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The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. SeeExample 1andExample 2.
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The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angl...
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The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions. SeeExample 4.
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The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the ...
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The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. SeeExample 6.
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The cofunction identities apply to complementary angles and pairs of reciprocal functions. SeeExample 7.
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Sum and difference formulas are useful in verifying identities. SeeExample 8andExample 9.
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Application problems are often easier to solve by using sum and difference formulas. SeeExample 10andExample 11.
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Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. SeeExample 1,Example 2,Example 3, andExample 4.
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Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. SeeExample 5andExample 6.
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Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. SeeExample 7,Example 8, andExample 9.
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From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine.
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We can use the product-to-sum formulas to rewrite products of sines, products of cosines, and products of sine and cosine as sums or differences of sines and cosines. SeeExample 1,Example 2, andExample 3.
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We can also derive the sum-to-product identities from the product-to-sum identities using substitution.
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We can use the sum-to-product formulas to rewrite sum or difference of sines, cosines, or products sine and cosine as products of sines and cosines. SeeExample 4.
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Trigonometric expressions are often simpler to evaluate using the formulas. SeeExample 5.
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The identities can be verified using other formulas or by converting the expressions to sines and cosines. To verify an identity, we choose the more complicated side of the equals sign and rewrite it until it is transformed into the other side. SeeExample 6andExample 7.
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When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution. SeeExample 1,Example 2, andExample 3.
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Equations involving a single trigonometric function can be solved or verified using the unit circle. SeeExample 4,Example 5, andExample 6, andExample 7.
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We can also solve trigonometric equations using a graphing calculator. SeeExample 8andExample 9.
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Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc. SeeExample 10,Example 11,Example 12, andExample 13.
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We can also use the identities to solve trigonometric equation. SeeExample 14,Example 15, andExample 16.
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We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval. SeeExample 17.
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Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions. SeeExample 18.
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cos 2 θ + sin 2 θ = 1 1 + cot 2 θ = csc 2 θ 1 + tan 2 θ = sec 2 θ cos 2 θ + sin 2 θ = 1 1 + cot 2 θ = csc 2 θ 1 + tan 2 θ = sec 2 θ
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tan ( − θ ) = − tan θ cot ( − θ ) = − cot θ sin ( − θ ) = − sin θ csc ( − θ ) = − csc θ cos ( − θ ) = cos θ sec ( − θ ) = sec θ tan ( − θ ) = − tan θ cot ( − θ ) = − cot θ sin ( − θ ) = − sin θ csc ( − θ ) = − csc θ cos ( − θ ) = cos θ sec ( − θ ) = sec θ
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sin θ = 1 csc θ cos θ = 1 sec θ tan θ = 1 cot θ csc θ = 1 sin θ sec θ = 1 cos θ cot θ = 1 tan θ sin θ = 1 csc θ cos θ = 1 sec θ tan θ = 1 cot θ csc θ = 1 sin θ sec θ = 1 cos θ cot θ = 1 tan θ
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tan θ = sin θ cos θ cot θ = cos θ sin θ tan θ = sin θ cos θ cot θ = cos θ sin θ
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cos ( α + β ) = cos α cos β − sin α sin β cos ( α + β ) = cos α cos β − sin α sin β
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cos ( α − β ) = cos α cos β + sin α sin β cos ( α − β ) = cos α cos β + sin α sin β
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sin ( α + β ) = sin α cos β + cos α sin β sin ( α + β ) = sin α cos β + cos α sin β
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sin ( α − β ) = sin α cos β − cos α sin β sin ( α − β ) = sin α cos β − cos α sin β
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tan ( α + β ) = tan α + tan β 1 − tan α tan β tan ( α + β ) = tan α + tan β 1 − tan α tan β
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tan ( α − β ) = tan α − tan β 1 + tan α tan β tan ( α − β ) = tan α − tan β 1 + tan α tan β
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sin θ = cos ( π 2 − θ ) cos θ = sin ( π 2 − θ ) tan θ = cot ( π 2 − θ ) cot θ = tan ( π 2 − θ ) sec θ = csc ( π 2 − θ ) csc θ = sec ( π 2 − θ ) sin θ = cos ( π 2 − θ ) cos θ = sin ( π 2 − θ ) tan θ = cot ( π 2 − θ ) cot θ = tan ( π 2 − θ ) sec θ = csc ( π 2 − θ ) c...
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sin ( 2 θ ) = 2 sin θ cos θ cos ( 2 θ ) = cos 2 θ − sin 2 θ = 1 − 2 sin 2 θ = 2 cos 2 θ − 1 tan ( 2 θ ) = 2 tan θ 1 − tan 2 θ sin ( 2 θ ) = 2 sin θ cos θ cos ( 2 θ ) = cos 2 θ − sin 2 θ = 1 − 2 sin 2 θ = 2 cos 2 θ − 1 tan ( 2 θ ) = 2 tan θ 1 − tan 2 θ
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sin 2 θ = 1 − cos ( 2 θ ) 2 cos 2 θ = 1 + cos ( 2 θ ) 2 tan 2 θ = 1 − cos ( 2 θ ) 1 + cos ( 2 θ ) sin 2 θ = 1 − cos ( 2 θ ) 2 cos 2 θ = 1 + cos ( 2 θ ) 2 tan 2 θ = 1 − cos ( 2 θ ) 1 + cos ( 2 θ )
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sin α 2 = ± 1 − cos α 2 cos α 2 = ± 1 + cos α 2 tan α 2 = ± 1 − cos α 1 + cos α = sin α 1 + cos α = 1 − cos α sin α sin α 2 = ± 1 − cos α 2 cos α 2 = ± 1 + cos α 2 tan α 2 = ± 1 − cos α 1 + cos α = sin α 1 + cos α = 1 − cos α sin α
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cos α cos β = 1 2 [ cos ( α − β ) + cos ( α + β ) ] sin α cos β = 1 2 [ sin ( α + β ) + sin ( α − β ) ] sin α sin β = 1 2 [ cos ( α − β ) − cos ( α + β ) ] cos α sin β = 1 2 [ sin ( α + β ) − sin ( α − β ) ] cos α cos β = 1 2 [ cos ( α − β ) + cos ( α + β ) ] sin α cos β = ...
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sin α + sin β = 2 sin ( α + β 2 ) cos ( α − β 2 ) sin α − sin β = 2 sin ( α − β 2 ) cos ( α + β 2 ) cos α − cos β = − 2 sin ( α + β 2 ) sin ( α − β 2 ) cos α + cos β = 2 cos ( α + β 2 ) cos ( α − β 2 ) sin α + sin β = 2 sin ( α + β 2 ) cos ( α − β 2 ) sin α − sin β = 2 ...
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double-angle formulas : identities derived from the sum formulas for sine, cosine, and tangent in which the angles are equal
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even-odd identities : set of equations involving trigonometric functions such that iff(−x)=−f(x),f(−x)=−f(x),the identity is odd, and iff(−x)=f(x),f(−x)=f(x),the identity is even
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half-angle formulas : identities derived from the reduction formulas and used to determine half-angle values of trigonometric functions
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product-to-sum formula : a trigonometric identity that allows the writing of a product of trigonometric functions as a sum or difference of trigonometric functions
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Pythagorean identities : set of equations involving trigonometric functions based on the right triangle properties
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quotient identities : pair of identities based on the fact that tangent is the ratio of sine and cosine, and cotangent is the ratio of cosine and sine
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reciprocal identities : set of equations involving the reciprocals of basic trigonometric definitions
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reduction formulas : identities derived from the double-angle formulas and used to reduce the power of a trigonometric function
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sum-to-product formula : a trigonometric identity that allows, by using substitution, the writing of a sum of trigonometric functions as a product of trigonometric functions
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The Law of Sines can be used to solve oblique triangles, which are non-right triangles.
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According to the Law of Sines, the ratio of the measurement of one of the angles to the length of its opposite side equals the other two ratios of angle measure to opposite side.
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There are three possible cases: ASA, AAS, SSA. Depending on the information given, we can choose the appropriate equation to find the requested solution. SeeExample 1.
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The ambiguous case arises when an oblique triangle can have different outcomes.
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There are three possible cases that arise from SSA arrangement—a single solution, two possible solutions, and no solution. SeeExample 2andExample 3.
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The Law of Sines can be used to solve triangles with given criteria. SeeExample 4.
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The general area formula for triangles translates to oblique triangles by first finding the appropriate height value. SeeExample 5.
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There are many trigonometric applications. They can often be solved by first drawing a diagram of the given information and then using the appropriate equation. SeeExample 6.
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The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.
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The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. SeeExample 1andExample 2.
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The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. SeeExample 3andExample 4.
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Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. SeeExample 5and SeeExample 6.
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The polar grid is represented as a series of concentric circles radiating out from the pole, or origin.
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To plot a point in the form(r,θ),θ>0,(r,θ),θ>0,move in a counterclockwise direction from the polar axis by an angle ofθ,θ,and then extend a directed line segment from the pole the length ofrrin the direction ofθ.θ.Ifθθis negative, move in a clockwise direction, and extend a directed line segment the length of...
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Ifrris negative, extend the directed line segment in the opposite direction ofθ.θ.SeeExample 2.
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To convert from polar coordinates to rectangular coordinates, use the formulasx=rcosθx=rcosθandy=rsinθ.y=rsinθ.SeeExample 3andExample 4.
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To convert from rectangular coordinates to polar coordinates, use one or more of the formulas:cosθ=xr,sinθ=yr,tanθ=yx,cosθ=xr,sinθ=yr,tanθ=yx,andr=x2+y2.r=x2+y2.SeeExample 5.
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Transforming equations between polar and rectangular forms means making the appropriate substitutions based on the available formulas, together with algebraic manipulations. SeeExample 6,Example 7, andExample 8.
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Using the appropriate substitutions makes it possible to rewrite a polar equation as a rectangular equation, and then graph it in the rectangular plane. SeeExample 9,Example 10, andExample 11.
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It is easier to graph polar equations if we can test the equations for symmetry with respect to the lineθ=π2,θ=π2,the polar axis, or the pole.
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There are three symmetry tests that indicate whether the graph of a polar equation will exhibit symmetry. If an equation fails a symmetry test, the graph may or may not exhibit symmetry. SeeExample 1.
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Polar equations may be graphed by making a table of values forθθandr.r.
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The maximum value of a polar equation is found by substituting the valueθθthat leads to the maximum value of the trigonometric expression.
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The zeros of a polar equation are found by settingr=0r=0and solving forθ.θ.SeeExample 2.
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Some formulas that produce the graph of a circle in polar coordinates are given byr=acosθr=acosθandr=asinθ.r=asinθ.SeeExample 3.
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The formulas that produce the graphs of a cardioid are given byr=a±bcosθr=a±bcosθandr=a±bsinθ,r=a±bsinθ,fora>0,a>0,b>0,b>0,andab=1.ab=1.SeeExample 4.
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The formulas that produce the graphs of a one-loop limaçon are given byr=a±bcosθr=a±bcosθandr=a±bsinθr=a±bsinθfor1<ab<2.1<ab<2.SeeExample 5.
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The formulas that produce the graphs of an inner-loop limaçon are given byr=a±bcosθr=a±bcosθandr=a±bsinθr=a±bsinθfora>0,a>0,b>0,b>0,anda<b.a<b.SeeExample 6.
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The formulas that produce the graphs of a lemniscates are given byr2=a2cos2θr2=a2cos2θandr2=a2sin2θ,r2=a2sin2θ,whereaâ‰0.aâ‰0.SeeExample 7.
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The formulas that produce the graphs of rose curves are given byr=acosnθr=acosnθandr=asinnθ,r=asinnθ,whereaâ‰0;aâ‰0;ifnnis even, there are2n2npetals, and ifnnis odd, there arennpetals. SeeExample 8andExample 9.
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The formula that produces the graph of an Archimedes’ spiral is given byr=θ,r=θ,θ≥0.θ≥0.SeeExample 10.
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Complex numbers in the forma+bia+biare plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Label thex-axis as therealaxis and they-axis as theimaginaryaxis. SeeExample 1.
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The absolute value of a complex number is the same as its magnitude. It is the distance from the origin to the point:|z|=a2+b2.|z|=a2+b2.SeeExample 2andExample 3.
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To write complex numbers in polar form, we use the formulasx=rcosθ,y=rsinθ,x=rcosθ,y=rsinθ,andr=x2+y2.r=x2+y2.Then,z=r(cosθ+isinθ).z=r(cosθ+isinθ).SeeExample 4andExample 5.
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To convert from polar form to rectangular form, first evaluate the trigonometric functions. Then, multiply through byr.r.SeeExample 6andExample 7.
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To find the product of two complex numbers, multiply the two moduli and add the two angles. Evaluate the trigonometric functions, and multiply using the distributive property. SeeExample 8.
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To find the quotient of two complex numbers in polar form, find the quotient of the two moduli and the difference of the two angles. SeeExample 9.
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To find the power of a complex numberzn,zn,raiserrto the powern,n,and multiplyθθbyn.n.SeeExample 10.
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Finding the roots of a complex number is the same as raising a complex number to a power, but using a rational exponent. SeeExample 11.
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Parameterizing a curve involves translating a rectangular equation in two variables,xxandy,y,into two equations in three variables,x,y, andt. Often, more information is obtained from a set of parametric equations. SeeExample 1,Example 2, andExample 3.
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Sometimes equations are simpler to graph when written in rectangular form. By eliminatingt,t,an equation inxxandyyis the result.
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To eliminatet,t,solve one of the equations fort,t,and substitute the expression into the second equation. SeeExample 4,Example 5,Example 6, andExample 7.
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Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve forttin one of the equations, and substitute the expression into the second equation. SeeExample 8.
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There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation.
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Find an expression forxxsuch that the domain of the set of parametric equations remains the same as the original rectangular equation. SeeExample 9.
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When there is a third variable, a third parameter on whichxxandyydepend, parametric equations can be used.
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To graph parametric equations by plotting points, make a table with three columns labeledt,x(t),t,x(t),andy(t).y(t).Choose values forttin increasing order. Plot the last two columns forxxandy.y.SeeExample 1andExample 2.
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When graphing a parametric curve by plotting points, note the associatedt-values and show arrows on the graph indicating the orientation of the curve. SeeExample 3andExample 4.
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Parametric equations allow the direction or the orientation of the curve to be shown on the graph. Equations that are not functions can be graphed and used in many applications involving motion. SeeExample 5.
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Projectile motion depends on two parametric equations:x=(v0cosθ)tx=(v0cosθ)tandy=−16t2+(v0sinθ)t+h.y=−16t2+(v0sinθ)t+h.Initial velocity is symbolized asv0.θv0.θrepresents the initial angle of the object when thrown, andhhrepresents the height at which the object is propelled.
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The position vector has its initial point at the origin. SeeExample 1.
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