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3.1: Basic Trigonometric Identities - Mathematics


So far we know a few relations between the trigonometric functions. For example, we know the reciprocal relations:

  1. (csc; heta ~=~ dfrac{1}{sin; heta} qquad ) when (sin; heta e 0)
  2. (sec; heta ~=~ dfrac{1}{cos; heta} qquad ) when (cos; heta e 0)
  3. (cot; heta ~=~ dfrac{1}{ an; heta} qquad ) when ( an; heta ) is defined and not (0)
  4. (sin; heta ~=~ dfrac{1}{csc; heta} qquad ) when (csc; heta ) is defined and not (0)
  5. (cos; heta ~=~ dfrac{1}{sec; heta} qquad ) when (sec; heta ) is defined and not (0)
  6. ( an; heta ~=~ dfrac{1}{cot; heta} qquad ) when (cot; heta ) is defined and not (0)

Notice that each of these equations is true for all angles ( heta ) for which both sides of the equation are defined. Such equations are called identities, and in this section we will discuss several trigonometric identities, i.e. identities involving the trigonometric functions. These identities are often used to simplify complicated expressions or equations. For example, one of the most useful trigonometric identities is the following:

[ an; heta ~=~ frac{sin; heta}{cos; heta} qquad ext{when } cos; heta e 0 label{3.1}]

To prove this identity, pick a point ((x,y) ) on the terminal side of ( heta ) a distance (r >0 ) from the origin, and suppose that (cos; heta e 0 ). Then (x e 0 ) (since (cos; heta = frac{x}{r})), so by definition

[ onumber
frac{sin; heta}{cos; heta} ~=~ dfrac{~dfrac{y}{r}~}{~dfrac{x}{r}~} ~=~ frac{y}{x} ~=~
an; heta ~.
]

Note how we proved the identity by expanding one of its sides ((frac{sin; heta}{cos; heta})) until we got an expression that was equal to the other side (( an; heta)). This is probably the most common technique for proving identities. Taking reciprocals in the above identity gives:

[ cot; heta ~=~ frac{cos; heta}{sin; heta} qquad ext{when } sin; heta e 0 label{3.2}]

We will now derive one of the most important trigonometric identities. Let ( heta ) be any angle with a point ((x,y) ) on its terminal side a distance (r>0 ) from the origin. By the Pythagorean Theorem, (r^2 = x^2 + y^2 ) (and hence (r=sqrt{x^2 + y^2})). For example, if ( heta ) is in QIII as in Figure 3.1.1, then the legs of the right triangle formed by the reference angle have lengths (|x| ) and (|y| ) (we use absolute values because (x ) and (y ) are negative in QIII). The same argument holds if ( heta) is in the other quadrants or on either axis. Thus,

[ onumber
r^2 ~=~ |{x}|^2 ~+~ |{y}|^2 ~=~ x^2 ~+~ y^2 ~,
]
so dividing both sides of the equation by (r^2 ) (which we can do since (r>0)) gives

[ onumber
frac{r^2}{r^2} ~=~ frac{x^2 ~+~ y^2}{r^2} ~=~ frac{x^2}{r^2} ~+~ frac{y^2}{r^2} ~=~
left(frac{x}{r} ight)^2 ~+~ left(frac{y}{r} ight)^2 ~.
]

Since (frac{r^2}{r^2} = 1 ), (frac{x}{r} = cos; heta ), and (frac{y}{r} = sin; heta ), we can rewrite this as:

[cos^2 ; heta ~+~ sin^2 ; heta ~=~ 1 label{3.3}]

You can think of this as sort of a trigonometric variant of the Pythagorean Theorem. Note that we use the notation (sin^2 ; heta ) to mean ((sin; heta)^2 ), likewise for cosine and the other trigonometric functions. We will use the same notation for other powers besides (2 ).

From the above identity we can derive more identities. For example:

[ sin^2 ; heta ~=~ 1 ~-~ cos^2 ; heta label{3.4} ]
[ cos^2 ; heta ~=~ 1 ~-~ sin^2 ; heta label{3.5} ]

from which we get (after taking square roots):

[ sin; heta ~=~ pm,sqrt{1 ~-~ cos^2 ; heta}label{3.6}]
[cos; heta ~=~ pm,sqrt{1 ~-~ sin^2 ; heta}label{3.7}]

Also, from the inequalities (0 le sin^2 ; heta = 1 ~-~ cos^2 ; heta le 1 ) and (0 le cos^2 ; heta = 1 ~-~ sin^2 ; heta le 1 ), taking square roots gives us the following bounds on sine and cosine:

[ -1 ~ le ~ sin; heta ~ le ~ 1 label{3.8}]
[-1 ~ le ~ cos; heta ~ le ~ 1 label{3.9}]

The above inequalities are not identities (since they are not equations), but they provide useful checks on calculations. Recall that we derived those inequalities from the definitions of sine and cosine in Section 1.4.

In Equation ef{3.3}, dividing both sides of the identity by (cos^2 ; heta ) gives

[ onumber
frac{cos^2 ; heta}{cos^2 ; heta} ~+~ frac{sin^2 ; heta}{cos^2 ; heta} ~=~
frac{1}{cos^2 ; heta} ~~,
]

so since ( an; heta = frac{sin; heta}{cos; heta} ) and (sec; heta = frac{1}{cos; heta} ), we get:

[1 ~+~ an^2 ; heta ~=~ sec^2 ; heta label{3.10}]

Likewise, dividing both sides of Equation ef{3.3} by (sin^2 ; heta ) gives

[ onumber
frac{cos^2 ; heta}{sin^2 ; heta} ~+~ frac{sin^2 ; heta}{sin^2 ; heta} ~=~
frac{1}{sin^2 ; heta} ~~,
]

so since (cot; heta = frac{cos; heta}{sin; heta} ) and (csc; heta = frac{1}{sin; heta} ), we get:

[cot^2 ; heta ~+~ 1 ~=~ csc^2 ; heta label{3.11}]

Example 3.1

Simplify (;cos^2 ; heta ~ an^2 ; heta; ).

Solution

We can use Equation ef{3.5} to simplify:

[ onumber egin{align*}
cos^2 ; heta~ an^2 ; heta ~ &= ~ cos^2 ; heta ~cdot~
frac{sin^2 ; heta}{cos^2 ; heta} onumber
&= ~ sin^2 ; heta
end{align*}]

Example 3.2

Simplify (;5sin^2 ; heta ~+~ 4cos^2 ; heta; ).

Solution

We can use Equation ef{3.1} to simplify:
[ onumber egin{align*}
5sin^2 ; heta ~+~ 4cos^2 ; heta ~ &= ~ 5sin^2 ; heta ~+~
4left( 1 ~-~ sin^2 ; heta ight) onumber
&= ~ 5sin^2 ; heta ~+~ 4 ~-~ 4sin^2 ; heta onumber
&= ~ sin^2 ; heta ~+~ 4
end{align*}]

Example 3.3

Prove that (; an ; heta ~+~ cot ; heta ~=~ sec ; heta ~ csc ; heta; ).

Solution

We will expand the left side and show that it equals the right side:

[ onumber egin{alignat*}{3}
an ; heta + cot ; heta ~ &= ~ frac{sin; heta}{cos; heta} ~+~
frac{cos; heta}{sin; heta} &{} qquad & ext{(by ef{3.1} and
ef{3.2})} onumber
&= ~ frac{sin; heta}{cos; heta} ;cdot; frac{sin; heta}{sin; heta} ~+~
frac{cos; heta}{sin; heta} ;cdot; frac{cos; heta}{cos; heta}
&{} qquad & ext{(multiply both fractions by (1))} onumber
&= ~ frac{sin^2 ; heta ~+~ cos^2 ; heta}{cos; heta ~ sin; heta} &{} qquad
& ext{(after getting a common denominator)} onumber
&= ~ frac{1}{cos; heta ~ sin; heta} &{} qquad & ext{(by ef{3.3})} onumber
&= ~ frac{1}{cos; heta} ~cdot~ frac{1}{sin; heta} onumber
&= ~ sec ; heta ~ csc ; heta
end{alignat*}]

In the above example, how did we know to expand the left side instead of the right side? In general, though this technique does not always work, the more complicated side of the identity is likely to be easier to expand. The reason is that, by its complexity, there will be more things that you can do with that expression. For example, if you were asked to prove that

[ onumber
sec; heta ~-~ sin; heta ~ an; heta ~=~ cos; heta ~,
]

there would not be much that you could do with the right side of that identity; it consists of a single term ((cos; heta)) that offers no obvious means of expansion.

Example 3.4

Prove that (;dfrac{1 ~+~ cot^2 ; heta}{sec; heta} ~=~ csc; heta ~ cot; heta; ).

Solution

Of the two sides, the left side looks more complicated, so we will expand that:

[ onumber egin{alignat*}{3}
frac{1 ~+~ cot^2 ; heta}{sec; heta} ~ &= ~ frac{csc^2 ; heta}{sec; heta}
&{} qquad & ext{(by ef{3.11})} onumber
&= ~ dfrac{csc; heta ~cdot~ dfrac{1}{sin; heta}}{dfrac{1}{cos; heta}} &{}
&{}[2mm] onumber
&= ~ csc; heta ~cdot~ frac{cos; heta}{sin; heta} &{} &{} onumber
&= ~ csc ; heta ~ cot ; heta &{} qquad & ext{(by ef{3.2})}
end{alignat*}]

When trying to prove an identity where at least one side is a ratio of expressions, cross-multiplying can be an effective technique:

[ onumber
frac{a}{b} ~=~ frac{c}{d} quad ext{if and only if}quad ad ~=~ bc
]

Example 3.6

Prove that (;dfrac{1 ~+~ sin; heta}{cos; heta} ~=~ dfrac{cos; heta}{1 ~-~ sin; heta}; ).

Solution

Cross-multiply and reduce both sides until it is clear that they are equal:

[ onumber egin{align*}
( 1 ~+~ sin; heta ) ( 1 ~-~ sin; heta ) ~ &= ~ cos; heta ~cdot~ cos; heta onumber
1 ~-~ sin^2 ; heta ~ &= ~ cos^2 ; heta
end{align*}]

By ef{3.5} both sides of the last equation are indeed equal. Thus, the original identity holds.

Example 3.7

Suppose that (;a,cos; heta = b; ) and (;c,sin; heta = d; ) for some angle ( heta ) and some constants (a ), (b ), (c ), and (d ). Show that (;a^2 c^2 = b^2 c^2 + a^2 d^2 ).

Solution

Multiply both sides of the first equation by (c ) and the second equation by (a):
[ onumber egin{align*}
ac,cos; heta ~ &= ~ bc onumber
ac,sin; heta ~ &= ~ ad
end{align*}]

Now square each of the above equations then add them together to get:

[ onumber egin{align*}
(ac,cos; heta)^2 ~+~ (ac,sin; heta)^2 ~ &= ~ (bc)^2 ~+~ (ad)^2 onumber
(ac)^2 left( cos^2 ; heta ~+~ sin^2 ; heta ight)~ &= ~ b^2 c^2 ~+~ a^2 d^2 onumber
a^2 c^2 ~ &= ~ b^2 c^2 ~+~ a^2 d^2 qquad ext{(by ef{3.3})}
end{align*}]

Notice how ( heta ) does not appear in our final result. The trick was to get a common coefficient ((ac)) for (;cos; heta; ) and (;sin; heta; ) so that we could use (;cos^2 ; heta + sin^2 ; heta = 1 ). This is a common technique for eliminating trigonometric functions from systems of equations.


The relationships between the sides and angles of triangles are related with trigonometric functions. There are two main ways in which trigonometric functions are typically discussed: in terms of right triangles and in terms of the unit circle. The right-angled triangle definition of trigonometric functions as described below is most often how they are introduced.

Right triangle definition

The output of a trigonometric function is a ratio of the lengths of two sides of a right triangle. The terms used to describe the sides of a right triangle are the hypotenuse, the adjacent side, and the opposite side, as shown in the figure below.

The sides of a right triangle are referenced as follows:

  • Adjacent: the side next to &theta that is not the hypotenuse
  • Opposite: the side opposite &theta.
  • Hypotenuse: the longest side of the triangle opposite the right angle.

The trigonometric functions are defined based on the ratios of two sides of the right triangle. There are six trigonometric functions: sine, cosine, tangent, cosecant, secant, and cotangent. These functions are often abbreviated as sin, cos, tan, csc, sec, and cot. Their definitions are shown below.

Sine, cosine, and tangent are the three most commonly used trigonometric functions. Cosecant, secant, and cotangent are the reciprocals of sine, cosine, and tangent, respectively. As such, as long as we remember the definitions of sine, cosine, and tangent, we can take their reciprocals to determine the definitions of cosecant, secant, and cotangent.

What is the value of sin(45°), cos(45°), and tan(45°)?

Bob walked 300 m straight up on a 30° hill, how high did Bob climb?

Height = distance walked × sin(30°)
= 300 × 0.5
= 150 m

Thus, Bob reached a height of 150 m after walking 300 m up the slope of the hill.


3.1: Basic Trigonometric Identities - Mathematics

The basic trigonometric identities come in several varieties. These include the reciprocal identities, ratio identities, Pythagorean identities, symmetric identities, and cofunction identities. Each of these identities follows directly from the definition.

Reciprocal Identities Theorem

The following identities are true for all values for which they are defined:

$sin t=dfrac<1>$ $ an t=dfrac<1>$ $sec t=dfrac<1>$
$cos t=dfrac<1>$ $cot t=dfrac<1>< an t>$ $csc t=dfrac<1>$

Proof: From the Unit Circle Definition, we have $csc t=dfrac<1>=dfrac<1>$. The proofs of the other five identities are similar.♦

Ratio Identities Theorem

The following identities are true for all values for which they are defined:

$sin t=dfrac$ $ an t=dfrac$ $sec t=dfrac< an t>$
$sin t=dfrac< an t>$ $ an t=dfrac$ $sec t=dfrac$
$cos t=dfrac< an t>$ $cot t=dfrac$ $csc t=dfrac$
$cos t=dfrac$ $cot t=dfrac$ $csc t=dfrac< an t>$

Proof: From the Unit Circle Definition, we have $ an t=dfrac=dfrac$. The proofs of the other eleven identities are similar.♦

Of the twelve ratio identities given, only two are commonly cited, namely the ones which involve ratios of sine and cosine. Besides these twelve ratios, other ratios can be produced and simplified, but either they will either equal the constant 1, or be a product of trig functions.

Pythagorean Identities Theorem

The following identities are true for all values for which they are defined:

$sin^2 t+cos^2 t=1$
$1+ an^2 t=sec^2 t$
$1+cot^2 t=csc^2 t$

Proof: The equation of the unit circle is $x^2+y^2=1$. Substituting using the Unit Circle Definitions, we obtain the first of the three identities. The other two identities can be obtained by dividing each side of the equation by an appropriate factor.♦

Note that the Pythagorean identities provide a way to express the square of every trigonometric function in an alternate form.

Symmetric Identities Theorem

The following identities are true for all values for which they are defined:

$sin (-t)=-sin t$ $ an (-t)=- an t$ $sec (-t)=+sec t$
$cos (-t)=+cos t$ $cot (-t)=-cot t$ $csc (-t)=-csc t$

Proof: Let the point $A$ on the unit circle have coordinates $(x,y)$ and arc length $t$, measured counterclockwise from the point $(1,0)$. Define point $B$ to be the point on the unit circle whose arc length is $t$, measured clockwise from the point $(1,0)$. In other words, measured counterclockwise, the arc length is $-t$. Then, by symmetry across the $x$-axis, the coordinates of point $B$ are $(x,-y)$. Therefore, we have $sin(-t)=-y=-sin t$.

The proof of the cosine identity is similar. For the tangent identity, we have $ an (-t)=dfrac=dfrac<-sin t><+cos t>=- an t$.

The proofs of the last three trigonometric identities are similar to the proof of the tangent identity.♦

Cofunction Theorem

The following identities are true for all values for which they are defined:

$sin t=cosleft(dfrac<2>-t ight)$ $cos t=sinleft(dfrac<2>-t ight)$
$ an t=cotleft(dfrac<2>-t ight)$ $cot t= anleft(dfrac<2>-t ight)$
$sec t=cscleft(dfrac<2>-t ight)$ $csc t=secleft(dfrac<2>-t ight)$

Proof: Let the point $A$ on the unit circle have coordinates $(x,y)$ and arc length $t$, measured counterclockwise from the point $(1,0)$. Define point $B$ to be the point on the unit circle whose arc length is $dfrac<2>-t$, measured counterclockwise from the same point. Then the coordinates of point $B$ are $(y,x)$. Therefore, we have $cosleft(dfrac<2>-t ight)=y=sin t$.


Quotient Identities

The definitions of the trig functions led us to the reciprocal identities, which can be seen in the Concept about that topic. They also lead us to another set of identities, the quotient identities.

Consider first the sine, cosine, and tangent functions. For angles of rotation (not necessarily in the unit circle) these functions are defined as follows:

Given these definitions, we can show that ( an heta =dfrac) , as long as (cos heta eq 0):

The equation ( an heta =dfrac) is therefore an identity that we can use to find the value of the tangent function, given the value of the sine and cosine.

Let's take a look at some problems involving quotient identities.

1. Find the value of ( an heta)?

If (cos heta =dfrac<5><13>) and (sin heta =dfrac<12><13>), what is the value of ( an heta )?

3. What is the value of (cot heta)?

If (cos heta =dfrac<7><25>) and (sin heta =dfrac<24><25>), what is the value of (cot heta)?

Earlier, you were asked if you can help your friend find the answer.

you can use this knowledge to help your friend with the sine and cosine values you measured for yourself earlier:

If (cos heta =dfrac<17><145>) and (sin heta =dfrac<144><145>), what is the value of ( an heta)?

( an heta =dfrac<144><17>). We can see this from the relationship for the tangent function:

If (sin heta =dfrac<63><65>) and (cos heta =dfrac<16><65>), what is the value of ( an heta)?

( an heta =dfrac<63><16>). We can see this from the relationship for the tangent function:

If ( an heta =dfrac<40><9>) and (cos heta =dfrac<9><41>), what is the value of (sin heta)?

(sin heta =dfrac<40><41>). We can see this from the relationship for the tangent function:


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Basic Trigonometric Functions

The trigonometric functions relate the angles in a right triangle to the ratios of the sides. Given the following triangle:

the basic trigonometric functions are defined for 0 < θ < π 2 0 < heta < frac <2>0 < θ < 2 π ​ as

sin ⁡ θ = opposite hypotenuse , cos ⁡ θ = adjacent hypotenuse , tan ⁡ θ = opposite adjacent . egin&sin heta = frac< ext>< ext>, &cos heta = frac< ext>< ext>, & an heta = frac< ext>< ext>. end ​ sin θ = hypotenuse opposite ​ , ​ cos θ = hypotenuse adjacent ​ , ​ tan θ = adjacent opposite ​ . ​

For a review of converting between degrees and radians, see Degrees and Radians. However, a more useful definition comes from the unit circle. If we consider a circle with a radius of 1 unit, centered at the origin, then the angle θ heta θ inside the circle describes a right triangle when we drop a perpendicular to the x x x -axis from the point of intersection with the circle.

Unit Circle

Notice that the right triangle so described has a hypotenuse equal to the radius of the circle, an adjacent side equal to the x x x -coordinate of the point ( x , y ) , (x, y), ( x , y ) , and an opposite side equal to the y y y -coordinate. This gives rise naturally to the following refined definitions:

These definitions have the advantage of being compatible with the triangle definition above, as well as allowing the evaluation of angles corresponding to any real number.

There are certain values of these functions which are useful to remember. They are:


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Basic and Pythagorean Identities

Notice how a "co-(something)" trig ratio is always the reciprocal of some "non-co" ratio. You can use this fact to help you keep straight that cosecant goes with sine and secant goes with cosine.

The following (particularly the first of the three below) are called "Pythagorean" identities.

Note that the three identities above all involve squaring and the number 1 . You can see the Pythagorean-Thereom relationship clearly if you consider the unit circle, where the angle is t, the "opposite" side is sin(t) = y, the "adjacent" side is cos(t) = x , and the hypotenuse is 1 .

We have additional identities related to the functional status of the trig ratios:

Notice in particular that sine and tangent are odd functions, being symmetric about the origin, while cosine is an even function, being symmetric about the y -axis. The fact that you can take the argument's "minus" sign outside (for sine and tangent) or eliminate it entirely (for cosine) can be helpful when working with complicated expressions.

Angle-Sum and -Difference Identities

By the way, in the above identities, the angles are denoted by Greek letters. The a-type letter, " &alpha ", is called "alpha", which is pronounced "AL-fuh". The b-type letter, " &beta ", is called "beta", which is pronounced "BAY-tuh".


Trigonometry - Sin, Cos, Tan, Cot

sin : R -> R
All trigonometric functions are periodic. The period of sin is 2$pi$ .
The range of the function is [-1,1].

The cosine function

cos : R -> R
The period of sin is 2$pi$ .
The range of the function is [-1,1].

The tangent function

tan : R -> R
The range of the function is R . Now, the period is $pi$ and the function is undefined at x = ($pi$/2) + k$pi$, k=0,1,2.
The graph of the tangent function on the interval 0 - $pi$

The cotangent function

cot : R -> R
The range of the function is R . The period is $pi$ and that the function is undefined at x = k$pi$, k=0,1,2.

The values of sin, cos, tan, cot at the angles of 0°, 30°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, 360°

$A^o$ ^o$ $30^o$ $45^o$ $60^o$ $90^o$ $120^o$ $135^o$ $150^o$ $180^o$ $210^o$ $225^o$ $240^o$ $270^o$ $300^o$ $315^o$ $330^o$ $360^o$
$A rad$ $ $frac<6>$ $frac<4>$ $frac<3>$ $frac<2>$ $frac<2pi><3>$ $frac<3pi><4>$ $frac<5pi><6>$ $pi$ $frac<7pi><6>$ $frac<5pi><4>$ $frac<4pi><3>$ $frac<3pi><2>$ $frac<5pi><3>$ $frac<7pi><4>$ $frac<11pi><6>$ $2pi$
$sin A$ $ $frac<1><2>$ $frac><2>$ $frac><2>$ $1$ $frac><2>$ $frac><2>$ $frac<1><2>$ $ $-frac<1><2>$ $-frac><2>$ $-frac><2>$ $-1$ $-frac><2>$ $-frac><2>$ $-frac<1><2>$ $
$cos A$ $1$ $frac><2>$ $frac><2>$ $frac<1><2>$ $ $-frac<1><2>$ $-frac><2>$ $-frac><2>$ $-1$ $-frac><2>$ $-frac><2>$ $-frac<1><2>$ $ $frac<1><2>$ $frac><2>$ $frac><2>$ $1$
$ an A$ $ $frac><3>$ $1$ $sqrt<3>$ $-$ $-sqrt<3>$ $-1$ $-frac><3>$ $ $frac><3>$ $1$ $sqrt<3>$ $-$ $-sqrt<3>$ $-1$ $-frac><3>$ $
$cot A$ $-$ $sqrt<3>$ $1$ $frac><3>$ $ $-frac><3>$ $-1$ $-sqrt<3>$ $-$ $sqrt<3>$ $1$ $frac><3>$ $ $-frac><3>$ $-1$ $-sqrt<3>$ $-$

The easiest way to remember the basic values of sin and cos at the angles of 0°, 30°, 60°, 90°:
sin([0, 30, 45, 60, 90]) = cos([90, 60, 45, 30, 0]) = sqrt([0, 1, 2, 3, 4]/4)

Basic Trigonometric Identities

For every angle A corresponds exactly one point P(cos(A),sin(A)) on the unit circle.


Very, very important formula:

$ egin &color<1>: sin alpha = sin eta &color<2>: cos alpha = cos eta Rightarrow alpha = eta + 2kpi or &hspace <4.5cm>alpha = - eta + 2kpi &color<3>: an alpha = an eta Rightarrow alpha = eta + kpi end $

Basic trigonometric equations:

Solution: We know that the $sinfrac<6>=frac<1><2>$ and therefore

Solution: We know that the $cos ( frac <4>) = frac<2>$ and therefore

Solution : We know that the $ an <3>> = sqrt<3>$ and therefore

Advanced trigonometric equations

Step 1: To solve for $x$, you must first isolate the sine term.

$ egin 2sin(2x) - 1 &= 0 2sin(2x) &= 1 sin(2x) &= frac<1> <2>end $

Step 2: We know that $sin(frac<6>) = frac<1><2>$ and therefore

$ egin sin (2x) = sin ( frac <6>)mathop o limits^> &2x = frac <6>+ kpi hspace <0.5cm> ext hspace <0.5cm> o x = frac <12>+ kpi hspace <0.5cm> ext &2x = frac <5pi> <6>+ 2kpi hspace <1.7cm>x = frac<5pi> <12>+ kpi end $

Step 1: To solve for $x$, firstly, you must isolate the tangent term.

Step 2: We know that $ an (frac<6>) = frac><3>$ and $ an (-frac<6>) = - frac><3>$, therefore

$ egin an (x) = an (frac<6>) &Rightarrow x = frac <6>+ kpi ext an (x) = an (-frac<6>) &Rightarrow x = -frac <6>+ kpi end $

Step 1: To solve for $x$, firstly, you must isolate the cosine term.

Step 2: We know that $cos (frac<3>) = frac<1><2>$, therefore

$ egin cos (2x - frac<3>) = cos (frac<3>)mathop o limits^> &2x - frac <3>+ 2kpi ext o 2x = frac <6>+ 2kpi ext o x = frac<12>+kpi ext &2x - frac <3>= -frac <3>+ 2kpi hspace <1cm>2x = 0 + 2kpi hspace <1cm>x = kpi end $


Quadrants in Trigonometry

The signs of the three basic trigonometric functions, sin, cos, and tan, vary based on which quadrant they are in. The sign of each trigonometric function in each quadrant can be determined using the signs of the coordinates along with basic trigonometric relationships. The diagram below shows the signs of these functions in different quadrants.


Did you know?

The first part of the word "quadrant" is from a Latin root meaning four.

A popular name for an all-terrain vehicle (ATV) is "quad," named for its four large tires.

Other words based on the same root include: quadrilateral, quadruplets, quadrillion.


Watch the video: Trigonometry Class 11. Formulas Trick. Trigonometric Functions. Chapter 3 (December 2021).