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Hello, I was hoping you could explain to me how to solve and show me the solution to this problem. I got this problem wrong on the homework today, but when I asked the teacher about it, I only got more confused (I have a sub, my teacher is on maternity leave).
5.5.3

I am giving a right triangle with a base of 24 and a height of 7 (the 90 degree angle is between those two). I am asked to first find the measure of the angle between the base (24) and the hypotenuse. Then, insert that value into the expression cos (theta/2)

Okay, thank you in advance.

Answer
Hi John,
The correction would be easier for you if you had shown me what you already did and how you went about it. Anyway, let's proceed without that.
If the angle between this base and the hypotenuse is θ, then we have, from trigonometry;
tanθ = 7/24
and the angle is easy to determine from there to then evaluate cos(θ/2).

The question, however, requires a direct evaluation of cos(θ/2) and so it's easier to work in cosine instead of tangent in this case.
Back to the triangle, we can solve the hypotenuse using the Pythagorean theorem.
H² = 7² + 24²
H = 25
and so
cosθ = 24/25
Now, from the double-angle trigonometric identity;
cos(2A) = 2cos²A - 1
2cos²A = 1 + cos(2A)
cos² A = [1 + cos(2A)] / 2

Letting A = θ/2, we have
cos²(θ/2) = (1 + cosθ) / 2
cos²(θ/2) = [1 + (24/25)] / 2
cos²(θ/2) = (49/25) / 2
cos²(θ/2) = 49/50

The value of cos(θ/2) is then the positive square root of 49/50 in this simple case, since we know that 0 < θ/2 < π/2.

I hope this is what you needed.

Regards

Calculus

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