L2.4 D1 Solving Equations using Identities - Part 1
Completion requirements
Unit 2
Trigonometry
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Part 2.4D corresponds to section 6.4, starting on page 316 of your Pre-Calculus 12 textbook.
Suppose you are cooking and have pulled out your imperial measuring spoons, but everything in the recipe is in metric. What should you do? One answer is to switch out your imperial measuring spoons for metric ones.
Identities can be used in a similar way. If an equation includes an expression that may be difficult to work with, sometimes using an identity will allow you to switch out the expression for something easier to work with. You have already seen this with reciprocal ratios when solving equations using a graphing calculator.
When an equation has two different trigonometric ratios, one way to solve it is to use identities to rewrite the equation in terms of a single ratio. A second method is to rewrite the equation as a product of factors equal to zero, where each factor contains only one type of ratio.
Identities can be used in a similar way. If an equation includes an expression that may be difficult to work with, sometimes using an identity will allow you to switch out the expression for something easier to work with. You have already seen this with reciprocal ratios when solving equations using a graphing calculator.
When an equation has two different trigonometric ratios, one way to solve it is to use identities to rewrite the equation in terms of a single ratio. A second method is to rewrite the equation as a product of factors equal to zero, where each factor contains only one type of ratio.
Algebraically solve «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»tan«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«mo»=«/mo»«mo»§#8722;«/mo»«mn»1«/mn»«/mrow»«/mstyle»«/math», «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»0«/mn»«mo»§#176;«/mo»«mo»§#8804;«/mo»«mi»x«/mi»«mo»§#60;«/mo»«mn»360«/mn»«mo»§#176;«/mo»«/mrow»«/mstyle»«/math».
It is a good habit to determine the non-permissible values of an equation before trying to solve it. For «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»tan«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» to be defined on the given domain, «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»x«/mi»«mo»§#8800;«/mo»«mn»90«/mn»«mo»§#176;«/mo»«mi mathvariant=¨normal¨»,«/mi»«mspace width=¨0.33em¨/»«msup»«mn»270«/mn»«mo accent=¨true¨»§#176;«/mo»«/msup»«/mrow»«/mstyle»«/math».
The equation includes both «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»tan«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» and «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math». It doesn’t look like it will factor easily, so try to rewrite the equation using identities.
Now, there is a single ratio. Use methods from Lesson 2.2 to solve for «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»x«/mi»«/mstyle»«/math».
The value «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»270«/mn»«mo»§#176;«/mo»«/mrow»«/mstyle»«/math» is not permissible, so it is not a solution. You can also verify «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»270«/mn»«mo»§#176;«/mo»«/mrow»«/mstyle»«/math» is not a solution using substitution.
The left side does not equal the right side, so «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»270«/mn»«mo»§#176;«/mo»«/mrow»«/mstyle»«/math» is not a solution.
The equation includes both «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»tan«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» and «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math». It doesn’t look like it will factor easily, so try to rewrite the equation using identities.
«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»tan«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»-«/mo»«mn»1«/mn»«/mtd»«/mtr»«mtr»«mtd»«mfrac»«mrow»«mi»sin«/mi»«mi»x«/mi»«/mrow»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mfrac»«mi»cos«/mi»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»-«/mo»«mn»1«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»sin«/mi»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»-«/mo»«mn»1«/mn»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»
Now, there is a single ratio. Use methods from Lesson 2.2 to solve for «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»x«/mi»«/mstyle»«/math».
«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»x«/mi»«mo»=«/mo»«mn»270«/mn»«mo»§#176;«/mo»«/mrow»«/mstyle»«/math»
The value «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»270«/mn»«mo»§#176;«/mo»«/mrow»«/mstyle»«/math» is not permissible, so it is not a solution. You can also verify «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»270«/mn»«mo»§#176;«/mo»«/mrow»«/mstyle»«/math» is not a solution using substitution.
| Left Side | Right Side |
| «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right¨»«mtr»«mtd»«mi»tan«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»tan«/mi»«mfenced»«mrow»«mn»270«/mn»«mo»§#176;«/mo»«/mrow»«/mfenced»«mi»cos«/mi»«mfenced»«mrow»«mn»270«/mn»«mo»§#176;«/mo»«/mrow»«/mfenced»«/mtd»«/mtr»«mtr»«mtd»«mi»undefined«/mi»«mo»§#8226;«/mo»«mn»0«/mn»«/mtd»«/mtr»«/mtable»«/mstyle»«/math» |
«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mo»-«/mo»«mn»1«/mn»«/mrow»«/mstyle»«/math»
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The left side does not equal the right side, so «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»270«/mn»«mo»§#176;«/mo»«/mrow»«/mstyle»«/math» is not a solution.
Algebraically solve «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sin«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«mo»=«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math», «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mn»0«/mn»«mo»§#8804;«/mo»«mi»x«/mi»«mo»§#60;«/mo»«mn»2«/mn»«mo»§#960;«/mo»«/mrow»«/mstyle»«/math».
Both «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sin«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» and «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»cos«/mi»«mi»x«/mi»«/mrow»«/mstyle»«/math» are defined for all real values of «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mi»x«/mi»«/mstyle»«/math», and there are no denominators, so the equation has no non-permissible values.
It doesn’t look like any identities will easily convert one ratio to another. Rearrange the equation so one side is zero, and try factoring.
«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»sin«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»sin«/mi»«mi»x«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»sin«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»sin«/mi»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»sin«/mi»«mi»x«/mi»«mfenced»«mrow»«mi»cos«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/mrow»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»
For the product to be equal to zero, at least one of the factors must be equal to zero, so you have either «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sin«/mi»«mi»x«/mi»«mo»=«/mo»«mn»0«/mn»«/mrow»«/mstyle»«/math» or «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»cos«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mn»0«/mn»«/mrow»«/mstyle»«/math». Solve each of these separately.
So, the entire solution is «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»x«/mi»«mo»=«/mo»«mn»0«/mn»«mi mathvariant=¨normal¨»,«/mi»«mspace width=¨0.33em¨/»«mfrac»«mo»§#960;«/mo»«mn»3«/mn»«/mfrac»«mi mathvariant=¨normal¨»,«/mi»«mspace width=¨0.33em¨/»«mo»§#960;«/mo»«mi mathvariant=¨normal¨»,«/mi»«mspace width=¨0.33em¨/»«mfrac»«mrow»«mn»5«/mn»«mo»§#960;«/mo»«/mrow»«mn»3«/mn»«/mfrac»«/mrow»«/mstyle»«/math».
Each of these solutions can be verified by substitution.
It doesn’t look like any identities will easily convert one ratio to another. Rearrange the equation so one side is zero, and try factoring.
«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»sin«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»sin«/mi»«mi»x«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»sin«/mi»«mi»x«/mi»«mi»cos«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»sin«/mi»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»sin«/mi»«mi»x«/mi»«mfenced»«mrow»«mi»cos«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/mrow»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»
For the product to be equal to zero, at least one of the factors must be equal to zero, so you have either «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»sin«/mi»«mi»x«/mi»«mo»=«/mo»«mn»0«/mn»«/mrow»«/mstyle»«/math» or «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»cos«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mn»0«/mn»«/mrow»«/mstyle»«/math». Solve each of these separately.
«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»sin«/mi»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mi mathvariant=¨normal¨»,«/mi»«mspace width=¨0.33em¨/»«mo»§#960;«/mo»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»
«math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mtable columnalign=¨right center left¨»«mtr»«mtd»«mi»cos«/mi»«mi»x«/mi»«mo»§#8722;«/mo»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»cos«/mi»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«/mtd»«/mtr»«mtr»«mtd»«mi»x«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mo»§#960;«/mo»«mn»3«/mn»«/mfrac»«mi mathvariant=¨normal¨»,«/mi»«mspace width=¨0.33em¨/»«mfrac»«mrow»«mn»5«/mn»«mo»§#960;«/mo»«/mrow»«mn»3«/mn»«/mfrac»«/mtd»«/mtr»«/mtable»«/mstyle»«/math»
So, the entire solution is «math style=¨font-family:Verdana¨ xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mstyle mathsize=¨14px¨»«mrow»«mi»x«/mi»«mo»=«/mo»«mn»0«/mn»«mi mathvariant=¨normal¨»,«/mi»«mspace width=¨0.33em¨/»«mfrac»«mo»§#960;«/mo»«mn»3«/mn»«/mfrac»«mi mathvariant=¨normal¨»,«/mi»«mspace width=¨0.33em¨/»«mo»§#960;«/mo»«mi mathvariant=¨normal¨»,«/mi»«mspace width=¨0.33em¨/»«mfrac»«mrow»«mn»5«/mn»«mo»§#960;«/mo»«/mrow»«mn»3«/mn»«/mfrac»«/mrow»«/mstyle»«/math».
Each of these solutions can be verified by substitution.