Video: Evaluating a Trigonometric Function for the Sum of Two Angles given the Trigonometric Functions and the Quadrants of Two Angles

Find sin (𝐴 + 𝐡) given sin 𝐴 = βˆ’24/25 where 270Β° < 𝐴 < 360Β° and cos 𝐡 = βˆ’4/5 where 180Β° < 𝐡 < 270Β°.

03:52

Video Transcript

Find sin 𝐴 plus 𝐡 given sin 𝐴 is equal to negative 24 over 25 where 𝐴 is greater than 270 degrees but less than 360 degrees and cos 𝐡 is equal to negative four-fifths where 𝐡 is greater than 180 degrees and less than 270 degrees.

We begin by recalling one of our compound angle identities. This states that sin 𝐴 plus 𝐡 is equal to sin 𝐴 cos 𝐡 plus cos 𝐴 sin 𝐡. We are given values in the question for sin 𝐴 and cos 𝐡. We need to use our knowledge of our cos diagram and Pythagorean triples to calculate cos 𝐴 and sin 𝐡. We’re told that angle 𝐴 lies between 270 and 360 degrees. This means it is in the fourth quadrant. The value of cos 𝐴 will be positive. However, the value of sin 𝐴 will be negative. This ties in with the value of sin 𝐴 we have been given, negative 24 over 25.

Angle 𝐡 lies in the third quadrant, as it is between 180 and 270 degrees. In this quadrant, only tan of the angle is positive. Therefore, cos 𝐡 and sin 𝐡 will both be negative. We already know that cos 𝐡 is negative four-fifths. We now recall two of our Pythagorean triples. These are three, four, five and seven, 24, 25, as three squared plus four squared is equal to five squared and seven squared plus 24 squared is equal to 25 squared. If sin 𝐴 was equal to 24 over 25, then cos 𝐴 would be equal to seven over 25. Likewise, if cos 𝐡 was equal to four-fifths, this would imply that sin 𝐡 is equal to three-fifths.

Using this information tells us that if sin 𝐴 is equal to negative 24 over 25, then cos 𝐴 is equal to seven over 25, when 𝐴 is between 270 and 360 degrees. If cos 𝐡 is equal to negative four-fifths, then sin 𝐡 is equal to negative three-fifths, when 𝐡 lies between 180 and 270 degrees. Sin 𝐴 plus 𝐡 is, therefore, equal to negative 24 over 25 multiplied by negative four over five plus seven over 25 multiplied by negative three over five.

Multiplying the first two fractions gives us 96 over 125. Multiplying the second two fractions gives us negative 21 over 125. This simplifies to 96 over 125 minus 21 over 125. 96 minus 21 is equal to 75. Both the numerator and denominator are divisible by 25. Sin 𝐴 plus 𝐡 in this question is, therefore, equal to three-fifths. 75 divided by 25 is three, and 125 divided by 25 is five.

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