If tan A = a / b, a (is greater than sign) 0, b (is greater than sign) 0, and 0 is (less than sign) A (is less than sign) pi / 2, then what is cos A?
a. a / b
B. b / a
C. a / radical(a^2 + b^2)
D. b / radical(a^2 + b^2)
E. radical(a^2 + b^2) / b
why all those words? Try using a little actual notation:
If tan A = a / b, a>0, b>0, and 0<A<pi/2, then what is cos A?
a. a / b
B. b / a
C. a / √(a^2 + b^2)
D. b / √(a^2 + b^2)
E. √(a^2 + b^2) / b
given tanA = a/b, you know that
opposite side is a
adjacent side is b
hypotenuse is √(a^2+b^2)
so, what is the definition of cosA?
To find cos A, we can use the identity: cos^2A + sin^2A = 1.
Given that tan A = a / b, we know that tan A = sin A / cos A. So we can substitute sin A / cos A for tan A in the identity mentioned above:
(sin A / cos A)^2 + sin^2 A = 1
Simplifying this equation, we get:
sin^2 A + cos^2 A = cos^2 A * (sin^2 A / cos^2 A) + cos^2 A = cos^2 A * (1 + tan^2 A)
Since 0 < A < pi/2, cos A is positive. Therefore, we can take the positive square root of cos^2 A in the equation:
cos A = sqrt(cos^2 A * (1 + tan^2 A))
Substituting tan A = a / b, we have:
cos A = sqrt(cos^2 A * (1 + (a / b)^2))
cos A = sqrt((1 / b^2) * (b^2 + a^2))
cos A = sqrt(a^2 + b^2) / b
Therefore, the correct answer is E. radical(a^2 + b^2) / b.
To find the value of cos A given the value of tan A, we can use the Pythagorean identity for trigonometric functions. The Pythagorean identity for tan A is:
tan^2 A = sin^2 A / cos^2 A
Since we are given that tan A = a / b, we can substitute it into the identity:
(a / b)^2 = sin^2 A / cos^2 A
We know that sin^2 A + cos^2 A = 1 from the Pythagorean identity for sine and cosine. Rearranging this equation, we get:
cos^2 A = 1 - sin^2 A
Substituting this into the previous equation:
(a / b)^2 = sin^2 A / (1 - sin^2 A)
We can simplify this equation by multiplying both sides by (1 - sin^2 A):
(a / b)^2 × (1 - sin^2 A) = sin^2 A
Expanding and rearranging the equation:
(1 - sin^2 A) = sin^2 A / (a / b)^2
Simplifying further:
1 - sin^2 A = sin^2 A × (b^2 / a^2)
Now, let's solve for sin^2 A:
sin^2 A = (b^2 / a^2) × (1 - sin^2 A)
Multiplying both sides by a^2:
a^2 × sin^2 A = b^2 × (1 - sin^2 A)
Now, let's solve for sin^2 A by moving all terms involving sin^2 A to one side:
a^2 × sin^2 A + b^2 × sin^2 A = b^2
Factoring out sin^2 A:
sin^2 A (a^2 + b^2) = b^2
Finally, solving for sin A:
sin A = ± √(b^2 / (a^2 + b^2))
Since 0 < A < π/2, sin A is positive. Therefore:
sin A = √(b^2 / (a^2 + b^2))
Now, recall that cos A = ± √(1 - sin^2 A). Substituting the value of sin A that we found:
cos A = ± √(1 - √(b^2 / (a^2 + b^2))^2)
Simplifying further:
cos A = ± √(1 - (b^2 / (a^2 + b^2)))
Since 0 < A < π/2, cos A is positive. Therefore:
cos A = √(1 - (b^2 / (a^2 + b^2)))
Simplifying the expression inside the square root:
cos A = √((a^2 + b^2 - b^2) / (a^2 + b^2))
cos A = √(a^2 / (a^2 + b^2))
Therefore, the answer is option C.