given that 1/2pi<theta<pi and sin theta=1/5, use appropriate trigonometric formulas to find the exact values of the following (i) cos(2theta) (ii) cos theta (iii) sin(2theta)
since we're in QII,
sinθ = 1/5
cosθ = -√24/5
now, just plug into the formulas. what do you get?
analyze given equation for following characteristics. (amplitude period, phrase shift)
y= 3sin (2x-(3.14/2))
To find the exact values of (i) cos(2θ), (ii) cosθ, and (iii) sin(2θ) using the trigonometric formula, we should first determine the value of θ.
Given that sinθ = 1/5, we can use the Pythagorean identity sin^2θ + cos^2θ = 1 to find the value of cosθ. Square both sides of the equation to get:
(1/5)^2 + cos^2θ = 1
1/25 + cos^2θ = 1
cos^2θ = 1 - 1/25
cos^2θ = 24/25
Taking the square root of both sides gives us:
cosθ = ±√(24/25)
Since θ lies in the range 1/2π < θ < π, we know that cosθ > 0. Therefore, cosθ = √(24/25).
Now, let's find the values of (i) cos(2θ), (ii) cosθ, and (iii) sin(2θ):
(i) cos(2θ):
Using the double-angle formula for cosine, cos(2θ) = cos^2θ - sin^2θ.
Substituting the values, we get:
cos(2θ) = (√(24/25))^2 - (1/5)^2
= 24/25 - 1/25
= 23/25
So, cos(2θ) = 23/25.
(ii) cosθ:
We have already determined that cosθ = √(24/25).
(iii) sin(2θ):
Using the double-angle formula for sine, sin(2θ) = 2sinθcosθ.
Substituting the values, we get:
sin(2θ) = 2 * (1/5) * (√(24/25))
= 2√(24/25) / 5
= (2/5) * √(24/25)
Therefore, sin(2θ) = (2/5) * √(24/25).
To summarize:
(i) cos(2θ) = 23/25
(ii) cosθ = √(24/25)
(iii) sin(2θ) = (2/5) * √(24/25)