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Calculusantiderivative
Find f. f ''(θ) = sin θ + cos θ, f(0) = 5, f '(0) = 3 My steps: f'(θ)=cosθsinθ+C When f'(0)=3, C=2, so f'(θ)=cosθsinθ2. f(θ)=sinθcosθ2x+D When f(0)=5, D=6, so f is sinθcosθ2x+6. How is that wrong?
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The identity below is significant because it relates 3 different kinds of products: a cross product and a dot product of 2 vectors on the left side, and the product of 2 real numbers on the right side. Prove the identity below. 
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How do I prove that secθ  tanθsinθ = cosθ So far I have LS 1/cosθ  sinθ/cosθ (sinθ) =1/cosθ sin^2/cosθ
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Hi. I am reposting this. Can you verify. Thanks. (cosθ) / (sinθ – 1) + (cosθ)/ (1 + sinθ) = 2tanθ (1 / cos^2θ) – (2) (1/cos^2θ) (sinθ / cosθ) + (sin^2θ / cos^2θ) = (1  sinθ) / (1 + sinθ)
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We have a qubit in the state ϕ⟩=3√20⟩+121⟩, which we want to measure in the {cosθ0⟩+sinθ1⟩,sinθ0⟩−cosθ1⟩} basis. In order for the two possible outcomes to be equiprobable, what should be the value of
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1. (P 15/17, 8/17) is found on the unit circle. Find sinΘ and cosΘ Work: P= (15/17, 8/17) cosΘ = a value P = (a,b) sinΘ = b value Answer: cosΘ = 15/17 sinΘ = 8/17 2. Should the triangle be solved beginning with Law of
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Trigonometry
Hi. Can you verify the following: (cosθ) / (sinθ – 1) + (cosθ)/ (1 + sinθ) (1 / cos^2θ) – (2) (1/cos^2θ) (sinθ / cosθ) + (sin^2θ / cos^2θ)
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Verify the identities. 1.) √1COSθ/1+COSθ= 1+SINθ/SINθ 2.) SEC X SIN(π/2X)= 1 3.) CSC X(CSC XSIN X)+SIN XCOS X/SIN X + COT X= CSC²X 4.) CSC^4 X2 CSC²X+1= COT^4 X 5.) CSC^4 θCOT^4 θ= 2 CSC²θ1 6.) TAN^5 X= TAN³X
asked by AwesomeGuy on February 17, 2013 
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You posted a question about having two cannon balls strike the same target simultaneously, with a time delay between shots: https://www.jiskha.com/display.cgi?id=1520556373 I have had some time to think about it, and have come up
asked by Steve on March 12, 2018