In Exercises 21–24, θ is an acute angle and sin θ is given. Use t...

Question

In Exercises 21–24, θ is an acute angle and sin θ is given. Use the Pythagorean identity sin²θ + cos²θ = 1 to find cos θ.sin θ = 6/7

Accepted Answer

Hello everybody. I hope we doing tonight today. Today we're going to begin this question that states the trigonometric functions sign of alpha is given. And they're asking us to find the value of cosine of alpha by using an appropriate Pythagoras is an acute angle. Now, they tell us that sign of alpha is equal to four divided by 17. Now the answers provided are a 11 divided by B open parentheses. Three multiplied by the squared of 19 closed parentheses divided by C, the squared of 191 divided by 17 or d the square root of 273 divided by 17. Now, as we start our question, we should see what Python we will be using. Now, let's recall what a Python identity is. So we have Python identity and this will state that sine squared X plus cosine squared X is equal to one. And why is this useful? Well, we can use this to have a cosine X equal something and we'll just get to that. So first we'll subtract the sine squared X from the, from both sides. So we'll have sine or we'll have cosine squared X is equal to one minus sine squared X. And now to have the cosine X, we take the squared of both sides and we'll have that cosine of X is equal to the squared of one minus sine squared X. And we can substitute X or change the X with alphas to apply it to our questions. So we'll have co of alpha is equal to one minus or the square root of one minus sine squared alpha. And now we can start plugging in. So we have cosine of alpha is equal to the screwed of one minus open parentheses or divided by 17, which is a sign of alpha that was given close parentheses squared. And now we can continue to solve and we'll have cosine of alpha is equal to the square root of one minus. And now we'll distribute the square to our fraction and we distribute it to both the numerator and the denominator. So we have one minus in the numerator. Now we'll have four squared which is 16, divided by 17 squared, which is 289. So then we'll have coin of alpha is equal to the square root of. Now, to turn our one into a fraction, we'll have, we see that our other fraction has a 289 in the denominator. So to turn our one into a fraction, we'll have 289 divided by 289. And we'll have that minus 16 divided by 289. And now we can subtract the fractions and we'll have cosine alpha is equal to the square root of 273 divided by 289. And I distribute the screw to both the numerator and the denominator. And we'll be left with cosine of alpha is equal to the square root of 273 in the numerator divided by 17 in the denominator. And that will be our final answer for this question. So we have that cosine of alpha is equal to numerator two uh square root of 273 divided by the denominator being or in this case, answer choice D so I hope that was helpful. And until next time.

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In Exercises 21–24, θ is an acute angle and sin θ is given. Use the Pythagorean identity sin²θ + cos²θ = 1 to find cos θ.sin θ = 6/7

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