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Exercise 1.2.23

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$\cos β = w_{1} ∕ ∥ w ∥$ and $\sin β = w_{2} ∕ ∥ w ∥ .$ Then $\cos (β - a) = \cos β \cos α + \sin β \sin α =$ $v_{1} w_{1} ∕ ∥ v ∥ ∥ w ∥ + v_{2} w_{2} ∕ ∥ v ∥ ∥ w ∥ = v \cdot w ∕ ∥ v ∥ ∥ w ∥ .$ This is $\cos 𝜃$ because $β - α = 𝜃$

mathchakra

2021-11-20 14:08

Exercise 1.2.23

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