a ⋅ b = | a | | b | cos ⁡ γSimplest way to show cos ⁡ γ = cos ⁡ ( φ − φ ′ ) sin ⁡ θ sin ⁡ θ ′ + cos ⁡ θ cos ⁡ θ ′ without using tangents?

Question

a  ⋅  b  =      |    a      |        |    b      |    cos  ⁡  γSimplest way to show   cos  ⁡  γ  =  cos  ⁡  (  φ  −      φ    ′    )  sin  ⁡  θ  sin  ⁡      θ    ′    +  cos  ⁡  θ  cos  ⁡      θ    ′    without using tangents?

mar1nerne · Accepted Answer

c  o  s  (  φ  −      φ    ′    )  =  c  o  s  φ  c  o  s      φ    ′    +  s  i  n  φ  s  i  n      φ    ′  Use unit vectors a,b i.e |a|=1,|b|=1  a  =  (  x  ,  y  ,  z  )  =  (  s  i  n  θ  c  o  s  φ  ,  s  i  n  θ  s  i  n  φ  ,  c  o  s  θ  )  b  =  (      x    ′    ,      y    ′    ,      z    ′    )  =  (  s  i  n      θ    ′    c  o  s      φ    ′    ,  s  i  n      θ    ′    s  i  n      φ    ′    ,  c  o  s      θ    ′    )dot product over the xy plane :  c  o  s  (  φ  −      φ    ′    )  =            x              x        ′            +      y              y        ′                                      x          2                +                  y          2                                      x                      ′            2                          +                  y                      ′            2                                =            x              x        ′            +      y              y        ′                    sin      ⁡      θ      sin      ⁡              θ        ′            dot product a.b :   cos  ⁡  γ  =  x      x    ′    +  y      y    ′    +  z      z    ′    =  x      x    ′    +  y      y    ′    +  c  o  s  θ  c  o  s      θ    ′    cos  ⁡  γ  =  c  o  s  (  φ  −      φ    ′    )  sin  ⁡  θ  sin  ⁡      θ    ′    +  c  o  s  θ  c  o  s  θ

a ⋅ b = <mrow class="MJX-TeXAtom-ORD"> | </mrow> a <m

Answered question

Answer & Explanation

New Questions in Precalculus