This is a proof of `lim_(θ -> 0) cos(θ) = 1`

Consider a right angled triangle ABC with right angle at B. Let angle C be equal to θ rad.

The cosine of angle θ is defined as the ratio of the adjacent side (also called the base) and the hypotenuse:

As angle θ gets smaller and smaller (while the other angles remain constant), the side hypotenuse and adjacent sides get more and more equal to each in length. As angle θ becomes zero, the adjacent and hypotenuse become overlapping sides and their lengths are equal.

Thus as `θ -> 0`, `AC -> BC`, so as `θ = 0` `AC = BC`. Hence in the cosine ratio, we can write

Consider a right angled triangle ABC with right angle at B. Let angle C be equal to θ rad.

The cosine of angle θ is defined as the ratio of the adjacent side (also called the base) and the hypotenuse:

`cos(θ) = "Adjacent"/"Hypotenuse" = "BC"/"AC"`Since this proof statement has limit as angle θ approaches zero, so we will consider the above cosine ratio as angle θ approaches zero:

As angle θ gets smaller and smaller (while the other angles remain constant), the side hypotenuse and adjacent sides get more and more equal to each in length. As angle θ becomes zero, the adjacent and hypotenuse become overlapping sides and their lengths are equal.

Thus as `θ -> 0`, `AC -> BC`, so as `θ = 0` `AC = BC`. Hence in the cosine ratio, we can write

As `θ -> 0`, `cos(θ) = "AC"/"BC" = "BC"/"BC"` or `"AC"/"AC"`Thus we can write the limit of `cos(θ)`, as `θ` approaches zero, as follows:

`lim_(θ -> 0) cos(θ) = "BC"/"BC" = "AC"/"AC" = 1`

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