4.4 Trigonometric equations - Revision history

Phil: "solution to" becomes "solution of"; some other minor changes

2009-01-06T16:46:03Z

"solution to" becomes "solution of"; some other minor changes

←Older revision		Revision as of 16:46, 6 January 2009
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	We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other point on the unit circle which has ''x''-coordinate <math>\frac{1}{2}</math> corresponds to the angle <math>-\pi/3</math>. Adding an integral number of revolutions to these angles we get the general solution		We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other point on the unit circle which has ''x''-coordinate <math>\frac{1}{2}</math> corresponds to the angle <math>-\pi/3</math>. Adding an integral number of revolutions to these angles we get the general solution

-	{{Displayed math\|\|<math>x = \pm \pi/3 + n \~~cdot~~ 2\pi\,\mbox{,}</math>}}	+	{{Displayed math\|\|<math>x = \pm \pi/3 + n \times 2\pi\,\mbox{,}</math>}}

	where <math>n</math> is an arbitrary integer.		where <math>n</math> is an arbitrary integer.
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	Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution, and so the general solution can be obtained from the solution <math>\pi/3</math> by adding or subtracting multiples of <math>\pi</math>:		Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution, and so the general solution can be obtained from the solution <math>\pi/3</math> by adding or subtracting multiples of <math>\pi</math>:
-	{{Displayed math\|\|<math>x = \pi/3 + n \~~cdot~~ \pi\,\mbox{,}</math>}}	+	{{Displayed math\|\|<math>x = \pi/3 + n \, \pi\,\mbox{,}</math>}}

	where <math>n</math> is an arbitrary integer.		where <math>n</math> is an arbitrary integer.
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	Factorising out <math>\sin x</math> we get		Factorising out <math>\sin x</math> we get
	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
-	\sin x~~\,\cdot~~\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}}	+	\sin x\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}}

	From this factorised form of the equation, we see that the solutions either have to satisfy <math>\sin x = 0</math> or <math>\sin x = -\tfrac{1}{2}</math>, which are two basic equations of the type <math>\sin x = a</math> and can be solved as in Example 1. The solutions turn out to be		From this factorised form of the equation, we see that the solutions either have to satisfy <math>\sin x = 0</math> or <math>\sin x = -\tfrac{1}{2}</math>, which are two basic equations of the type <math>\sin x = a</math> and can be solved as in Example 1. The solutions turn out to be
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	Dividing both sides by 2 and factorising out <math>\cos x</math>, gives		Dividing both sides by 2 and factorising out <math>\cos x</math>, gives
-	{{Displayed math\|\|<math>\cos x~~\,\cdot~~\,( \sin x – 2) = 0\,\mbox{.}</math>}}	+	{{Displayed math\|\|<math>\cos x\,( \sin x – 2) = 0\,\mbox{.}</math>}}

	The left-hand side can only be zero if one of the factors is zero, and we have reduced the original equation into two basic equations:		The left-hand side can only be zero if one of the factors is zero, and we have reduced the original equation into two basic equations:

Jdinglis at 20:23, 10 November 2008

2008-11-10T20:23:10Z

←Older revision		Revision as of 20:23, 10 November 2008
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	<center>{{:4.4 - Figure - Two unit circles with angles π/3 and π+π/3, respectively}}</center>		<center>{{:4.4 - Figure - Two unit circles with angles π/3 and π+π/3, respectively}}</center>

-	Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution, and so the general solution can be obtained from the solution <math>\pi/3</math>by adding or subtracting multiples of <math>\pi</math>:	+	Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution, and so the general solution can be obtained from the solution <math>\pi/3</math> by adding or subtracting multiples of <math>\pi</math>:
	{{Displayed math\|\|<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}}		{{Displayed math\|\|<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}}

Jdinglis at 20:22, 10 November 2008

2008-11-10T20:22:06Z

←Older revision		Revision as of 20:22, 10 November 2008
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	<center>{{:4.4 - Figure - Two unit circles with angles π/3 and -π/3, respectively}}</center>		<center>{{:4.4 - Figure - Two unit circles with angles π/3 and -π/3, respectively}}</center>

-	We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other point on the unit circle which has "x"-coordinate <math>\frac{1}{2}</math> corresponds to the angle <math>-\pi/3</math>. Adding an integral number of revolutions to these angles we get the general solution	+	We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other point on the unit circle which has ''x''-coordinate <math>\frac{1}{2}</math> corresponds to the angle <math>-\pi/3</math>. Adding an integral number of revolutions to these angles we get the general solution

	{{Displayed math\|\|<math>x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}</math>}}		{{Displayed math\|\|<math>x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}</math>}}

Jdinglis at 19:57, 10 November 2008

2008-11-10T19:57:59Z

←Older revision		Revision as of 19:57, 10 November 2008
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	4 (1 – \cos^2\!x)\ –\ 4 \cos x &= 1\,\\		4 (1 – \cos^2\!x)\ –\ 4 \cos x &= 1\,\\
	\Rightarrow 4\ –\ 4 \cos^2\!x\ –\ 4 \cos x &= 1\,\\		\Rightarrow 4\ –\ 4 \cos^2\!x\ –\ 4 \cos x &= 1\,\\
-	\Rightarrow –4\cos^2\!\x –\ 4 \cos x\ +\ 4\ –\ 1 &= 0\,\\	+	\Rightarrow –4\cos^2\!x\ –\ 4 \cos x\ +\ 4\ –\ 1 &= 0\,\\
	\Rightarrow \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\		\Rightarrow \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
	\end{align*}</math>}}		\end{align*}</math>}}

Jdinglis at 19:56, 10 November 2008

2008-11-10T19:56:07Z

←Older revision		Revision as of 19:56, 10 November 2008
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	Using the Pythagorean identity we can replace <math>\sin^2\!x</math> by <math>1 – \cos^2\!x</math>. Then{{Displayed math\|\|<math>		Using the Pythagorean identity we can replace <math>\sin^2\!x</math> by <math>1 – \cos^2\!x</math>. Then{{Displayed math\|\|<math>
	\begin{align*}		\begin{align*}
-	4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\\	+	4 (1 – \cos^2\!x)\ –\ 4 \cos x &= 1\,\\
-	\Rightarrow 4 – 4 \cos^2\!x – 4 \cos x &= 1\,\\	+	\Rightarrow 4\ –\ 4 \cos^2\!x\ –\ 4 \cos x &= 1\,\\
-	\Rightarrow –4\cos^2\!x – 4 \cos x + 4 – 1 &= 0\,\\	+	\Rightarrow –4\cos^2\!\x –\ 4 \cos x\ +\ 4\ –\ 1 &= 0\,\\
	\Rightarrow \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\		\Rightarrow \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
	\end{align*}</math>}}		\end{align*}</math>}}

Jdinglis at 19:53, 10 November 2008

2008-11-10T19:53:34Z

←Older revision		Revision as of 19:53, 10 November 2008
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	* <math>\sin x = 2</math>.		* <math>\sin x = 2</math>.

-	However, <math>\sin x</math> can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions. That just leaves <math>\cos x = 0</math>, and using the unit circle we see that the general solution is <math>x = \pi / 2 + n ~~\cdot~~ \pi</math>.	+	However, <math>\sin x</math> can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions. That just leaves <math>\cos x = 0</math>, and using the unit circle we see that the general solution is
		+	{{Displayed math\|\|<math>x = \frac{\pi}{2} + n\pi\mbox{,}</math>}}
		+	for <math>n</math> an arbitrary integer.
		+
	</div>		</div>

Jdinglis at 19:50, 10 November 2008

2008-11-10T19:50:18Z

←Older revision		Revision as of 19:50, 10 November 2008
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	* <math>\sin x = 2</math>.		* <math>\sin x = 2</math>.

-	However, can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions.	+	However, <math>\sin x</math> can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions. That just leaves <math>\cos x = 0</math>, and using the unit circle we see that the general solution is <math>x = \pi / 2 + n \cdot \pi</math>.
-		+
-	That ~~leaves~~ just	+
-	<math>\cos x = 0</math>, and using the unit circle we see that the general solution is <math>x = \pi / 2 + n \cdot \pi</math>.	+
	</div>		</div>

Jdinglis at 19:48, 10 November 2008

2008-11-10T19:48:22Z

←Older revision		Revision as of 19:48, 10 November 2008
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	* <math>\sin x = 2</math>.		* <math>\sin x = 2</math>.

-	However, ~~<math>\sin x</math>~~ can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions. That leaves just	+	However, can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions.
		+
		+	That leaves just
	<math>\cos x = 0</math>, and using the unit circle we see that the general solution is <math>x = \pi / 2 + n \cdot \pi</math>.		<math>\cos x = 0</math>, and using the unit circle we see that the general solution is <math>x = \pi / 2 + n \cdot \pi</math>.
	</div>		</div>
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	\cos x = \tfrac{1}{2}\,\mbox{.}</math>}}		\cos x = \tfrac{1}{2}\,\mbox{.}</math>}}

-	Since the value of <math>\cos x</math> is between <math>–1</math> and <math>1</math> the equation <math>\cos x=-\tfrac{3}{2}</math> has no solutions. That leaves only the basic equation	+	Since the value of <math>\cos x</math> is between <math>–1</math> and <math>1</math>, the equation <math>\cos x=-\tfrac{3}{2}</math> has no solutions. That leaves only the basic equation
	{{Displayed math\|\|<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}}		{{Displayed math\|\|<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}}

-	~~that~~ may be solved as in Example 2.	+	which may be solved as in Example 2.
	</div>		</div>

Jdinglis at 19:46, 10 November 2008

2008-11-10T19:46:09Z

←Older revision		Revision as of 19:46, 10 November 2008
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	\Rightarrow 4 – 4 \cos^2\!x – 4 \cos x &= 1\,\\		\Rightarrow 4 – 4 \cos^2\!x – 4 \cos x &= 1\,\\
	\Rightarrow –4\cos^2\!x – 4 \cos x + 4 – 1 &= 0\,\\		\Rightarrow –4\cos^2\!x – 4 \cos x + 4 – 1 &= 0\,\\
-	\~~RIghtarrow~~ \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\	+	\Rightarrow \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
	\end{align*}</math>}}		\end{align*}</math>}}

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-	'''Keep in mind that...''	+	'''Keep in mind that...'''

	It is a good idea to learn the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.		It is a good idea to learn the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.

Jdinglis at 19:44, 10 November 2008

2008-11-10T19:44:58Z

←Older revision		Revision as of 19:44, 10 November 2008
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	The figure illustrates the two points on the circle which have ''y''-coordinate <math>\tfrac{1}{2}</math>, i.e. the two points whose corresponding angles have sine value <math>\tfrac{1}{2}</math>. The first is the standard angle <math>30^\circ = \pi / 6</math> and by symmetry the other angle makes <math>30^\circ</math> with the negative ''x''-axis. Therefore the other angle is <math>180^\circ – 30^\circ = 150^\circ</math> or in radians <math>\pi – \pi / 6 = 5\pi / 6</math>. These are the only solutions to the equation <math>\sin x = \tfrac{1}{2}</math> between <math>0</math> and <math>2\pi</math>.		The figure illustrates the two points on the circle which have ''y''-coordinate <math>\tfrac{1}{2}</math>, i.e. the two points whose corresponding angles have sine value <math>\tfrac{1}{2}</math>. The first is the standard angle <math>30^\circ = \pi / 6</math> and by symmetry the other angle makes <math>30^\circ</math> with the negative ''x''-axis. Therefore the other angle is <math>180^\circ – 30^\circ = 150^\circ</math> or in radians <math>\pi – \pi / 6 = 5\pi / 6</math>. These are the only solutions to the equation <math>\sin x = \tfrac{1}{2}</math> between <math>0</math> and <math>2\pi</math>.

-	However, we can add an arbitrary number of revolutions to these two angles and still get the same value for the sine . Thus all angles <math>x</math> ~~such that~~ <math>\sin x = \tfrac{1}{2}</math> are	+	However, we can add an arbitrary number of revolutions to these two angles and still get the same value for the sine . Thus all angles <math>x</math> for which <math>\sin x = \tfrac{1}{2}</math> are
	{{Displayed math\|\|<math>\begin{cases}		{{Displayed math\|\|<math>\begin{cases}
	x &= \dfrac{\pi}{6} + 2n\pi\\		x &= \dfrac{\pi}{6} + 2n\pi\\
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	<center>{{:4.4 - Figure - Two unit circles with angles π/3 and -π/3, respectively}}</center>		<center>{{:4.4 - Figure - Two unit circles with angles π/3 and -π/3, respectively}}</center>

-	We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other ~~direction in~~ the unit circle, which ~~produces the same value for the cosine is~~ the angle <math>-\pi/3</math>. Adding an integral number of revolutions to these angles we get the general solution	+	We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other point on the unit circle which has "x"-coordinate <math>\frac{1}{2}</math> corresponds to the angle <math>-\pi/3</math>. Adding an integral number of revolutions to these angles we get the general solution

	{{Displayed math\|\|<math>x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}</math>}}		{{Displayed math\|\|<math>x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}</math>}}
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-	A solution to the equation is the standard angle <math>x=\pi/3</math>.	+	One solution to the equation is the standard angle <math>x=\pi/3</math>.

-	If we study the unit circle then we see that tangent of an angle is equal to the slope of the straight line through the origin ~~making~~ an angle <math>x</math> with the positive ''x''-axis .	+	If we study the unit circle then we see that the tangent of an angle is equal to the slope of the straight line through the origin which makes an angle <math>x</math> with the positive ''x''-axis .

	<center>{{:4.4 - Figure - Two unit circles with angles π/3 and π+π/3, respectively}}</center>		<center>{{:4.4 - Figure - Two unit circles with angles π/3 and π+π/3, respectively}}</center>

-	Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution ~~<math>\pi/3</math>~~, ~~<math>\pi/3 +\pi</math>, <math>\pi/3+ \pi +\pi</math>~~ and so ~~on. The~~ general solution can be obtained ~~by using~~ the solution <math>\pi/3</math> ~~and~~ adding or subtracting multiples of <math>\pi</math>,	+	Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution, and so the general solution can be obtained from the solution <math>\pi/3</math>by adding or subtracting multiples of <math>\pi</math>:
	{{Displayed math\|\|<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}}		{{Displayed math\|\|<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}}

-	where <math>n</math> s an arbitrary integer.	+	where <math>n</math> is an arbitrary integer.
	</div>		</div>

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	Trigonometric equations can vary in many ways, and it is impossible to give a full catalogue of all possible equations. But let us study some examples where we can use our knowledge of solving basic equations.		Trigonometric equations can vary in many ways, and it is impossible to give a full catalogue of all possible equations. But let us study some examples where we can use our knowledge of solving basic equations.

-	Some trigonometric equations can be simplified by ~~being rewritten~~ with the help of trigonometric relationships. This, for example, could lead to a quadratic equation, as in the example below ~~where one uses <math>\cos 2x = 2 \cos^2\!x – 1</math>~~.	+	Some trigonometric equations can be simplified by rewriting them with the help of the trigonometric relationships. This, for example, could lead to a quadratic equation, as in the example below.

	<div class="exempel">		<div class="exempel">
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-	Rewrite by using the formula <math>\cos 2x = 2 \cos^2\!x – 1</math> ~~giving~~	+	Rewrite by using the formula <math>\cos 2x = 2 \cos^2\!x – 1</math>, so that
-	{{Displayed math\|\|<math>(2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{,}</math>}}	+	{{Displayed math\|\|<math>(2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{.}</math>}}

-	~~which can be simplified to the equation (after division by 2)~~	+	Simplifying, we see that

	{{Displayed math\|\|<math>\cos^2\!x - 2 \cos x +1 =0\,\mbox{.}</math>}}		{{Displayed math\|\|<math>\cos^2\!x - 2 \cos x +1 =0\,\mbox{.}</math>}}
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	{{Displayed math\|\|<math>(\cos x-1)^2 = 0\,\mbox{.}</math>}}		{{Displayed math\|\|<math>(\cos x-1)^2 = 0\,\mbox{.}</math>}}

-	This equation can only be satisfied if <math>\cos x = 1</math>. The basic equation <math>\cos x=1</math> can be solved in the normal way and the complete solution is	+	This equation can only be satisfied if <math>\cos x = 1</math>. The basic equation <math>\cos x=1</math> can be solved in the normal way, and thus the complete solution is

	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
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-	According to the Pythagorean identity <math>\sin^2\!x + \cos^2\!x = 1</math>, i.e. <math>1 – \cos^2\!x = \sin^2\!x</math>, the equation can be written as	+	According to the Pythagorean identity <math>\sin^2\!x + \cos^2\!x = 1</math>, i.e. <math>1 – \cos^2\!x = \sin^2\!x</math>. Thus the equation can be written as
	{{Displayed math\|\|<math>\tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.}</math>}}		{{Displayed math\|\|<math>\tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.}</math>}}

-	Factorising out <math>\sin x</math> ~~one gets~~	+	Factorising out <math>\sin x</math> we get
	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	\sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}}		\sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}}

-	From this factorised form of the equation, we see that the solutions either have to satisfy <math>\sin x = 0</math> or <math>\sin x = -\tfrac{1}{2}</math>, which are two basic equations of the type <math>\sin x = a</math> and can be solved as in ~~example~~ 1. The solutions turn out to be	+	From this factorised form of the equation, we see that the solutions either have to satisfy <math>\sin x = 0</math> or <math>\sin x = -\tfrac{1}{2}</math>, which are two basic equations of the type <math>\sin x = a</math> and can be solved as in Example 1. The solutions turn out to be
	{{Displayed math\|\|<math>		{{Displayed math\|\|<math>
	\begin{cases}		\begin{cases}
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-	By rewriting the equation using the formula for double-angles ~~one gets~~	+	By rewriting the equation using the formula for double-angles we get
	{{Displayed math\|\|<math>2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}</math>}}		{{Displayed math\|\|<math>2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}</math>}}

-	~~We divide~~ both sides ~~with~~ 2 and ~~factorise~~ out <math>\cos x</math>, ~~which~~ gives	+	Dividing both sides by 2 and factorising out <math>\cos x</math>, gives
	{{Displayed math\|\|<math>\cos x\,\cdot\,( \sin x – 2) = 0\,\mbox{.}</math>}}		{{Displayed math\|\|<math>\cos x\,\cdot\,( \sin x – 2) = 0\,\mbox{.}</math>}}

-	~~As the product of factors on the~~ left-hand side can only be zero if one of the factors is zero, we have reduced the original equation into two basic equations	+	The left-hand side can only be zero if one of the factors is zero, and we have reduced the original equation into two basic equations:
	* <math>\cos x = 0</math>,		* <math>\cos x = 0</math>,
	* <math>\sin x = 2</math>.		* <math>\sin x = 2</math>.

-	~~But~~ <math>\sin x</math> can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions. That leaves just	+	However, <math>\sin x</math> can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions. That leaves just
-	<math>\cos x = 0</math>, and using the unit circle ~~gives~~ the general solution <math>x = \pi / 2 + n \cdot \pi</math>.	+	<math>\cos x = 0</math>, and using the unit circle we see that the general solution is <math>x = \pi / 2 + n \cdot \pi</math>.
	</div>		</div>

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-	Using the Pythagorean identity ~~one~~ can replace <math>\sin^2\!x</math> by <math>1 – \cos^2\!x</math>. Then ~~we will have~~{{Displayed math\|\|<math>	+	Using the Pythagorean identity we can replace <math>\sin^2\!x</math> by <math>1 – \cos^2\!x</math>. Then{{Displayed math\|\|<math>
	\begin{align*}		\begin{align*}
-	4 (1 – \cos^2\!x) – 4 \cos x &= 1\,~~\mbox{,}~~\\	+	4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\\
-	4 – 4 \cos^2\!x – 4 \cos x &= 1\,~~\mbox{,}~~\\	+	\Rightarrow 4 – 4 \cos^2\!x – 4 \cos x &= 1\,\\
-	–4\cos^2\!x – 4 \cos x + 4 – 1 &= 0\,~~\mbox{,}~~\\	+	\Rightarrow –4\cos^2\!x – 4 \cos x + 4 – 1 &= 0\,\\
-	\cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\	+	\RIghtarrow \cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
	\end{align*}</math>}}		\end{align*}</math>}}

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	{{Displayed math\|\|<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}}		{{Displayed math\|\|<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}}

-	that may be solved as in ~~example~~ 2.	+	that may be solved as in Example 2.
	</div>		</div>

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-	'''~~Remember:'~~''	+	'''Keep in mind that...''

	It is a good idea to learn the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.		It is a good idea to learn the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.