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Expressing sin(x) and cos(x) in terms of tan(x)

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In this video, I show how with a right angled triangle with hypotenuse 1, sides (a) and (b), and using Pythagoras' Theorem, that:
cos(x) = 1 / sqrt( 1 + tan^2(x)) and
sin(x) = tan(x) / sqrt( 1 + tan^2(x))
From the Pythagorean Identity: cos^2(x) + sin^2(x) = 1. If we divide this by cos^2x, we get...
1 + tan^2(x) = 1/cos^2(x)
And if we flip this over, we get...
cos^2(x) = 1/[1 + tan^2(x)]
And taking the square root of both sides...
cos(x) = 1/√[1 + tan^2(x)]
For sin(x), we note that sin(x)/cos(x) = tan(x), which we can rearrange to sin(x) = tan(x)cos(x). And thus...
sin(x) = tan(x)/√[1 + tan^2(x)]
These are useful trigonometric identities that can help us when performing integrals of trig functions.
Thanks for watching. Please give me a "thumbs up" if you have found this video helpful.
Please ask me a maths question by commenting below and I will try to help you in future videos.
cos(x) = 1 / sqrt( 1 + tan^2(x)) and
sin(x) = tan(x) / sqrt( 1 + tan^2(x))
From the Pythagorean Identity: cos^2(x) + sin^2(x) = 1. If we divide this by cos^2x, we get...
1 + tan^2(x) = 1/cos^2(x)
And if we flip this over, we get...
cos^2(x) = 1/[1 + tan^2(x)]
And taking the square root of both sides...
cos(x) = 1/√[1 + tan^2(x)]
For sin(x), we note that sin(x)/cos(x) = tan(x), which we can rearrange to sin(x) = tan(x)cos(x). And thus...
sin(x) = tan(x)/√[1 + tan^2(x)]
These are useful trigonometric identities that can help us when performing integrals of trig functions.
Thanks for watching. Please give me a "thumbs up" if you have found this video helpful.
Please ask me a maths question by commenting below and I will try to help you in future videos.
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