find the exact value of tan(2sin-1 3/5)
let sin^-1 (3/5) = Ø
then sinØ = 3/5 and cosØ = 4/5
so you want tan 2Ø
= sin2Ø/cos2Ø
= 2sinØcosØ/(2cos^2Ø - 1)
= 2(3/5)(4/5)/(32/25 - 1)
=(24/25) / (7/25)
= (24/25)(25/7)
= 24/7
Well, if you let me put on my mathematician clown nose, I can give it a shot!
Let's start by finding the value of sine inverse of 3/5. So, sin^(-1)(3/5) is the angle whose sine is 3/5. When you brave the math circus, you'll find that this angle is approximately 36.87 degrees.
Now, hold on tight as we calculate tan(2*sin^(-1)(3/5)).
Tan (2*sin^(-1)(3/5)) = tan(2*36.87)
But wait, tan(2θ) has a nifty formula: tan(2θ) = (2*tan(θ))/(1-tan^2(θ)).
So, we can use this formula to solve it:
tan(2*sin^(-1)(3/5)) = (2*tan(36.87))/(1-tan^2(36.87)).
Just keep in mind that this equation is more of a mathematical joke than an exact value. Unfortunately, it's not as simple as pi(e).
To find the exact value of tan(2sin^(-1)(3/5)), we can use trigonometric identities.
First, let's find the value of sin^(-1)(3/5):
sin^(-1)(3/5) is the angle whose sine is 3/5.
Let's assume this angle is denoted as θ.
Therefore, sin(θ) = 3/5.
Using Pythagorean identity, we can find the value of cos(θ):
cos^2(θ) = 1 - sin^2(θ)
cos^2(θ) = 1 - (3/5)^2
cos^2(θ) = 1 - 9/25
cos^2(θ) = 16/25
Now, we can determine the sign of cos(θ):
Since sin(θ) is positive (3/5), and sin is positive in quadrant I and II, we conclude that cos(θ) is positive.
Taking the square root of both sides, we get:
cos(θ) = sqrt(16/25)
cos(θ) = 4/5
Now, we can find the value of tan(2sin^(-1)(3/5)):
First, we determine the value of 2sin^(-1)(3/5):
2θ = 2sin^(-1)(3/5)
θ = sin^(-1)(3/5)
Using the double angle formula for tangent, we have:
tan(2θ) = (2tan(θ))/(1 - tan^2(θ))
Substituting θ=sin^(-1)(3/5), we get:
tan(2sin^(-1)(3/5)) = (2tan(sin^(-1)(3/5)))/(1 - tan^2(sin^(-1)(3/5)))
Now, we need to find the value of tan(sin^(-1)(3/5)):
Let's assume sin^(-1)(3/5) = α, so sin(α) = 3/5.
We can use right triangles to find the values of tan(α).
Using Pythagorean theorem, we can determine the value of the third side:
c^2 = a^2 + b^2
1^2 = (3/5)^2 + b^2
1 = 9/25 + b^2
b^2 = 1 - 9/25
b^2 = 25/25 - 9/25
b^2 = 16/25
b = 4/5
Therefore, tan(α) = (opposite side / adjacent side) = (4/5) / (3/5) = 4/3.
Now, we can substitute tan(sin^(-1)(3/5)) = 4/3 into the equation:
tan(2sin^(-1)(3/5)) = (2tan(α))/(1 - tan^2(α))
tan(2sin^(-1)(3/5)) = (2(4/3))/(1 - (4/3)^2)
tan(2sin^(-1)(3/5)) = (8/3)/(1 - 16/9)
tan(2sin^(-1)(3/5)) = (8/3)/(9/9 - 16/9)
tan(2sin^(-1)(3/5)) = (8/3)/(-7/9)
tan(2sin^(-1)(3/5)) = (-8/3)*(9/7)
tan(2sin^(-1)(3/5)) = -24/7
Therefore, the exact value of tan(2sin^(-1)(3/5)) is -24/7.
To find the exact value of tan(2sin^(-1)(3/5)), we can follow these steps:
Step 1: Recall the identity for sin(2θ):
sin(2θ) = 2sin(θ)cos(θ)
Step 2: Substitute θ with sin^(-1)(3/5):
sin(2sin^(-1)(3/5)) = 2sin(sin^(-1)(3/5))cos(sin^(-1)(3/5))
Step 3: Simplify sin(sin^(-1)(3/5)):
sin(sin^(-1)(x)) = x
sin(sin^(-1)(3/5)) = 3/5
Step 4: Simplify cos(sin^(-1)(3/5)):
cos²(θ) + sin²(θ) = 1
cos²(sin^(-1)(3/5)) + sin²(sin^(-1)(3/5)) = 1
cos²(sin^(-1)(3/5)) + (3/5)² = 1
cos²(sin^(-1)(3/5)) + 9/25 = 1
cos²(sin^(-1)(3/5)) = 16/25
cos(sin^(-1)(3/5)) = ±4/5
Since θ is an acute angle and sin(θ) = 3/5 (positive), the value of cos(sin^(-1)(3/5)) is positive, so we take the positive value: cos(sin^(-1)(3/5)) = 4/5.
Step 5: Substitute the values back into the original equation:
tan(2sin^(-1)(3/5)) = 2 * (3/5) * (4/5)
Step 6: Simplify the expression:
tan(2sin^(-1)(3/5)) = 24/25
Therefore, the exact value of tan(2sin^(-1)(3/5)) is 24/25.