Lesson Explainer: Evaluating Trigonometric Ratios given the Value of Another Ratio | Nagwa Lesson Explainer: Evaluating Trigonometric Ratios given the Value of Another Ratio | Nagwa

Lesson Explainer: Evaluating Trigonometric Ratios given the Value of Another Ratio Mathematics • First Year of Secondary School

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In this explainer, we will learn how to find the value of a trigonometric function from a given value of another trigonometric function.

Let us begin by recalling the trigonometric ratios and the unit circle. The unit circle is a circle with a radius of 1 whose center lies at the origin of a coordinate plane. For any point (𝑥,𝑦) on the unit circle, a right triangle can be formed as shown in the following diagram. The hypotenuse of this right triangle makes an angle 𝜃 with the positive 𝑥-axis.

Using right-angled trigonometry, we can define the trigonometric functions in terms of the unit circle: sinoppositehypotenusesosincosadjacenthypotenusesocostanoppositeadjacentsotan𝜃==𝑦1,𝑦=𝜃,𝜃==𝑥1,𝑥=𝜃,𝜃==𝑦𝑥,𝑦𝑥=𝜃.

We note that tan𝜃 is not defined when 𝑥=0.

The 𝑥- and 𝑦-coordinates of a point on the unit circle given by an angle 𝜃 are defined by 𝑥=𝜃cos and 𝑦=𝜃sin.

We also observe that while we have derived these definitions for an angle 𝜃 in quadrant 1, they hold for an angle in any quadrant. We know that to the right of the origin, the 𝑥-values are positive, and to the left of the origin, the 𝑥-values are negative. Similarly, above the origin, the 𝑦-values are positive, and below the origin, the 𝑦-values are negative.

If we add four points to our grid, (𝑥,𝑦), (−𝑥,𝑦), (−𝑥,−𝑦), (𝑥,−𝑦), where 𝑥 and 𝑦 are positive values, we see that they lie in each of the four quadrants.

At this stage, we also recall that the trigonometric functions cosecant 𝜃, secant 𝜃, and cotangent 𝜃 are the reciprocals of sine 𝜃, cosine 𝜃, and tangent 𝜃 such that cscsinseccoscottan𝜃=1𝜃,𝜃=1𝜃,𝜃=1𝜃.

From the definition of the unit circle, we have cschypotenuseoppositesechypotenuseadjacentcotadjacentopposite𝜃==1𝑦,𝜃==1𝑥,𝜃==𝑥𝑦.

The Quadrant in Which the Terminal Side of the Angle LiesThe Interval in Which the Measure of the Angle BelongsSigns of Trigonometric Functions
csc, sincos, sectan, cot
First0,𝜋2+++
Second𝜋2,𝜋+−−
Third𝜋,3𝜋2−−+
Fourth3𝜋2,2𝜋−+−

We now consider a series of examples where we need to find the value of a trigonometric function from a given value of another trigonometric function.

In our first example, we need to calculate the sine of an angle, given both its cosine and tangent.

Example 1: Finding the Value of the Sine of an Angle given the Values of the Tangent and Cosine

Find sin𝐵 given tan𝐵=43 and cos𝐵=35.

Answer

We begin by recalling that since the tangent and cosine of our angle are both positive, then the angle must lie in the first quadrant such that 0<𝐵<90∘∘. From our CAST diagram, we observe that the sine of the angle must also be positive.

As cosandtan𝐵=35𝐵=43, we can sketch a right triangle in the first quadrant. The right triangle is a Pythagorean triple consisting of three positive integers 3, 4, and 5 such that 3+4=5.

Since sinoppositehypotenuse𝐵=, then, from the diagram, sin𝐵=45.

We can therefore conclude that if tan𝐵=43 and cos𝐵=35, then sin𝐵=45.

For the remainder of the examples, we will need to use the reciprocal trigonometric identities.

Example 2: Finding the Value of the Cosecant of an Angle given the Cotangent

If cot𝜃=−43 and cos𝜃>0, find csc𝜃.

Answer

We begin by recalling that the cotangent of an angle is the reciprocal of the tangent of that angle such that cottan𝜃=1𝜃.

Since cot𝜃=−43, then tan𝜃=−34.

Using our knowledge of the CAST diagram, as cos𝜃>0 and tan𝜃<0, we know that our angle lies in the fourth quadrant.

Since sin𝜃<0 in the fourth quadrant and as the cosecant of an angle is the reciprocal of the sine of that angle, then csc𝜃<0.

As tan𝜃=−34, we can sketch a right triangle in the fourth quadrant. The right triangle is a Pythagorean triple consisting of three positive integers 3, 4, and 5 such that 3+4=5.

Since sinoppositehypotenuse𝛼=, then, from the diagram, sin𝛼=35.

From the diagram, we see that 𝜃=360−𝛼∘. Using the properties of related angles, we know that sinsin(360−𝛼)=−𝛼.∘

So, sinsinsin𝜃=−𝛼𝜃=−35.

As cscsin𝜃=1𝜃, then, csc𝜃=−53.

We can therefore conclude that if cot𝜃=−43 and cos𝜃>0, then csc𝜃=−53.

Example 3: Finding the Cotangent of an Angle in a Specified Range given the Value of Sine

Find cot𝜃 given sin𝜃=35, where 90<𝜃<180∘∘.

Answer

Using our knowledge of the CAST diagram, as 90<𝜃<180∘∘, we know that our angle lies in the second quadrant.

Since tan𝜃<0 in the second quadrant and as the cotangent of an angle is the reciprocal of the tangent of that angle, then cot𝜃<0.

As sin𝜃=35, we can sketch a right triangle in the second quadrant. The right triangle is a Pythagorean triple consisting of three positive integers 3, 4, and 5 such that 3+4=5.

Since tanoppositeadjacent𝛼=, then, from the diagram, tan𝛼=34.

From the diagram, we see that 𝜃=180−𝛼∘ and, using the properties of related angles, we know that tantan(180−𝛼)=−𝛼.∘

So, tantantan𝜃=−𝛼𝜃=−34.

As cottan𝜃=1𝜃, then, cot𝜃=−43.

We can therefore conclude that if sin𝜃=35, where 90<𝜃<180∘∘, then cot𝜃=−43.

Example 4: Finding the Value of an Expression Using Trigonometric Equivalences

Find the value of sincoscossin𝛼𝛽−𝛼𝛽, given tan𝛼=34, where 𝛼 is the smallest positive angle, and tan𝛽=158, where 180<𝛽<270∘∘.

Answer

Using our knowledge of the CAST diagram, as tan𝛼=34, where 𝛼 is the smallest positive angle, we know that 𝛼 lies in the first quadrant. Since tan𝛽=158, where 180<𝛽<270∘∘, then 𝛽 lies in the third quadrant.

As tan𝛼=34, we can sketch a right triangle in the first quadrant. The right triangle is a Pythagorean triple consisting of three positive integers 3, 4, and 5 such that 3+4=5.

Since sinoppositehypotenuseandcosadjacenthypotenuse𝛼=𝛼=, then, from the diagram, sinandcos𝛼=35𝛼=45.

As tan𝛽=158, we can sketch a right triangle in the third quadrant. The right triangle is a Pythagorean triple consisting of three positive integers 8, 15, and 17 such that 8+15=17.

Since sinoppositehypotenuseandcosadjacenthypotenuse𝜃=𝜃=, then, from the diagram, sinandcos𝜃=1517𝜃=817.

From the diagram, we see that 𝛽=180+𝜃∘ and, using the properties of related angles, we know that sinsin(180+𝜃)=−𝜃.∘

So, sinsinsin𝛽=−𝜃𝛽=−1517.

Also, coscos(180+𝜃)=−𝜃.∘

So, coscoscos𝛽=−𝜃𝛽=−817.

Substituting these values into our expression, we have sincoscossin𝛼𝛽−𝛼𝛽=35−817−45−1517=−2485−−6085=3685.

We can therefore conclude that if tan𝛼=34, where 𝛼 is the smallest positive angle, and tan𝛽=158, where 180<𝛽<270∘∘, then sincoscossin𝛼𝛽−𝛼𝛽=3685.

Example 5: Finding the Value of an Expression Using Trigonometric Equivalences

Find the value of seccsccot𝜃𝜃−𝜃, given that 180<𝜃<270∘∘ and sin𝜃=−35.

Answer

Using our knowledge of the CAST diagram, as 180<𝜃<270∘∘, we know that 𝜃 lies in the third quadrant.

As sin𝜃=−35, we can sketch a right triangle in the third quadrant. The right triangle is a Pythagorean triple consisting of three positive integers 3, 4, and 5 such that 3+4=5.

Since cosadjacenthypotenuseandtanoppositeadjacent𝛼=𝛼=, then, from the diagram, cosandtan𝛼=45𝛼=34.

From the diagram, we see that 𝜃=180+𝛼∘ and, using the properties of related angles, we know that coscos(180+𝛼)=−𝛼.∘

So, coscoscos𝜃=−𝛼𝜃=−45.

Also, tantan(180+𝛼)=𝛼.∘

So, tantantan𝜃=𝛼𝜃=34.

The reciprocal trigonometric identities cosecant 𝜃, secant 𝜃, and cotangent 𝜃 are the reciprocals of sine 𝜃, cosine 𝜃, and tangent 𝜃 such that cscsinseccoscottan𝜃=1𝜃,𝜃=1𝜃,𝜃=1𝜃.

So, cscseccot𝜃=−53,𝜃=−54,𝜃=43.

Substituting these values into our expression, we have seccsccot𝜃𝜃−𝜃=−54−53−43=2512−1612=912=34.

We can therefore conclude that if 180<𝜃<270∘∘ and sin𝜃=−35, then seccsccot𝜃𝜃−𝜃=34.

Example 6: Finding the Value of an Expression Involving Reciprocal Trigonometric Functions Using Trigonometric Equivalences

Find the value of cotcsctansec𝜃−𝜃𝜃−𝜃 given 𝜃∈3𝜋2,2𝜋 and sin𝜃=−45.

Answer

Using our knowledge of the CAST diagram, as 𝜃∈3𝜋2,2𝜋, where 3𝜋2=270radians∘ and 2𝜋=360radians∘, we know that 𝜃 lies in the fourth quadrant.

As sin𝜃=−45, we can sketch a right triangle in the fourth quadrant. The right triangle is a Pythagorean triple consisting of three positive integers 3, 4, and 5 such that 3+4=5.

Since cosadjacenthypotenuseandtanoppositeadjacent𝛼=𝛼=, then, from the diagram, cosandtan𝛼=35𝛼=43.

From the diagram, we see that 𝜃=360−𝛼∘ and, using the properties of related angles, we know that coscos(360−𝛼)=𝛼.∘

So, coscoscos𝜃=𝛼𝜃=35.

Also, tantan(360−𝛼)=−𝛼.∘

So, tantantan𝜃=−𝛼𝜃=−43.

The reciprocal trigonometric identities cosecant 𝜃, secant 𝜃, and cotangent 𝜃 are the reciprocals of sine 𝜃, cosine 𝜃, and tangent 𝜃 such that cscsinseccoscottan𝜃=1𝜃,𝜃=1𝜃,𝜃=1𝜃.

So, cscseccot𝜃=−54,𝜃=53,𝜃=−34.

Substituting these values into our expression, we have cotcsctansec𝜃−𝜃𝜃−𝜃=−−==−3=−16.

We can therefore conclude that if 𝜃∈3𝜋2,2𝜋 and sin𝜃=−45, then cotcsctansec𝜃−𝜃𝜃−𝜃=−16.

We will finish this explainer by recapping some of the key points.

Key Points

  • We can find the value of a trigonometric function from a given value of another trigonometric function by recalling the 6 trigonometric functions and the CAST diagram.
  • Using the symmetries in the unit circle, we can also use the related-angle properties: sinsincoscostantansinsincoscostantansinsincoscostantan(180−𝜃)=𝜃,(180−𝜃)=−𝜃,(180−𝜃)=−𝜃,(180+𝜃)=−𝜃,(180+𝜃)=−𝜃,(180+𝜃)=𝜃,(360−𝜃)=−𝜃,(360−𝜃)=𝜃,(360−𝜃)=−𝜃.∘∘∘∘∘∘∘∘∘
  • These properties can be seen on the following diagram of the unit circle, where ±𝜃cos is the 𝑥-coordinate and ±𝜃sin is the 𝑦-coordinate.

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