Video: Using the Pythagorean Identities to Evaluate a Trigonometric Function Involving Cofunction Identities given the Trigonometric Function and Quadrant of an Angle

Find the value of sec (90Β° + πœƒ) given csc πœƒ = 17/8 where 0Β° < πœƒ < 90Β°.

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Video Transcript

Find the value of sec 90 degrees plus πœƒ given that csc πœƒ equals 17 over eight, where πœƒ is between zero degrees and 90 degrees.

There are various ways that we can solve this question. We can use the fact that csc πœƒ is 17 over eight and πœƒ is between zero degrees and 90 degrees to find the value of πœƒ. And once we have this value of πœƒ, we can just substitute it into the expression sec 90 degrees plus πœƒ that we wanted to evaluate.

So let’s do this. Csc πœƒ is equal to 17 over eight; csc πœƒ is a one over sin πœƒ, and so we have that one over sin πœƒ is 17 over eight. Taking the reciprocal on both sides of the equation, or just rearranging the equation to make the subject sin πœƒ, we see that sin πœƒ is equal to eight over 17, and hence that πœƒ is the inverse sin or arcsin of eight over 17.

With our calculator in degree mode, we find that the πœƒ is equal to 28.072 dot dot dot degrees. This value that are calculator gives us is in the range that we want; it’s between zero degrees and 90 degrees. And so we don’t have to modify it in any way. We just substitute this value of πœƒ into sec 90 degrees plus πœƒ, remembering that sec π‘₯ is one over cos π‘₯ if our calculator doesn’t have a sec button to get a value of negative 17 over eight.

But perhaps your calculator doesn’t give you this exact value, negative 17 over eight, as a fraction or perhaps you can’t use a calculator at all. Luckily we can answer this question without using a calculator. We go back a few steps to where we discovered that sin πœƒ is equal to eight over 17, and we turn our attention to sec 90 degrees plus πœƒ.

Sec π‘₯ is one over cos π‘₯ and so sec 90 degrees plus πœƒ is one over cos 90 degrees plus πœƒ. We can use the angle sum formula for cosine, setting π‘₯ equal to 90 degrees and 𝑦 equal to πœƒ. 90 degrees is a special angle, so we know the values of cos 90 degrees and sin 90 degrees.

Cos 90 degrees is zero and sin 90 degrees is one. The zero cos πœƒ term disappears, and we’re left with one over negative one sin πœƒ, which is negative one over sin πœƒ. And we know that one over sin πœƒ is equal to csc πœƒ, so we find this is negative csc πœƒ. And we’re told in the question that csc πœƒ is 17 over eight, so the value of sec 90 degrees plus πœƒ is negative 17 over eight.

So that method preceded by simplifying sec 90 degrees plus πœƒ using an angle sum identity. There is a third method which I’ll show you for fun, which involves doing something slightly nonobvious. We write down the expression we wish to evaluate again, sec 90 degrees plus πœƒ, and we notice that it looks very much like the left-hand side of the cofunction identity sec 90 degrees minus π‘₯ equals csc π‘₯.

The only difference is that we have a 90 degrees plus something and not 90 degrees minus something. But we can fix that by writing 90 degrees plus πœƒ as 90 degrees minus negative πœƒ. We set π‘₯ equal to negative πœƒ in the identity to get csc negative πœƒ. But csc is an odd function, and if you don’t believe me, we’ll prove this at the end of the video.

And so csc of negative πœƒ is negative csc πœƒ. Using the value of csc πœƒ we’re given in the question, we get a value of negative 17 over eight. That was a much quicker way of finding the value of sec 90 degrees plus πœƒ using the cofunction identity sec 90 degrees minus π‘₯ equals csc π‘₯ and the fact that csc is an odd function.

This last fact is something we should prove csc negative π‘₯ is equal to one over sin negative π‘₯ by definition and certainly sine is an odd function. We can prove this, but we’re not going to in this video. And so sine of negative π‘₯ is negative of sine π‘₯. Multiplying the numerator and denominator of this fraction by negative one, we get that this is equal to negative one over sine π‘₯. And one over sine π‘₯ is csc π‘₯, so we get that this is equal to negative csc π‘₯.

We have proved that csc negative π‘₯ is equal to negative csc π‘₯, and hence that the csc function is odd.

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