Express cos x + 3 sin x in the form Rcos(x − alpha), where R > 0 and 0< alpha < 90, giving the exact value of R and the value of alpha correct to 2 decimal places.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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`cosx+3sinx`

`=sqrt(1^2+3^2) {[1/sqrt(1^2+3^2)]cosx+[3/sqrt(1^2+3^2)]sinx}`

`= sqrt(10) {[1/sqrt(10)]cosx+[3/sqrt(10)]sinx}`

If 1,3 and `sqrt(10)` are legs of a triangle they have the relationship denoted by Pythagoras theorem.

`1^2+3^2=10`

`cos(alpha) =1/sqrt(10)`

`sin(alpha)=3/sqrt(10)`

`cosx+3sinx`

`=sqrt(10)(cosx xx cosalpha+sinx xx sinalpha)`

`=sqrt(10)(cos(x-alpha))`

`cosx+3sinx=sqrt(10)cos(x-alpha)`

This is in the form of `Rcos(x-alpha) ` where;

`R=sqrt(10)=3.16`

`alpha=cos^-1 (1/sqrt(10)) = 71.57^0`

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