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Taking a closer look at how the trigonometric functions are defined, it is possible to find a range of identities relating the different trigonometric functions. These are highly useful when rewriting and manipulating trigonometric expressions.

The trigonometric ratios cosecant, secant, and cotangent are reciprocals of sine, cosine, and tangent, respectively. Therefore, they can be defined using their respective reciprocal.

$cscθ=sinθ1 $

$secθ=cosθ1 $

$cotθ=tanθ1 $

Consider a right triangle with the three sides labeled with respect to an acute angle $θ.$

Next, the sine, cosine, tangent, cosecant, secant, and cotangent ratios are written. $sinθ=hypopp cosθ=hypadj tanθ=adjopp cscθ=opphyp secθ=adjhyp cotθ=oppadj $ The reciprocal of the sine ratio will now be calculated.$sinθ=hypopp $

Solve for $sinθ1 $

MultEqn

$LHS⋅sinθ1 =RHS⋅sinθ1 $

$1=hypopp (sinθ1 )$

MultEqn

$LHS⋅opphyp =RHS⋅opphyp $

$opphyp =sinθ1 $

RearrangeEqn

Rearrange equation

$sinθ1 =opphyp $

Both tangent and cotangent can be alternatively defined using sine and cosine.

Tangent is defined as $tan(θ)=ba $ in the following triangle.

By manipulating the right-hand side, it can be expressed as sine over cosine instead.

$tan(θ)=ba $

ReduceFrac

$ba =b/ca/c $

$tan(θ)=b/ca/c $

Substitute

$a/c=sin(θ)$

$tan(θ)=b/csin(θ) $

Substitute

$b/c=cos(θ)$

$tan(θ)=cos(θ)sin(θ) $

Cotangent is the reciprocal of tangent, leading to the following identity.

$cot(θ)=sin(θ)cos(θ) $

$sin_{2}(θ)+cos_{2}(θ)=1$

This 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, it follows that $x_{2}+y_{2}=1.$ In fact, this is true for every point on the Unit Circle, not only for points in the first quadrant. Recall that, for points $(x,y)$ on the Unit Circle corresponding to the angle $θ,$ $x=cos(θ)andy=sin(θ).$ Substituting these into the previous equality gives the Pythagorean Identity, short of rearranging the terms: $cos_{2}(θ)+sin_{2}(θ)=1.$ Note that $cos_{2}(θ)$ represents $(cos(θ))_{2},$ and similarly for $sin_{2}(θ).$ Dividing both sides by either $cos_{2}(θ)$ or $sin_{2}(θ)$ leads to two variations of the Pythagorean Identity.

$1+tan_{2}(θ)=sec_{2}(θ)$

$1+cot_{2}(θ)=csc_{2}(θ)$

Studying angles and their trigonometric values in a right triangle more closely reveals further relationships between sine and cosine, and between tangent and cotangent.

The angles in a right triangle, expressed in radians, is $2π ,$ the unknown $θ,$ and $2π −θ.$ The third angle comes from that the sum of the angles in a triangle is $π.$

Cosine of $θ$ can now be expressed using the angle's adjacent side and the hypotenuse: $cos(θ)=cb .$ Sine of the opposite angle can be similarly expressed as $sin(2π −θ)=cb .$ As the right-hand sides are equal, so must the left-hand sides be, leading to the identity.

$sin(2π −θ)=cos(θ)$

This identity is true for all angles, not just those possible to construct in a right triangle.

Using similar reasoning, the following two identities can also be found.

$cos(2π −θ)=sin(θ)$

$tan(2π −θ)=cot(θ)$

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

$sin(-θ)=-sin(θ)$

$cos(-θ)=cos(θ)$

Using these two identities, it can be shown that $y=tan(x)$ has odd symmetry.

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

$tan(-θ)$

$tan(θ)=cos(θ)sin(θ) $

$cos(-θ)sin(-θ) $

$sin(-θ)=-sin(θ)$

$cos(-θ)-sin(θ) $

$cos(-θ)=cos(θ)$

$cos(θ)-sin(θ) $

MoveNegNumToFrac

Put minus sign in front of fraction

$-cos(θ)sin(θ) $

$cos(θ)sin(θ) =tan(θ)$

$-tan(θ)$

Given that $cos(θ)=21 $ and $0<θ<2π ,$ find the values of $sin(θ),$ $tan(θ),$ $csc(θ),$ $sec(θ),$ and $cot(θ)$ without determining $θ.$

Show Solution

To begin, we can find the value of $sin(θ)$ using the Pythagorean Identity. By substituting $cos(θ)=21 ,$ we get an equation that is solvable for the value of $sin(θ).$

$sin_{2}(θ)+cos_{2}(θ)=1$

Substitute

$cos(θ)=21 $

$sin_{2}(θ)+(21 )_{2}=1$

PowQuot

$(ba )_{m}=b_{m}a_{m} $

$sin_{2}(θ)+41 =1$

SubEqn

$LHS−41 =RHS−41 $

$sin_{2}(θ)=43 $

SqrtEqn

$LHS =RHS $

$sin(θ)=±43 $

SqrtQuot

$ba =b a $

$sin(θ)=±23 $

Notice that there are two values for $sinθ.$ From the prompt, we know that $θ$ must be between $0$ and $2π ,$ corresponding to the first quadrant on the Unit Circle. For these angles, $sin(θ)$ is positive. Thus, the negative solution can be discarded. $sin(θ)=23 $ Using the sine and cosine values, $tanθ$ can be found.

$tan(θ)=cos(θ)sin(θ) $

SubstituteII

$sin(θ)=23 $, $cos(θ)=21 $

$tan(θ)=23 /21 $

DivFracByFracSL

$ba /dc =ba ⋅cd $

$tan(θ)=23 ⋅12 $

MultFrac

Multiply fractions

$tan(θ)=23 ⋅2 $

SimpQuot

Simplify quotient

$tan(θ)=3 $

The remaining values are the reciprocals of the ones we've found so far. Thus, we can swap the numerator with the denominator for the fractions to find the desired values. $csc(θ)sec(θ) =sin(θ)1 =3 2 =cos(θ)1 =12 =2 $ As the tangent value isn't expressed as a fraction, it can simply be substituted. $cot(θ)=tan(θ)1 =3 1 $ We have now found all the desired values.

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