{3})}^{2}} = \frac{\frac{25}{9} \times \frac{25}{16}}{\frac{25}{16} - \frac{25}{9}} = \frac{\frac{625}{144}}{\frac{225 - 400}{144}} = \frac{625}{- 175} = - \frac{25}{7}$

c s c 2 θ = \frac{1}{\sin 2 θ} = \frac{1}{2 \cos θ \sin θ} = \frac{1}{2 (- \frac{3}{5}) (\frac{4}{5})} = \frac{1}{- \frac{24}{25}} = - \frac{25}{24}

c o t 2 θ = \frac{1 - \tan^{2} θ}{2 \tan θ}

, but

\tan θ = \frac{\sin θ}{\cos θ} = \frac{\frac{4}{5}}{\frac{- 3}{5}} = \frac{- 4}{3}

, thus we have

c o t 2 θ = \frac{1 - {(\frac{- 4}{3})}^{2}}{2 (\frac{- 4}{3})} = \frac{1 - \frac{16}{9}}{- \frac{8}{3}} = \frac{- \frac{7}{9}}{- \frac{8}{3}} = - \frac{7}{9} \times (- \frac{3}{8}) = \frac{7}{24} .

Half-angle formulas

Trigonometric functions can be halved but not in the same manner normal numbers are halved. If you have the expression 6y and you are to half it, it is easy to multiply 6y by 0.5 to get 3y. Note that sin30° is 0.5 and halving the angle gives 15 degrees, but sin15° would not give you 0.25. As normal mathematical operations would multiply 0.5 by 0.5 (half) to give 0.25, trigonometric identities would require their own formula to half its function.

Deriving the half-angle formula for Sine

To find $\sin \frac{θ}{2}$ , we recall first that

$\cos 2 θ = 1 - 2 \sin^{2} θ$

Let $θ = \frac{\emptyset}{2}$ , thus

$\cos (2 \times \frac{ϕ}{2}) = 1 - 2 \sin^{2} \frac{ϕ}{2} \cos ϕ = 1 - 2 \sin^{2} \frac{ϕ}{2}$

In order to isolate $\sin^{2} \frac{ϕ}{2}$ , subtract 1 from both sides, to get

$co ϕ - 1 = 1 - 2 \sin^{2} \frac{ϕ}{2} - 1 - 2 \sin^{2} \frac{ϕ}{2} = \cos ϕ - 1$

We divide both sides of the equation by -2, we get

$\sin^{2} \frac{ϕ}{2} = \frac{1 - \cos ϕ}{2}$

Taking the square root of both sides of the equation, we get

$\sin \frac{ϕ}{2} = \pm \sqrt{\frac{1 - \cos ϕ}{2}}$

The half-angle formula for the sine function is given by,

$\sin \frac{ϕ}{2} = \pm \sqrt{\frac{1 - \cos ϕ}{2}}$

If $\sin θ = \frac{2}{3}$ , and 90°<θ<180°, find $\sin \frac{θ}{2}$ .

Solution:

Since $90 ° < θ < 180 °,$ $45 ° < \frac{θ}{2} < 90 °$ , thus $\sin θ > 0 .$ Hence,

\sin \frac{θ}{2} = \sqrt{\frac{1 - \cos θ}{2}}

Thus, to find $\sin \frac{θ}{2}$ , we need to find cosθ. Recall that

$\cos^{2} θ = 1 - \sin^{2} θ \cos θ = \sqrt{1 - \sin^{2} θ}$

Since

$\sin θ = \frac{2}{3}$

Then,

$\cos θ = \sqrt{1 - {(\frac{2}{3})}^{2}} \cos θ = \sqrt{1 - \frac{4}{9}} \cos θ = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$

Now we can substitute the value of cosθ into our equation

$\sin \frac{θ}{2} = \sqrt{\frac{1 - \cos θ}{2}} \sin \frac{θ}{2} = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{2}} \sin \frac{θ}{2} = \sqrt{\frac{\frac{3 - \sqrt{5}}{3}}{2}} \sin \frac{θ}{2} = \sqrt{\frac{3 - \sqrt{5}}{3} \times \frac{1}{2}} \sin \frac{θ}{2} = \sqrt{\frac{3 - \sqrt{5}}{6}}$

Deriving the half-angle formula for cosine

Recall that

$\cos 2 θ = 2 \cos^{2} θ - 1$

Where,

$θ = \frac{\emptyset}{2}$

Therefore

$\cos (2 \times \frac{\emptyset}{2}) = (2 \times \cos^{2} \frac{\emptyset}{2}) - 1 \cos \emptyset = (2 \times \cos^{2} \frac{\emptyset}{2}) - 1$

Add 1 to both sides of the equation

$\cos \emptyset + 1 = (2 \times \cos^{2} \frac{\emptyset}{2}) - 1 + 1 \cos \emptyset + 1 = 2 \times \cos^{2} \frac{\emptyset}{2}$

Divide both sides by 2

$\frac{\cos \emptyset + 1}{2} = \cos^{2} \frac{\emptyset}{2}$

Find the square root of both sides of the equation

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

Thus

$\cos \frac{θ}{2} = \pm \sqrt{\frac{\cos θ + 1}{2}}$

Given that

$\sin θ = - \frac{3}{4}$

for $180 ° < θ < 270 °$ , find $\cos \frac{θ}{2}$ .

Solution:

To begin, get the value of cosθ.

Note that

$\cos^{2} θ = 1 - \sin^{2} θ$

Thus

$\cos^{2} θ = 1 - {(- \frac{3}{4})}^{2} \cos^{2} θ = 1 - \frac{9}{16} \cos^{2} θ = \frac{7}{16} \cos θ = \pm \frac{\sqrt{7}}{4}$

Recall from the question that θ falls within the third quadrant, hence cosine values would be negative. Thus

$\cos θ = - \frac{\sqrt{7}}{4}$

Note before

$\cos \frac{θ}{2} = \pm \sqrt{\frac{\cos θ + 1}{2}}$

So by substituting the value of cosθ we get

$\cos \frac{θ}{2} = \pm \sqrt{\frac{1 - \frac{\sqrt{7}}{4}}{2}} \cos \frac{θ}{2} = \pm \sqrt{\frac{\frac{4 - \sqrt{7}}{4}}{2}} \cos \frac{θ}{2} = \pm \sqrt{\frac{4 - \sqrt{7}}{4} \times \frac{1}{2}} \cos \frac{θ}{2} = \pm \sqrt{\frac{4 - \sqrt{7}}{8}} \cos \frac{θ}{2} = \pm \frac{\sqrt{4 - \sqrt{7}}}{2 \sqrt{2}}$

Multiply the right hand side of the equation by $\frac{\sqrt{2}}{\sqrt{2}}$ (rationalization of surds)

$\cos \frac{θ}{2} = \pm \frac{\sqrt{4 - \sqrt{7}}}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \cos \frac{θ}{2} = \pm \frac{\sqrt{8 - 2 \sqrt{7}}}{4}$

Now θ has been halved, the conditions would change too by

$180 ° < θ < 270 °$

Angles here fall in the third quadrant.

Dividing that by 2 you have

$90 ° < \frac{θ}{2} < 135 °$

$\frac{θ}{2}$ falls in the second quadrant and cosθ is negative in the second quadrant.

$\cos \frac{θ}{2} = - \frac{\sqrt{8 - 2 \sqrt{7}}}{4}$

Deriving the half-angle formula for tangents

Knowing that

$\tan θ = \frac{\sin θ}{\cos θ} \sin \frac{θ}{2} = \pm \sqrt{\frac{1 - \cos θ}{2}} \cos \frac{θ}{2} = \pm \sqrt{\frac{\cos θ + 1}{2}}$

Then

$\tan \frac{θ}{2} = \frac{\sqrt{\frac{1 - \cos θ}{2}}}{\sqrt{\frac{\cos θ + 1}{2}}} \tan \frac{θ}{2} = \frac{\frac{\sqrt{1 - \cos θ}}{\sqrt{2}}}{\frac{\sqrt{\cos θ + 1}}{\sqrt{2}}} \tan \frac{θ}{2} = \frac{\sqrt{1 - \cos θ}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{\cos θ + 1}} \tan \frac{θ}{2} = \frac{\sqrt{1 - \cos θ}}{\sqrt{\cos θ + 1}}$

Multiply the right-hand side of the equation by $\frac{\sqrt{\cos θ + 1}}{\sqrt{\cos θ + 1}}$ and you would have

$\tan \frac{θ}{2} = \frac{\sqrt{1 - \cos^{2} θ}}{\cos θ + 1}$

Recall that

$1 - \cos^{2} θ = \sin^{2} θ$

Then

$\tan \frac{θ}{2} = \frac{\sqrt{\sin^{2} θ}}{\cos θ + 1} t an \frac{θ}{2} = \frac{\sin θ}{\cos θ + 1}$

Find $\tan \frac{\emptyset}{2}$ when $\tan \emptyset = \frac{4}{3}$ .

Solution:

With the value given, the opposite and adjacent are 4 and 3 respectively. Using the Pythagoras theorem we shall arrive at a value for the hypotenuse.

$h y p o t e n u s e^{2} = o p p o s i t e^{2} + a d j a c e n t^{2} h y p o t e n u s e^{2} = 4^{2} + 3^{2} h y p o t e n u s e^{2} = 25 h y p o t e n u s e = 5$

Now we have the value of the hypotenuse, then

$\sin \emptyset = \frac{4}{5} \cos \emptyset = \frac{3}{5}$

You can now apply the formula

$t an \frac{θ}{2} = \frac{\sin θ}{\cos θ + 1} θ = \emptyset \tan \frac{\emptyset}{2} = \frac{\sin \emptyset}{\cos \emptyset + 1} \tan \frac{\emptyset}{2} = \frac{\frac{4}{5}}{\frac{3}{5} + 1} \tan \frac{\emptyset}{2} = \frac{\frac{4}{5}}{\frac{8}{5}} \tan \frac{\emptyset}{2} = \frac{4}{5} \times \frac{5}{8} \tan \frac{\emptyset}{2} = \frac{1}{2}$

Deriving the half-angle formula for secant, cosecant and cotangent

As mentioned earlier, secant, cosecant, cotangent are the inverse of cosine, sine and tangent respectively. So as to derive their half-angle formulas, you just need to find the multiplicative inverse of the corresponding half-angle formulas. Thus the half-angle formula of secant becomes:

$s e c θ = \frac{1}{\cos θ} \cos \frac{θ}{2} = \pm \sqrt{\frac{\cos θ + 1}{2}} s e c \frac{θ}{2} = \pm \sqrt{\frac{2}{\cos θ + 1}}$

the half-angle formula of cosecant becomes:

$\cos e c θ = \frac{1}{\sin θ} \sin \frac{θ}{2} = \pm \sqrt{\frac{1 - \cos θ}{2}} \cos e c \frac{θ}{2} = \pm \sqrt{\frac{2}{1 - \cos θ}}$

and the half-angle formula of cotangent becomes:

$c o t θ = \frac{1}{\tan θ} \tan \frac{θ}{2} = \frac{\sin θ}{\cos θ + 1} c o t \frac{θ}{2} = \frac{\cos θ + 1}{\sin θ}$

This is the same as

$c o t \frac{θ}{2} = \frac{\cos θ + 1}{\sin θ} c o t \frac{θ}{2} = \frac{\cos θ}{\sin θ} + \frac{1}{\sin θ} c o t \frac{θ}{2} = c o t θ + \cos e c θ$

If $s e c θ = \frac{13}{12}$ , find values for $s e c \frac{θ}{2}$ , $\cos e c \frac{θ}{2}$ and $c o t \frac{θ}{2}$ .

Solution:

Since,

$s e c θ = \frac{13}{12}$

and

$s e c θ = \frac{1}{\cos θ}$

Then,

$\cos θ = \frac{12}{13}$

Knowing that;

$\sin^{2} θ = 1 - \cos^{2} θ \sin θ = \sqrt{1 - \cos^{2} θ}$

Therefore;

$\sin θ = \sqrt{1 - {(\frac{12}{13})}^{2}} \sin θ = \sqrt{1 - \frac{144}{169}} \sin θ = \sqrt{\frac{25}{169}} \sin θ = \frac{5}{13}$

Since the values for cosθ as well as sinθ have been found, it is easier to find the half angles of sec, cosec and cot. Thus half-angle of sec becomes:

$s e c \frac{θ}{2} = \pm \sqrt{\frac{2}{\cos θ + 1}} s e c \frac{θ}{2} = \sqrt{\frac{2}{\frac{12}{13} + 1}} s e c \frac{θ}{2} = \sqrt{\frac{2}{\frac{25}{13}}} s e c \frac{θ}{2} = \sqrt{2 \times \frac{13}{25}} s e c \frac{θ}{2} = \sqrt{\frac{26}{25}} s e c \frac{θ}{2} = \frac{\sqrt{26}}{5}$

For half-angle of cosec

$\cos e c \frac{θ}{2} = \pm \sqrt{\frac{2}{1 - \cos θ}} \cos e c \frac{θ}{2} = \sqrt{\frac{2}{1 - \frac{12}{13}}} \cos e c \frac{θ}{2} = \sqrt{\frac{2}{\frac{1}{13}}} \cos e c \frac{θ}{2} = \sqrt{2 \times \frac{13}{1}} \cos e c \frac{θ}{2} = \sqrt{26}$

And for the half-angle of cot

$c o t \frac{θ}{2} = \frac{\cos θ + 1}{\sin θ} c o t \frac{θ}{2} = \frac{\frac{12}{13} + 1}{\frac{5}{13}} c o t \frac{θ}{2} = \frac{\frac{25}{13}}{\frac{5}{13}} c o t \frac{θ}{2} = \frac{25}{13} \times \frac{13}{5} c o t \frac{θ}{2} = 5$

Applications of double-angle and half-angle formulas

Here are a few examples that show the application of double-angle and half-angle formulas.

Solve the for θ in

$\sin (2 θ) + 4 \sin θ + 2 \cos θ = - 4$

Solution:

Recall that

$\sin 2 θ = 2 \sin θ \cos θ$

Substitute into the equation. Therefore,

$2 \sin θ \cos θ + 4 \sin θ + 2 \cos θ = - 4 2 \sin θ \cos θ + 4 \sin θ + 2 \cos θ + 4 = 0 (2 \sin θ \cos θ + 4 \sin θ) + (2 \cos θ + 4) = 0 2 \sin θ (\cos θ + 2) + 2 (\cos θ + 2) = 0 (2 \sin θ + 2) (\cos θ + 2) = 0$

This means that

$2 \sin θ + 2 = 0 2 \sin θ = - 2 \frac{2 \sin θ}{2} = \frac{- 2}{2} \sin θ = - 1$

$\cos θ + 2 = 0 \cos θ = - 2$

Now in finding θ, we have to find both arcsinθ and arccosθ. Therefore,

$θ = \sin^{- 1} (- 1) θ = 270 °$

However, -2 goes beyond the possible values for arccosθ. Hence,

$\cos θ = - 2$

is invalid

Thus the value of θ is 270°.

$\sin ϕ = \sqrt{\frac{1}{5}}$

find

$\cos \frac{ϕ}{2}$

Solution:

The first thing to do is to find cosϕ. Knowing that

$\cos^{2} ϕ + \sin^{2} ϕ = 1 \cos^{2} ϕ = 1 - \sin^{2} ϕ \cos ϕ = \sqrt{1 - \sin^{2} ϕ}$

Now, substitute the value of sinϕ to find cosϕ.

$\cos ϕ = \sqrt{1 - \sin^{2} ϕ} \cos ϕ = \sqrt{1 - \frac{1}{5}} \cos ϕ = \sqrt{\frac{4}{5}} \cos ϕ = \frac{2}{\sqrt{5}} \cos ϕ = \frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \cos ϕ = \frac{2 \sqrt{5}}{5}$

Recall that

$\cos \frac{ϕ}{2} = \pm \sqrt{\frac{\cos θ + 1}{2}}$

Hence,

$\cos \frac{ϕ}{2} = \pm \sqrt{\frac{\frac{2 \sqrt{5}}{5} + 1}{2}} \cos \frac{ϕ}{2} = \pm \sqrt{\frac{\frac{2 \sqrt{5} + 5}{5}}{2}} \cos \frac{ϕ}{2} = \pm \sqrt{\frac{2 \sqrt{5} + 5}{5} \times \frac{1}{2}} \cos \frac{ϕ}{2} = \pm \sqrt{\frac{2 \sqrt{5} + 5}{10}}$

Double-Angle and Half-Angle Formulas - Key takeaways

A trigonometric function cannot be halved or doubled using normal arithmetic methods. Rather, some formulas are needed to carry out such operations.
To double the angle of sine functions: $si n 2 θ = 2 \sin θ \cos θ$
To double the angle of cosine functions, any of the following formulas can be used. $\cos 2 θ = \cos^{2} θ - \sin^{2} θ$ or $\cos 2 θ = 1 - 2 \sin^{2} θ$ or $\cos 2 θ = 2 \cos^{2} θ - 1$ .
To double the angle of tangent functions: $\tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ}$ .
To double the angle of secant functions: $s e c 2 θ = \frac{1}{\cos^{2} θ - \sin^{2} θ}$ .
To double the angle of cosecant functions: $\cos e c 2 θ = \frac{1}{2 \sin θ \cos θ}$ .
To double the angle of cotangent functions: $c o t 2 θ = \frac{1 - \tan^{2} θ}{2 \tan θ}$ .
To find the half-angle of sine functions use: $\sin \frac{θ}{2} = \pm \sqrt{\frac{1 - \cos θ}{2}}$ .
To find the half-angle of cosine function use: $\cos \frac{θ}{2} = \pm \sqrt{\frac{\cos θ + 1}{2}}$
To find the half-angle of tangent functions use: $t an \frac{θ}{2} = \frac{\sin θ}{\cos θ + 1}$ .
To find the half-angle of secant functions use: $s e c \frac{θ}{2} = \pm \sqrt{\frac{2}{\cos θ + 1}}$ .
To find the half-angle of cosecant functions use: $\cos e c \frac{θ}{2} = \pm \sqrt{\frac{2}{1 - \cos θ}}$ .
To find the half-angle of cotangent functions use: $c o t \frac{θ}{2} = \frac{\cos θ + 1}{\sin θ}$ .

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Frequently Asked Questions about Double Angle and Half Angle Formulas

What are double-angle and half-angle formulas?

Double-angle and half-angle formulas are formulas used in finding the trigonometric values for angles that are doubled or halved.

How are double angle and half angle formulas used?

Double-angle and half-angle formulas are used by applying directly the formulas when finding double/half-angle trigonometric identities.

What is an example of double-angle and half-angle formulas?

An example of a double angle formula is sin2A = 2sinAcosA, while that of half-angle is sinA/2 = square root ((1-cosA)/2).

How do you derive double angle formulas?

To derive double angle formulas you would need to apply the sum of trigonometric function formulas.

What are the types of double angle formulas?

The types of double angle formulas are those of sin, cos, sec, cosec, tan and cot.

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Double Angle and Half Angle Formulas

StudySmarter Editorial Team

Double-angle formulas

Deriving the double-angle formula for Sine function

Deriving the double-angle formula for Cosine function

Deriving the double-angle formula for Tangent function

Deriving the double-angle formulas for Secant, Cosecant and Cotangent functions

Double angle formula for Secant

Double angle formula for Cosecant

Double angle formula for Cotangent

Half-angle formulas

Deriving the half-angle formula for Sine

Deriving the half-angle formula for cosine

Deriving the half-angle formula for tangents

Deriving the half-angle formula for secant, cosecant and cotangent

Applications of double-angle and half-angle formulas

Double-Angle and Half-Angle Formulas - Key takeaways

Learn with 0 Double Angle and Half Angle Formulas flashcards in the free StudySmarter app

Frequently Asked Questions about Double Angle and Half Angle Formulas

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Double Angle and Half Angle Formulas

StudySmarter Editorial Team

Double-angle formulas

Deriving the double-angle formula for Sine function

Deriving the double-angle formula for Cosine function

Deriving the double-angle formula for Tangent function

Deriving the double-angle formulas for Secant, Cosecant and Cotangent functions

Double angle formula for Secant

Double angle formula for Cosecant

Double angle formula for Cotangent

Half-angle formulas

Deriving the half-angle formula for Sine

Deriving the half-angle formula for cosine

Deriving the half-angle formula for tangents

Deriving the half-angle formula for secant, cosecant and cotangent

Applications of double-angle and half-angle formulas

Double-Angle and Half-Angle Formulas - Key takeaways

Learn with 0 Double Angle and Half Angle Formulas flashcards in the free StudySmarter app

Frequently Asked Questions about Double Angle and Half Angle Formulas

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