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In this video, we will learn to solve trigonometric equations using the double-angle identity.
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We will find solutions given a particular range in degrees or radians.
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The equations that we are dealing with will be able to be simplified using the double-angle identities or the half-angle identities.
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We will begin by recalling how we can solve simple equations using the CAST method or trig graphs.
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As already mentioned, two units for measuring angles are degrees and radians.
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We recall that 180 degrees is equal to 𝜋 radians.
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This means that 90 degrees is equal to 𝜋 over two radians and 360 degrees is equal to two 𝜋 radians.
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We can find the solutions to a trig equation by drawing the appropriate graph or by using the CAST diagram.
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Whilst there are an infinite number of solutions to our equations, in this video, we will focus on solutions between zero and 360 degrees or zero and two 𝜋 radians.
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We measure the angles in a counterclockwise direction from the positive 𝑥-axis as shown.
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We can replace the values in degrees by the corresponding values in radians.
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In the first quadrant, labeled A, the values of sin 𝜃, cos of 𝜃, and tan of 𝜃 are all positive.
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In the second quadrant, labeled S, the value of sin 𝜃 is positive, whereas the value of cos 𝜃 and tan 𝜃 are negative.
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In the third quadrant, the tan of 𝜃 is positive, whereas the sin of 𝜃 and cos of 𝜃 are negative.
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This is therefore true when 𝜃 lies between 180 and 270 degrees.
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Finally, when 𝜃 lies between 270 and 360 degrees, known as the fourth quadrant, the value of cos 𝜃 is positive, whereas the values of sin 𝜃 and tan 𝜃 are negative.
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We can use this information to solve simple trig equations.
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Find the set of all possible solutions of cos 𝜃 equals one-half, given that 𝜃 exists between zero and 360 degrees inclusive.
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The square brackets here mean that 𝜃 must be greater than or equal to zero degrees and less than or equal to 360 degrees.
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If we had curved brackets or parentheses, then 𝜃 would be strictly greater than or less than the value.
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To solve the equation cos 𝜃 equals one-half, we take the inverse cosine of both sides of the equation.
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Solving this gives us a value of 𝜃 equal to 60 degrees.
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At this stage, it is worth sketching our CAST diagram, as we can then see which quadrants our solutions lie in.
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As the cos of 𝜃 is positive, there will be solutions in the first and fourth quadrants.
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We already have the solution in the first quadrant.
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This is equal to 60 degrees.
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We can then calculate the solution in the fourth quadrant by subtracting 60 degrees from 360 degrees.
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This gives us an answer of 300 degrees.
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The set of solutions of cos 𝜃 equals one-half between zero and 360 degrees are 60 degrees and 300 degrees.
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We will now consider some more complicated problems using the double-angle identities.
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Our double-angle identities are as follows.
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Firstly, the sin of two 𝜃 is equal to two sin 𝜃 cos 𝜃.
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Note that identities are also sometimes written with a triple line, which shows they are true for all values of the variable.
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The cos of two 𝜃 is equal to cos squared 𝜃 minus sin squared 𝜃.
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Using the fact that sin squared 𝜃 plus cos squared 𝜃 equals one, cos two 𝜃 is also equal to two cos squared 𝜃 minus one and one minus two sin squared 𝜃.
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Finally, we have the tan of two 𝜃 is equal to two tan 𝜃 divided by one minus tan squared 𝜃.
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In this video, we will not prove these identities.
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We will simply quote them to solve our equations.
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In our next question, we will use the fact that sin two 𝜃 is equal to two sin 𝜃 cos 𝜃.
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Find the set of possible solutions of two sin 𝜃 cos 𝜃 equals zero, given 𝜃 is greater than or equal to zero and less than 360 degrees.
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The square bracket tells us that 𝜃 is greater than or equal to zero degrees, whereas the curved bracket tells us that 𝜃 is strictly less than 360 degrees.
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We can solve our equation using our knowledge of the double-angle identities.
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We know that sin of two 𝜃 is equal to two sin 𝜃 cos 𝜃.
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This means that we need to solve sin of two 𝜃 is equal to zero.
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Taking the inverse sine of both sides of this equation, we get two 𝜃 is equal to the inverse sin of zero.
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Whilst solving this equation would give us one solution, it is worth sketching the graphs of sin 𝜃 and sin two 𝜃 before proceeding.
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We know that sin 𝜃 is a periodic function, and we are interested in values between zero and 360 degrees.
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It has a maximum value of one and a minimum value of negative one, as shown.
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The graph of sin two 𝜃 will be a dilation or enlargement.
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The graph will be stretched by a scale factor of one-half in the horizontal direction.
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This means that it will look as shown.
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This has a maximum value of one when 𝜃 is 45 degrees and a minimum value of negative one when 𝜃 is 135 degrees.
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We are interested in the points at which the graph is equal to zero.
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There are five such points on our graph, at zero, 90, 180, 270, and 360 degrees.
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We recall though that 𝜃 must be less than 360 degrees.
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So our last point is not a solution.
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The four solutions of the equation two sin 𝜃 cos 𝜃 equals zero, where 𝜃 is greater than or equal to zero degrees and less than 360 degrees, are zero, 90, 180, and 270 degrees.
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In our next example, we will use one of the double-angle identities involving cos of two 𝜃.
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Find the solution set for 𝑥 given the cos of two 𝑥 plus 13 root three multiplied by the cos of 𝑥 is equal to negative 19, where 𝑥 exists between zero and two 𝜋.
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The parentheses or curved brackets tell us that 𝑥 is strictly greater than zero and less than two 𝜋.
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This also indicates that we need to give our solutions in radians.
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We can solve the equation in this range by using our double-angle identities.
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One of these identities states that the cos of two 𝑥 is equal to two cos squared 𝑥 minus one.
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We can therefore rewrite our equation as shown.
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We can then add 19 to both sides of the equation, giving us two cos squared 𝑥 plus 13 root three cos 𝑥 plus 18 is equal to zero.
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If we let 𝑦 equal cos of 𝑥, this can be rewritten as two 𝑦 squared plus 13 root three 𝑦 plus 18 equals zero.
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We now have a quadratic equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, which we can solve using the quadratic formula as shown.
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Our values of 𝑎, 𝑏, and 𝑐 are two, 13 root three, and 18, respectively.
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Substituting these values into the formula, we get two values for 𝑦.
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Either 𝑦 is equal to negative root three over two or negative six root three.
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This means that the cos of 𝑥 must be equal to negative root three over two or negative six root three.
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As the value of the cos of 𝑥 must lie between negative one and one inclusive, the second option has no solutions.
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We can now solve the equation the cos of 𝑥 is equal to negative root three over two using our CAST diagram.
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As the cos of our angle is negative, we will have solutions in the second and third quadrants.
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One of our solutions will be between 𝜋 over two and 𝜋 and our second between 𝜋 and three 𝜋 over two.
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We know that the cos of 30 degrees or 𝜋 over six radians is equal to root three over two.
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This means that our two solutions will be equal to 𝜋 minus 𝜋 over six and 𝜋 plus 𝜋 over six.
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This gives us two solutions: five 𝜋 over six and seven 𝜋 over six.
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The solution set of the equation cos two 𝑥 plus 13 root three cos 𝑥 equals negative 19 are five 𝜋 over six and seven 𝜋 over six.
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Before answering our final question, we will consider the half-angle identities.
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Firstly, we have the sin of 𝜃 over two is equal to the positive or negative of the square root of one minus cos 𝜃 all divided by two.
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The second identity is very similar.
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The cos of 𝜃 over two is equal to the positive or negative square root of one plus cos 𝜃 all divided by two.
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Finally, we have the tan of 𝜃 over two is equal to the sin of 𝜃 divided by one plus the cos of 𝜃.
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Alternatively, this is equal to one minus the cos of 𝜃 divided by the sin of 𝜃.
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We will now consider a question where we need to use one of these identities.
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Solve tan of 𝑥 over two is equal to sin 𝑥, where 𝑥 is greater than or equal to zero and less than two 𝜋 radians.
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We are told in the question that our solutions must be less than two 𝜋 and greater than or equal to zero.
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This means that we will be able to solve the equations using the CAST diagram or by sketching our trig graphs.
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Using one of our half-angle identities, we know that tan of 𝑥 over two is equal to one minus cos 𝑥 divided by sin 𝑥.
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This means that we can rewrite our equation as one minus cos 𝑥 divided by sin 𝑥 is equal to sin 𝑥
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Multiplying both sides of the equation by sin 𝑥 gives us one minus cos 𝑥 is equal to sin squared 𝑥.
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As sin squared 𝑥 is equal to one minus cos squared 𝑥, we have one minus cos 𝑥 is equal to one minus cos squared 𝑥.
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We can then add cos squared 𝑥 and subtract one from both sides of our equation.
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This gives us cos squared 𝑥 minus cos 𝑥 is equal to zero.
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Factoring out cos 𝑥 gives us cos 𝑥 multiplied by cos 𝑥 minus one is equal to zero.
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This means that either cos of 𝑥 equals zero or cos of 𝑥 minus one equals zero.
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The second equation can be rewritten as the cos of 𝑥 equals one.
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We now have two equations that can be solved either by using the CAST diagram or by drawing the graph of cos 𝑥.
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The cosine graph between zero and two 𝜋 looks as shown.
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Remembering that our values need to be greater than or equal to zero and less than two 𝜋, we see that the graph is equal to zero when 𝑥 is equal to 𝜋 over two and three 𝜋 over two.
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cos of 𝑥 is equal to one when 𝑥 is equal to zero or two 𝜋.
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However, two 𝜋 is not within the range of values for 𝑥.
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This means that there are three solutions to our equation.
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The tan of 𝑥 over two is equal to the sin of 𝑥 when 𝑥 is equal to zero, 𝜋 over two, and three 𝜋 over two.
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We will now summarize the key points from this video.
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We saw in this video that we can solve trig equations by using the double-angle identities.
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We can also use adaptions of these equations known as the half-angle identities.
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In both of these cases, we will usually be given a range of values often between zero and 360 degrees or zero and two 𝜋 radians.
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We can then use our CAST diagram or the trig graphs to help us solve the equations.