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Hello,
I'm having trouble with a few trig. problems. If someone could help me out it would be great!
1. Find the value of x given 2 sin^-1 x= -pi
2. suppose tan(angle) = t . Prove that sin2(angle) = [2t] / [1+t^2]
3. Given sin(angle)=0.3, find the value of sec(angle)
Thats all for now.
All help would be appreciated.
Thanks
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1. This one isn't too hard, it's mainly just rearranging.
2 sin^-1 x = -π
sin^-1 x = -π/2
x = sin -π/2
∴ x = -1.
2. I'm not sure that this is allowed, but I'd find it easier to prove that [2t]/[1+t²] = sin2θ.
Using standard identities, 1+tan²θ = sec²θ = 1/cos²θ.
That means that [2t]/[1+t²] = [2t]/[1/cos²θ] = 2t*cos²θ.
t = sinθ/cosθ, so 2t*cos²θ = 2sinθcosθ, and another standard identity is that that is equal to sin2θ.
3. sin²θ + cos²θ = 1, for all θ.
So if we know sinθ, that means that we can find cosθ by rearranging.
In this case, cosθ = √(1-0.3sup2) = √0.91 ≈ 0.95
secθ = 1/cosθ, and so it would be 1/0.95 = 1.05.
Why did the vector cross the road?
It wanted to be normal.
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