Answer:
The answer is below
Step-by-step explanation:
Show that f(x) f(y) = f(x+y)
From trigonometric:
sin(x + y) = sinxcosy + cosxsiny
sin(x - y) = sinxcosy - cosxsiny
cos(x + y) = cosxcosy - sinxsiny
cos(x - y) = cosxcosy + sinxsiny
![f(x)=\left[\begin{array}{ccc}cosx&-sinx&0\\sinx&cosx&0\\0&0&1\end{array}\right] ,f(y)=\left[\begin{array}{ccc}cosy&-siny&0\\siny&cosy&0\\0&0&1\end{array}\right] \\\\\\f(x)f(y)=\left[\begin{array}{ccc}cosxcosy-sinxsiny&-cosxsiny-sinxcosy&0\\sinxcosy+cosxsiny&-sinxsiny+cosxcosy&0\\0&0&1\end{array}\right] \\\\\\f(x)f(y)=\left[\begin{array}{ccc}cos(x+y)&-sin(x+y)&0\\sin(x+y)&cos(x+y)&0\\0&0&1\end{array}\right] \\\\\\](https://tex.z-dn.net/?f=f%28x%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcosx%26-sinx%260%5C%5Csinx%26cosx%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%2Cf%28y%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcosy%26-siny%260%5C%5Csiny%26cosy%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cf%28x%29f%28y%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcosxcosy-sinxsiny%26-cosxsiny-sinxcosy%260%5C%5Csinxcosy%2Bcosxsiny%26-sinxsiny%2Bcosxcosy%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cf%28x%29f%28y%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%28x%2By%29%26-sin%28x%2By%29%260%5C%5Csin%28x%2By%29%26cos%28x%2By%29%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C)
![f(x+y)=\left[\begin{array}{ccc}cos(x+y)&-sin(x+y)&0\\sin(x+y)&cos(x+y)&0\\0&0&1\end{array}\right] \\\\\\Therefore\ f(x)f(y)=f(x+y)](https://tex.z-dn.net/?f=f%28x%2By%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%28x%2By%29%26-sin%28x%2By%29%260%5C%5Csin%28x%2By%29%26cos%28x%2By%29%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5CTherefore%5C%20f%28x%29f%28y%29%3Df%28x%2By%29)
Answer: Sure!
Step-by-step explanation:
:)
Answer:
I think it's graph 2 hope that helps
Explain whether the points (−13,4), (−7,3), (−1,2), (5,1), (11,0), (17,−1) represent the set of all the solutions for the equati
ivann1987 [24]
Answer:
No, because the set of all solutions of y=−16x+116 is represented by the line of the equation.
Step-by-step explanation:
Using pythagorean identities,tan(sin^-1(-5/13))= tan(arcsin(-5/13))
Let A = arcsin(-5/13) = -arcsin(5/13)
Thus, sin A = -5/13 and cos A = sqrt(1 - (5/13)^2) = 12/13.= tan(A)= sin(A) / cos(A)= -5/13 / (12/13)= -5/12
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