Trigonometric Identities

LHS * 1/sin θ=RHS* 1/sin θ

1=opp/hyp (1/sin θ)

LHS * hyp/opp=RHS* hyp/opp

hyp/opp=1/sin θ

Rearrange equation

1/sin θ=hyp/opp

It has been found that 1sin θ, which is the reciprocal of sin θ, is equal to hypopp. By the definition, the cosecant of θ is also the ratio of the lengths of the hypotenuse and the opposite side to ∠ θ. Therefore, by the Transitive Property of Equality, 1sin θ is equal to csc θ. 1/sin θ= hyp/opp csc θ= hyp/opp ⇓ csc θ=1/sin θ By following a similar procedure, the other two identities for secant and cotangent can be proven.

Rule

Tangent and Cotangent Identities

The tangent of an angle θ can be expressed as the ratio of the sine of θ to the cosine of θ.

tanθ = sinθ/cosθ

Similarly, the cotangent of θ can be expressed as the ratio of the cosine of θ to the sine of θ.

cotθ = cosθ/sinθ

Proof

Tangent Identity

Two proofs will be written for this identity, one using a right triangle and the other using a unit circle.

Right Triangle

In a right triangle, the tangent of an angle θ is defined as the ratio of the length of the opposite side k to the length of the adjacent side l.

At the same time, the sine and cosine of θ can be written as follows. sinθ=k/m cosθ=l/m By manipulating the right-hand side of the equation tan θ = kl, the tangent can be expressed as the sine over the cosine of θ.

tanθ = k/l

ReduceFrac

a/b=.a /m./.b /m.

tanθ = k/m/l/m

SubstituteII

k/m= sin(θ), l/m= cos(θ)

tanθ = sinθ/cosθ

The proof of the identity is complete.

Unit Circle

Consider a unit circle and an angle θ in standard position.

It is known that the point of intersection P of the terminal side of the angle and the unit circle has coordinates (cos θ , sin θ).

Draw a right triangle using the origin and P(cos θ , sin θ) as two of its vertices. The length of the hypotenuse is 1 and the lengths of the legs are sin θ and cos θ.

As shown previously, the tangent of a right triangle is defined as the ratio of the length of the opposite side — in this case, sin θ — to the length of the adjacent side, which here is cos θ. tan θ =sin θ/cos θ ✓

Proof

Cotangent Identity

Two more proofs will be written for this identity, one of them using just a right triangle and the other using a unit circle.

Right Triangle

In a right triangle, the cotangent of an angle θ is defined as the ratio of the length of the adjacent side l to the length of the opposite side k.

Additionally, the sine and cosine of θ can be written as follows. sinθ=k/m cosθ=l/m By manipulating the right-hand side of the equation cot θ = lk, the cotangent can be expressed as the cosine over the sine of θ.

cotθ = l/k

ReduceFrac

a/b=.a /m./.b /m.

cotθ = l/m/k/m

SubstituteII

l/m= cos(θ), k/m= sin(θ)

cotθ = cosθ/sinθ

This proof is complete.

Unit Circle

Using a unit circle, it has been already proven that the tangent of an angle θ is the ratio of the sine to the cosine of the angle θ. tan θ = sin θ/cos θ By manipulating the above equation, it can be shown that the cotangent of θ is the ratio of the cosine of θ to the sine of θ.

tan θ = sin θ/cos θ

LHS * cos θ/sin θ=RHS* cos θ/sin θ

tan θ (cos θ/sin θ)=1

.LHS /tan θ.=.RHS /tan θ.

cos θ/sin θ=1/tan θ

cot(θ) = 1/tan(θ)

cos θ/sin θ=cot θ

Rearrange equation

cot θ =cos θ/sin θ ✓

This proof is complete.

Rule

Pythagorean Identities

For any angle θ, the following trigonometric identities hold true.

sin^2 θ + cos^2 θ = 1

1 + tan^2 θ = sec^2 θ

1 + cot^2 θ = csc^2 θ

Proof

For Acute Angles

Consider a right triangle with a hypotenuse of 1.

By recalling the sine and cosine ratios, the lengths of the opposite and adjacent sides to ∠ θ can be expressed in terms of the angle.

	Definition	Substitute	Simplify
sin θ	Length of oppositeside to∠ θ/Hypotenuse	opp/1	opp
cos θ	Length of adjacentside to∠ θ/Hypotenuse	adj/1	adj

It can be seen that if the hypotenuse of a right triangle is 1, the sine of an acute angle is equal to the length of its opposite side. Similarly, the cosine of the angle is equal to the length of its adjacent side.

By the Pythagorean Theorem, the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Therefore, for the above triangle, the sum of the squares of sin θ and cos θ is equal to the square of 1.

sin^2 θ+ cos^2 θ&= 1^2 ⇓ & sin^2 θ+cos^2 θ&=1

Since cos θ represents a side length, it is not 0. Therefore, by diving both sides of the above equation by cos^2θ, the second identity can be obtained.

sin^2 θ+cos^2 θ=1

.LHS /cos^2θ.=.RHS /cos^2θ.

sin^2 θ+cos^2 θ/cos^2θ=1/cos^2θ

▼

Simplify

WriteSumFrac

Write as a sum of fractions

sin^2 θ/cos^2θ+cos^2 θ/cos^2θ=1/cos^2θ

QuotOne

a/a=1

sin^2 θ/cos^2θ+1=1/cos^2θ

WritePow

Write as a power

sin^2 θ/cos^2θ+1=1^2/cos^2θ

a^m/b^m=(a/b)^m

(sin θ/cosθ)^2+1=(1/cosθ)^2

sin θ/cos θ=tan θ

tan ^2 θ+1=(1/cosθ)^2

1/cos θ=sec θ

tan ^2 θ+1=sec ^2 θ

CommutativePropAdd

Commutative Property of Addition

1+tan ^2 θ=sec ^2 θ

The second identity was obtained.

1+tan ^2 θ=sec ^2 θ

Since sin θ represents a side length, it is not 0. Therefore, by dividing both sides of sin ^2 θ +cos ^2 θ = 1 by sin ^2 θ, the third identity can be proven.

sin^2 θ+cos^2 θ=1

.LHS /sin^2θ.=.RHS /sin^2θ.

sin^2 θ+cos^2 θ/sin^2θ=1/sin^2θ

▼

Simplify

WriteSumFrac

Write as a sum of fractions

sin^2 θ/sin^2θ+cos^2 θ/sin^2θ=1/sin^2θ

QuotOne

a/a=1

1+cos^2 θ/sin^2θ=1/sin^2θ

WritePow

Write as a power

1+cos^2 θ/sin^2θ=1^2/sin^2θ

a^m/b^m=(a/b)^m

1+(cos θ/sinθ)^2=(1/sinθ)^2

cos θ/sin θ=cot θ

1+cot ^2 θ=(1/sinθ)^2

1/sin θ=csc θ

1+cot ^2 θ=csc ^2 θ

Finally, the third identity was obtained.

1+cot ^2 θ=csc ^2 θ

Proof

For Any Angle

The first identity can be shown using the unit circle and the Pythagorean Theorem. Consider a point (x,y) on the unit circle in the first quadrant, corresponding to the angle θ. A right triangle can be constructed with θ.

By the Pythagorean Theorem, the sum of the squares of x and y equals 1. x^2 + y^2 = 1 In fact, this is true not only for points in the first quadrant, but for every point on the unit circle. Recall that, for points (x,y) on the unit circle corresponding to angle θ, it is known that x = cos θ and that y = sin θ. By substituting these expressions into the equation, the first identity can be obtained.

cos^2 θ + sin^2 θ = 1

Dividing both sides by either cos^2 θ or sin^2 θ leads to two variations of the Pythagorean Identity.

Rule

Negative Angle Identities

The function y = sin(x) has odd symmetry, and y = cos(x) has even symmetry, which can be seen from their graphs. As a result, the corresponding identities hold true.

sin(- θ) = - sinθ cos(- θ) = cosθ tan(- θ) = - tanθ

Proof

sin(- θ) = - sinθ and cos(- θ) = cosθ

The identities will be proven using a unit circle. Consider an arbitrary ∠ θ on a unit circle. Let A be the point that the angle forms on the circle.

Recall that the values on the x-axis are represented by cosine and the value on the y-axis are represented by sine. Therefore, the coordinates of A are (cosθ,sinθ).

Next, the point A can be reflected over the x-axis.

Since A' is the same distance from the y-axis as A, its x-coordinate is also cosθ. Note that a reflection is a congruent transformation, so A and A' are equidistant from the x-axis too. However, A' is below the x-axis, so its y-coordinate is a negative sinθ.

Additionally, after the reflection, the angle between A', the origin, and the x-axis is - θ. This means that the lengths of the newly created horizontal and vertical segments are cos(- θ) and sin(- θ). Therefore, the coordinates of A' can also be written as (cos(- θ),sin(- θ)).

This way there are two pairs of coordinates of the same point A'. This allows to conclude the following. A'( cos(- θ), sin(- θ)) &and A'( cosθ, - sinθ) & ⇓ cos(- θ)&= cos θ sin(- θ)&= - sinθ

Proof

tan(- θ) = - tanθ

By expressing tan(- θ) using sine and cosine, this identity can be shown.

tan(- θ)

tan(θ)=sin(θ)/cos(θ)

sin(- θ)/cos(- θ)

sin(- θ)=- sin(θ)

- sin(θ)/cos(- θ)

cos(- θ)=cos(θ)

- sin(θ)/cos(θ)

MoveNegNumToFrac

Put minus sign in front of fraction

- sin(θ)/cos(θ)

sin(θ)/cos(θ)=tan(θ)

- tan(θ)

Rule

Cofunction Identities

For any angle θ, the following trigonometric identities hold true.

sin ( π/2 - θ ) = cosθ

\cos\left(\dfrac \pi 2 - \theta\right) = \sin\theta

\tan\left(\dfrac \pi 2 - \theta\right) = \cot\theta

Proof

For Acute Angles

Consider a right triangle. The measure of its right angle is 90^(∘) or π2 radians. Let θ be the radian measure of one of the acute angles. Since the sum of two acute angles in a right triangle is π2, the measure of the third acute angle must be π2-θ.

$A right triangle with the side lengths a, b, and c, and two acute angles theta and pi\2-\theta$

Let also a, b, and c represent the side lengths of the triangle. In this case, cosine of θ can be expressed as the ratio of the lengths of the angle's adjacent side and the hypotenuse. cosθ = b/c At the same time, sine of the opposite angle can be expressed as the ratio of the lengths of the angle's opposite side and the hypotenuse. sin ( π/2 - θ ) = b/c Since the right-hand sides of the equations are equal, by the Transitive Property of Equality, the left-hand sides are also equal.

sin ( π/2 - θ ) = cosθ

This identity is true for all angles, not just those that make it possible to construct a right triangle. Using similar reasoning, the corresponding identities for cosine and tangent can be proven.

Proof

For Any Angle

The given identities can be proven using the Angle Sum and Difference Identities for any angle measure. Consider the identity for the sine of the difference between two angles. sin(x-y)=sinxcosy-cosxsiny This identity will be applied to the left-hand side of the first identity.

sin ( π/2 - θ )

sin(α-β)=sin(α)cos(β)-cos(α)sin(β)

sin ( π/2)cos(θ)-cos ( π/2)sin(θ)

\ifnumequal{90}{0}{\sin\left(0\right)=0}{}\ifnumequal{90}{30}{\sin\left(\dfrac{\pi}{6}\right)=\dfrac{1}{2}}{}\ifnumequal{90}{45}{\sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{90}{60}{\sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{90}{90}{\sin\left(\dfrac{\pi}{2}\right)=1}{}\ifnumequal{90}{120}{\sin\left(\dfrac{2\pi}{3}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{90}{135}{\sin\left(\dfrac{3\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{90}{150}{\sin\left(\dfrac{5\pi}{6}\right)=\dfrac{1}{2}}{}\ifnumequal{90}{180}{\sin\left(\pi\right)=0}{}\ifnumequal{90}{210}{\sin\left(\dfrac{7\pi}6\right)=\text{-} \dfrac 1 2}{}\ifnumequal{90}{225}{\sin\left(\dfrac{5\pi}{4}\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{90}{240}{\sin\left(\dfrac{4\pi}3\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{90}{270}{\sin\left(\dfrac{3\pi}{2}\right)=\text{-} 1}{}\ifnumequal{90}{300}{\sin\left(\dfrac{5\pi}3\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{90}{315}{\sin\left(\dfrac{7\pi}4\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{90}{330}{\sin\left(\dfrac{11\pi}6\right)=\text{-} \dfrac 1 2}{}\ifnumequal{90}{360}{\sin\left(2\pi\right)=0}{}

1*cos(θ)-cos ( π/2)sin(θ)

MultByOne

a * 1=a

cos(θ)-cos ( π/2)sin(θ)

\ifnumequal{90}{0}{\cos\left(0\right)=1}{}\ifnumequal{90}{30}{\cos\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{90}{45}{\cos\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{90}{60}{\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}}{}\ifnumequal{90}{90}{\cos\left(\dfrac{\pi}{2}\right)=0}{}\ifnumequal{90}{120}{\cos\left(\dfrac{2\pi}{3}\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{90}{135}{\cos\left(\dfrac{3\pi}{4}\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{90}{150}{\cos\left(\dfrac{5\pi}{6}\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{90}{180}{\cos\left(\pi\right)=\text{-} 1}{}\ifnumequal{90}{210}{\cos\left(\dfrac{7\pi}6\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{90}{225}{\cos\left(\dfrac{5\pi}{4}\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{90}{240}{\cos\left(\dfrac{4\pi}3\right)=\text{-} \dfrac {1}2}{}\ifnumequal{90}{270}{\cos\left(\dfrac{3\pi}{2}\right)=0}{}\ifnumequal{90}{300}{\cos\left(\dfrac{5\pi}3\right)=\dfrac{1}2}{}\ifnumequal{90}{315}{\cos\left(\dfrac{7\pi}4\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{90}{330}{\cos\left(\dfrac{11\pi}6\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{90}{360}{\cos\left(2\pi\right)=1}{}

cos(θ)-0*sin(θ)

ZeroPropMult

Zero Property of Multiplication

cos(θ)-0

IdPropAdd

Identity Property of Addition

cos(θ)

Therefore, the Cofunction Identity for sine was obtained.

sin ( π/2 - θ )=cos(θ)

Rule

Angle Sum and Difference Identities

To evaluate trigonometric functions of the sum of two angles, the following identities can be applied.

sin(x+y) &= sin x cos y + cos x sin y [0.8em] cos(x+y) &= cos x cos y - sin x sin y [0.8em] tan(x+y) &= tan x + tan y/1 - tan x tan y

There are also similar identities for the difference of two angles.

sin(x-y) &= sin x cos y - cos x sin y [0.8em] cos(x-y) &= cos x cos y + sin x sin y [0.8em] tan(x-y) &= tan x - tan y/1 + tan x tan y

These identities are useful when finding the exact value of the sine, cosine, or tangent of a given angle.

Proof

Let △ AFD be a right triangle with hypotenuse 1 and an acute angle with measure x+y.

By definition, the sine of an angle is the ratio between the lengths of the opposite side and the hypotenuse. sin(x+y) = DF/1 ⇓ DF = sin(x+y) The idea now is to rewrite DF in terms of sin x, sin y, cos x, and cos y. To do it, draw a ray so that ∠ A is divided into two angles with measures x and y. Let C be a point on this ray such that △ ACD and △ ABC are right triangles.

Consider △ ACD. By calculating the sine and cosine of x, the legs of this triangle can be rewritten. sin x = DC/1 ⇒ DC = sin x [0.8em] cos x = AC/1 ⇒ AC = cos x Now consider △ ABC. Knowing that AC=cos x, the sine of y can be used to write BC in terms of x and y.

sin y = BC/AC

▼

Solve for BC

Substitute

AC= cos x

sin y = BC/cos x

LHS * cos x=RHS* cos x

cos xsin y=BC

Rearrange equation

BC=cos xsin y

Let G be the point of intersection between FD and AC. Notice that ∠ AGF ≅ ∠ DGC by the Vertical Angles Theorem.

By the Third Angle Theorem, it is known that ∠ GAF ≅ ∠ GDC. Therefore, m∠ GDC = y.

Since the purpose is to rewrite DF, plot a point E on DF such that EC ∥ AB. This way a rectangle ECBF is formed. The opposite sides of a rectangle have the same length, so EF and CB are equal. Also, CE⊥ DF makes △ CED a right triangle.

Consequently, EF = cos x sin y and DE can be written in terms of sin x and cos y using the cosine ratio. cos y = DE/sin x ⇓ DE = sin x cos y Finally, by the Segment Addition Postulate, DF is equal to the sum of DE and EF. All these lengths have been rewritten in terms of the sine and cosine of x and y. DF = DE+ EF ⇓ sin(x+y) = sin x cos y + cos xsin y This concludes the proof of the first identity. The other identities can be proven using similar reasoning.

Extra

Calculating sin 120^(∘)

Consider the following process for calculating the exact value of sin 120^(∘).

To be able to use the angle sum identities, the angle 120^(∘) needs to be rewritten as the sum of two angles for which the sine and cosine are known. For example, 120^(∘) can be rewritten as 90^(∘)+ 30^(∘).
Use the first formula for the angle sum.
Based on the trigonometric ratios of common angles, it is known that sin 90^(∘)=1, sin 30^(∘)= 12, cos 90^(∘)=0, and cos 30^(∘)= sqrt(3)2.

Following these three steps, the value of sin 120^(∘) can be found.

sin 120^(∘)

Rewrite 120^(∘) as 90^(∘)+30^(∘)

sin(90^(∘)+30^(∘))

sin(x+y) = sin x cos y + cos x sin y

sin 90^(∘) * cos 30^(∘) + cos 90^(∘) * sin 30^(∘)

SubstituteValues

Substitute values

1 * sqrt(3)/2 + 0 * 1/2

▼

Simplify

OneMult

1* a=a

sqrt(3)/2 + 0 * 1/2

ZeroPropMult

Zero Property of Multiplication

sqrt(3)/2 + 0

IdPropAdd

Identity Property of Addition

sqrt(3)/2

Notice that 120^(∘) could also be rewritten as 60^(∘) + 60^(∘), because sin 60^(∘) and cos 60^(∘) are known values.

Rule

Double-Angle Identities

The double-angle identities materialize when two angles with the same measure are substituted into the angle sum identities.

cl Sine: & sin 2θ=2sin θ cos θ [0.7em] & cos 2θ =cos^2 θ - sin^2 θ Cosine: & cos 2θ = 2cos^2 θ - 1 & cos 2θ = 1 - 2sin^2 θ [0.2cm] Tangent: & tan 2θ =2tanθ/1-tan^2 θ

These identities simplify calculations when evaluating trigonometric functions of twice an angle measure.

Proof

Double-Angle Identities

Start by writing the Angle Sum Identity for sine and cosine. sin(x + y) = sin x cos y + cos x sin y cos(x + y) = cos x cos y - sin x sin y

Let x=θ and y=θ. With this, x+y becomes 2θ. Then, these two formulas can be rewritten in terms of θ. sin 2θ = sinθ cosθ + cosθ sinθ cos 2θ = cosθ cosθ - sinθ sinθ

Sine Identity

Approaching the first equation, the Commutative Property of Multiplication can be applied to the second term of its right-hand side. Then, by adding the terms on the right-hand side of this equation, the formula for sin 2θ is obtained.

sin 2θ = sinθ cosθ + sinθ cosθ ⇓ sin 2θ = 2sinθ cosθ ✓

Cosine Identities

Approaching the second equation, the Product of Powers Property can be used to rewrite its right-hand side. By doing this, the first identity for the cosine of the double of an angle is obtained.

cos 2θ = cosθ cosθ - sinθ sinθ ⇓ cos 2θ = cos^2θ - sin^2θ ✓

Now, recall that, by the Pythagorean Identity, the sine square plus the cosine square of the same angle equals 1. From this identity, two different equations can be set. sin^2 θ + cos^2 θ = 1 ⇓ sin^2 θ = 1-cos^2 θ & (I) cos^2 θ = 1-sin^2 θ & (II) Next, substitute Equation (I) into the first identity for the cosine.

cos 2θ = cos^2θ - sin^2θ

▼

Substitute 1-cos^2 θ for sin^2 θ and simplify

Substitute

sin^2 θ= 1-cos^2 θ

cos 2θ = cos^2θ - ( 1-cos^2 θ)

Distr

Distribute -1

cos 2θ = cos^2θ - 1 + cos^2 θ

AddTerms

Add terms

cos 2θ = 2cos^2θ - 1 ✓

That way, the second identity for the cosine has been obtained. To obtain the third cosine identity, substitute Equation (II) into the first identity for the cosine.

cos 2θ = cos^2θ - sin^2θ

▼

Substitute 1-sin^2 θ for cos^2 θ and simplify

Substitute

cos^2 θ= 1-sin^2 θ

cos 2θ = 1-sin^2 θ - sin^2θ

SubTerms

Subtract terms

cos 2θ = 1-2sin^2 θ ✓

Tangent Identity

To prove the tangent identity, start by rewriting tan 2θ in terms of sine and cosine. tan 2θ = sin 2θ/cos 2θ Next, substitute the first sine identity in the numerator and the first cosine identity in the denominator. tan 2θ = 2sin θ cos θ/cos^2 θ - sin^2 θ Then, divide the numerator and denominator by cos^2θ. tan 2θ = 2sin θ cos θcos^2θ/cos^2 θ - sin^2 θcos^2θ Finally, simplifying the right-hand side the tangent identity will be obtained.

tan 2θ = 2sin θ cos θcos^2θ/cos^2 θ - sin^2 θcos^2θ

▼

Simplify right-hand side

WriteDiffFrac

Write as a difference of fractions

tan 2θ = 2sin θ cos θcos^2θ/cos^2 θcos^2θ - sin^2 θcos^2θ

CrossCommonFac

Cross out common factors

tan 2θ = 2sin θ cos θcos^2θ/cos^2 θcos^2θ - sin^2 θcos^2θ

CancelCommonFac

Cancel out common factors

tan 2θ = 2sin θcosθ/1 - sin^2 θcos^2θ

a^m/b^m=(a/b)^m

tan 2θ = 2sin θcosθ/1 - ( sin θcosθ)^2

MovePartNumLeft

a* b/c=a*b/c

tan 2θ = 2* sin θcosθ/1 - ( sin θcosθ)^2

sin(θ)/cos(θ)=tan(θ)

tan 2θ = 2tanθ/1-tan^2θ ✓

Extra

Calculating cos 120^(∘)

To calculate the exact value of cos 120^(∘), these steps can be followed.

To be able to use the double-angle identities, the angle 120^(∘) needs to be rewritten as 2 multiplied by another angle. Therefore, rewrite 120^(∘) as 2* 60^(∘).
Use the second formula for the cosine of twice an angle.
Based on the trigonometric ratios of common angles, it is known that cos 60^(∘) = 12.

Following these three steps, the value of cos 120^(∘) can be found.

cos 120^(∘)

Rewrite 120^(∘) as 2* 60^(∘)

cos (2* 60^(∘))

cos 2θ = 2cos^2 θ - 1

2cos^2 (60^(∘)) - 1

cos 60^(∘) = 1/2

2(1/2)^2 - 1

▼

Simplify

CalcPow

Calculate power

2* 1/4 - 1

Multiply

1/2 - 1

Rewrite 1 as 2/2

1/2 - 2/2

SubFrac

Subtract fractions

1-2/2

SubTerms

Subtract terms

-1/2

MoveNegNumToFrac

Put minus sign in front of fraction

-1/2

Rule

Half-Angle Identities

The half-angle identities are special cases of angle difference identities. To evaluate trigonometric functions of half an angle, the following identities can be applied.

sin θ/2 &= ±sqrt(1 - cosθ/2) [0.75em] cos θ/2 &= ±sqrt(1 + cosθ/2) [0.75em] tan θ/2 &= ±sqrt(1 - cosθ/1 + cosθ)

The sign of each formula is determined by the quadrant where the angle θ2 lies.

These identities are useful when finding the exact value of the sine, cosine, or tangent at a given angle.

Proof

Half-Angle Identities

First, write two of the Double-Angle Identities for cosine. cos 2x &= 1 - 2sin^2 x cos 2x &= 2cos^2 x - 1

Sine Identity

Start by solving the first identity written above for sin x.

cos 2x = 1 - 2sin^2 x

▼

Solve for sin x

SubEqn

LHS-1=RHS-1

cos 2x - 1 = -2sin^2 x

.LHS /(-2).=.RHS /(-2).

cos 2x - 1/-2 = sin^2 x

Rearrange equation

sin^2 x = cos 2x - 1/-2

MoveNegDenomToFrac

Put minus sign in front of fraction

sin^2 x = -cos 2x - 1/2

DistrNegSignSwap

-(b-a)=a-b

sin^2 x = 1-cos 2x/2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sin x = ±sqrt(1-cos 2x/2)

Next, substitute θ2 for x to obtain the half-angle identity for the sine.

sin θ/2 &= ±sqrt(1-cos(2* θ2)/2) & ⇓ sin θ/2 &= ±sqrt(1-cosθ/2) ✓

Cosine Identity

Start by solving the second identity written at the beginning for cos x.

cos 2x = 2cos^2 x - 1

▼

Solve for cos x

AddEqn

LHS+1=RHS+1

1+cos 2x = 2cos^2 x

.LHS /2.=.RHS /2.

1+cos 2x/2 = cos^2 x

Rearrange equation

cos^2 x = 1+cos 2x/2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

cos x = ±sqrt(1+cos 2x/2)

Next, substitute θ2 for x to obtain the half-angle identity for the cosine.

cos θ/2 &= ±sqrt(1+cos (2* θ2)/2) & ⇓ cos θ/2 &= ±sqrt(1+cosθ/2) ✓

Tangent Identity

To derive the tangent identity, start by recalling the definition of the tangent ratio. tan x = sin x/cos x Next, substitute θ2 for x. tan θ2 = sin ( θ2)/cos ( θ2) Finally, substitute the half-identities for the sine and cosine into the equation above and simplify the right-hand side.

tan θ/2 = sin ( θ2)/cos ( θ2)

SubstituteII

sin θ2= ±sqrt(1 - cosθ/2), cos θ/2= ±sqrt(1 + cosθ/2)

tan θ/2 = ±sqrt(1 - cosθ2)/±sqrt(1 + cosθ2)

▼

Simplify right-hand side

DivSqrt

sqrt(a)/sqrt(b)=sqrt(a/b)

tan θ/2 = ±sqrt(1-cosθ2/1+cosθ2)

DivFracByFracD

.a /b./.c /d.=a/b*d/c

tan θ/2 = ±sqrt(1-cosθ/2 * 2/1+cosθ)

CrossCommonFac

Cross out common factors

tan θ/2 = ±sqrt(1-cosθ/2 * 2/1+cosθ)

CancelCommonFac

Cancel out common factors

tan θ/2 = ±sqrt(1-cosθ/1 * 1/1+cosθ)

MultFrac

Multiply fractions

tan θ/2 = ±sqrt(1-cosθ/1+cosθ) ✓

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Calculating cos 15^(∘)

Consider the calculation of the exact value of cos 15^(∘).

To be able to use the half-angle identities, the angle 15^(∘) needs to be rewritten as a certain angle divided by 2. Therefore, rewrite 15^(∘) as 30^(∘)2.
Based on the trigonometric ratios of common angles, it is known that cos 30^(∘) = sqrt(3)2.
According to the diagram of the quadrants, an angle that measures 15^(∘) is in the first quadrant. Therefore, the cosine ratio is positive.

With these three steps and the second identity in mind, the value of cos 15^(∘) can be found.

cos 15^(∘)