how to write cos 2x in terms of cos x?
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calendarEducator since 2013
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The most straightforward way to obtain the expression for cos(2x) is by using the "cosine of the sum" formula: cos(x + y) = cosx*cosy - sinx*siny.
To get cos(2x), write 2x = x + x. Then,
cos(2x) = cos(x + x) = cosx*cosx - sinx*sinx = cos² (x) - sin² (x)
Then, from Pythagorean Identity, we can get express sine in terms of cosine:
sin² (x) + cos² (x) = 1
sin² (x) = 1 - cos² (x)
Plugging this into the formula for cos(2x), we get
cos(2x) = cos² (x) - (1 - cos²(x)) = 2cos²(x) - 1.
Alternatively, we could get the expression for cos(2x) in terms of sin(x):
cos(2x) = (1 - sin²(x)) - sin²(x) = 1 - 2sin²(x)
So, there are three formulas for the cosine of the double angle:
cos(2x) = cos²(x) - sin²(x)
= 1 - 2sin²(x)
= 2cos²(x) - 1.
The last one writes the cosine of 2x in terms of cosine of x.
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calendarEducator since 2010
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We know that the expression for cos ( x + y) is:
cos (x + y) = cos x * cos y - sin x* sin y
Now substituting x for both x and y we get
cos ( x + x) = cos x * cos x - sin x * sin x
=> cos 2x = (cos x)² - (sin x)²
Now we use the relation (cos x)² + (sin x)² = 1 which gives (sin x)² = 1 - (cos x)²
So we can eliminate (sin x)² and get
=> cos 2x = (cos x)² - 1 + (cos x)²
=> cos 2x = 2*( cos x)² - 1
Therefore in terms of cos x , cos 2x = 2*( cos x)² - 1
cos2x = cos (x+x)
cosxcosx - sinxsinx
(cosx)^2-(sinx)^2
Since (sinx)^2 = 1-(cosx)^2
Therefore: = (cosx)^2 -(1-(cosx)^2)
= (cosx)^2 -1+(cosx)^2
= 2(cosx)^2 -1
We'll write cos 2x as the cosine of the sum of 2 like angles:
cos(x+x) = cos x*cos x - sin x*sin x
cos(x+x) = (cos x)^2 - (sin x)^2 (1)
We'll write sin x in terms of cos x, applyingthe fundamental formula of trigonometry:
(sin x)^2 + (cosx)^2 = 1
(sin x)^2 = 1 - (cos x)^2 (2)
We'll substitute (2) in (1):
cos(x+x) = (cos x)^2 - [1 - (cos x)^2]]
We'll remove the brackets:
cos 2x = (cos x)^2 - 1+ (cos x)^2]
We'll combine like terms:
cos 2x = 2(cos x)^2 - 1
So,the expression of cos 2x, written in terms of cos x, is:
cos 2x = 2(cos x)^2 - 1
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