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In this video, we will learn how to simplify trigonometric expressions by applying trigonometric identities.
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We begin by recalling that an identity is an equation that is true no matter what values are chosen.
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We will use combinations of these trigonometric identities, for example, the cofunction identities, shift identities, and Pythagorean identities.
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Before looking at some specific examples, letβs consider the properties of the unit circle.
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We recall that the unit circle is a circle of radius one as shown.
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It enables us to measure the sin, cos, or tan of any angle π, where π is measured in the counterclockwise direction from the positive π₯-axis.
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The π₯-coordinate of any point on the unit circle is equal to cos π, and the π¦-coordinate is equal to sin π.
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The right triangle on our diagram together with the Pythagorean theorem leads us to the first Pythagorean identity.
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sin squared π plus cos squared π is equal to one.
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Recalling the reciprocal trigonometric identities enables us to form two further Pythagorean identities.
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The three reciprocal functions are the cosecant, secant, and cotangent such that csc π is equal to one over sin π, sec π is equal to one over cos π, and cot π is equal to one over tan π.
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It is also worth noting that since tan π is equal to sin π over cos π, then cot π is equal to cos π over sin π.
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Dividing both sides of our first Pythagorean identity by cos squared π, we have sin squared π over cos squared π plus cos squared π over cos squared π is equal to one over cos squared π.
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Using the reciprocal identities, this simplifies to tan squared π plus one is equal to sec squared π.
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In the same way, we can divide both sides of the first identity by sin squared π.
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This simplifies to one plus cot squared π is equal to csc squared π.
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We now have a set of three Pythagorean identities, which we will use together with the reciprocal identities to solve a couple of examples.
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Simplify sin π multiplied by csc π minus cos squared π.
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In this question, we are asked to simplify a trigonometric expression.
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One way of doing this is using the reciprocal and Pythagorean identities.
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In questions of this type, it is not always clear what to do first.
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However, as a general rule, it is worth replacing any reciprocal functions with the sine, cosine, or tangent function.
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We know that csc π is equal to one over sin π.
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Substituting this into our expression, we have sin π multiplied by one over sin π minus cos squared π.
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The sin π on the numerator and denominator of our first term cancels, leaving us with one minus cos squared π.
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Next, we recall one of the Pythagorean identities: sin squared π plus cos squared π is equal to one.
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Subtracting cos squared π from both sides, this can be rewritten as sin squared π is equal to one minus cos squared π.
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This means that our expression can be rewritten as sin squared π.
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sin π multiplied by csc π minus cos squared π written in its simplest form is sin squared π.
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We will now consider a second example of this type.
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Simplify sin squared π plus cos squared π divided by csc squared π minus cot squared π.
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In order to answer this question, we need to recall the Pythagorean identities.
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Firstly, we have sin squared π plus cos squared π is equal to one.
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Dividing each term by sin squared π and using our knowledge of the reciprocal trigonometric functions, we have one plus cot squared π is equal to csc squared π.
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We notice that the left-hand side of the first identity is identical to the numerator of our expression.
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Subtracting cot squared π from both sides of the second identity, we have one is equal to csc squared π minus cot squared π.
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The right-hand side of this is the same as the denominator of our expression.
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As sin squared π plus cos squared π equals one and csc squared π minus cot squared π also equals one, our expression simplifies to one divided by one.
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And this is equal to one.
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The expression sin squared π plus cos squared π over csc squared π minus cot squared π is equal to one.
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Before looking at one final example, we will recall the cofunction and shift identities.
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Once again, we begin by considering the unit circle.
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Since the angles in a triangle sum to 180 degrees, the third angle in our right triangle is equal to 90 degrees minus π.
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Letβs consider what happens if we redraw this triangle such that the angle between the positive π₯-axis and the hypotenuse is 90 degrees minus π.
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The coordinates of the point marked on the unit circle will be cos of 90 degrees minus π, sin of 90 degrees minus π.
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We notice that the distance in the π₯-direction here is the same as the distance in the π¦-direction in our first triangle.
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This means that the cos of 90 degrees minus π must be equal to sin π.
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Likewise, the sin of 90 degrees minus π is equal to cos π.
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Since sin π over cos π is equal to tan π, the tan of 90 degrees minus π is equal to cos π over sin π.
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And using our knowledge of the reciprocal functions, this is equal to cot π.
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It follows that the sec of 90 degrees minus π is equal to csc π.
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The csc of 90 degrees minus π is equal to sec π.
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And the cot of 90 degrees minus π is equal to tan π.
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These six identities are known as the cofunction identities.
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We can also use the unit circle to find identities involving angles such as 180 degrees minus π, 180 degrees plus π, and 360 degrees minus π.
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In our final example, we will use these identities together with the Pythagorean identities to simplify an expression.
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Simplify one plus cot squared three π over two minus π over one plus tan squared π over two minus π.
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In order to answer this question, we will need to use a variety of trigonometric identities.
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There are many ways to start here.
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However, we will begin by trying to rewrite the expression simply in terms of π.
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By firstly sketching the unit circle, we recall that π radians is equal to 180 degrees.
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This means that π over two radians is equal to 90 degrees.
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The denominator of our expression can therefore be rewritten as one plus tan squared of 90 degrees minus π.
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One of our cofunction identities states that tan of 90 degrees minus π is equal to cot π.
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This means that tan squared of 90 degrees minus π is equal to cot squared π.
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And the denominator of our expression is therefore equal to one plus cot squared π.
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Letβs now consider the angle three π over two minus π.
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Once again, we can see from the unit circle that three π over two radians is equal to 270 degrees.
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This means that the numerator of our expression is equal to one plus cot of 270 degrees minus π.
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If π lies in the first quadrant, as shown in our right triangle, then three π over two minus π, or 270 degrees minus π, lies in the third quadrant.
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It is clear from the diagram that cos of three π over two minus π is equal to negative sin π and sin of three π over two minus π is equal to negative cos π.
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Since sin π over cos π is tan π and cos π over sin π is cot π, then cot of 270 degrees minus π is equal to tan π.
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Squaring both sides of this identity, we can rewrite the numerator of our expression as one plus tan squared π.
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Our next step is to recall two of the Pythagorean identities.
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Firstly, tan squared π plus one is equal to sec squared π.
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And secondly, one plus cot squared π is equal to csc squared π.
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Our expression simplifies to sec squared π over csc squared π.
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And this can be rewritten as sec squared π multiplied by one over csc squared π.
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Recalling the reciprocal identities sec π is equal to one over cos π and csc π is equal to one over sin π, we have one over cos squared π multiplied by sin squared π, which can be rewritten as sin squared π over cos squared π and, in turn, is equal to tan squared π.
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The expression one plus cot squared three π over two minus π over one plus tan squared π over two minus π written in its simplest form is tan squared π.
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We will now summarize the key points from this video.
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In this video, we simplified trigonometric expressions using a variety of trigonometric identities.
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We used the three Pythagorean identities sin squared π plus cos squared π is equal to one, tan squared π plus one is equal to sec squared π, and one plus cot squared π is equal to csc squared π.
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We also used the reciprocal identities csc π is equal to one over sin π, sec π is equal to one over cos π, and cot π is equal to one over tan π.
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Since tan π is equal to sin π over cos π, we also saw that cot π is equal to cos π over sin π.
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We also recalled the cofunction identities sin of 90 degrees minus π is equal to cos π and cos of 90 degrees minus π is equal to sin π.
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And using the reciprocal identities above, these led us to four further cofunction identities.
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Finally, we used the unit circle to determine other related angle identities as demonstrated in our final example.
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In many examples, we need to apply more than one identity or type of identity to simplify a trigonometric expression.