Express each of the following in the form Rsin(x+a), where r>0 and 0< a <2pi

Question

Express each of the following in the form Rsin(x+a), where r>0 and 0< a <2pi

(a) 5cosx + 12sinx
(b) 12cosx + 5sinx

Answer 1

12cosx+5sinx=13sin(x+67.38)

Answer 2

I will do the first, and you try the second:

let Rsin(x+a) 5cosx + 12sinx

Rsin(x+a) = R(sinxcosa + cosxsina)
= Rsinxcosa + Rcosxsina

so

Rsinxcosa + Rcosxsina = 5cosx + 12sinx
if this is an identity it must be true for all values of x

Pick any two values of x which you can easily find

let x = 0
Rsin0cosa + Rcos0sina = 5cos0 + 12 sin0
but sin0 = 0 , and cos0 = 1
Rsina = 5
sina = 5/R

let x = 90° or π/2
Rsin90cosa + Rcosasin90 = 5cos90 + 12sin90
sin90 = 1 , and cos90 = 0
Rcosa = 12
cosa = 12/R

but sin^2 a + cos^2 a = 1
25/R^2 + 144/R^2 = 1
R^2 = 169
R = 13

also sina/cosa = (5/R)/(12/R)
tana = 5/12
a = 22.6199°

So 5cosx + 12sinx = 13sin(x + 22.6199°)
or if you want radians, set your calculator to radians when you find the tangent inverse
5sinx + 12cosx = 13sin(x + .3948) in radians

check for any angle
e.g. x = 53°
5cos53 + 12sin53 = 12.5927..
13sin(53+22.6199) = 13sin 75.6199 = 12.5927..

not bad

Answer 3

I noticed a small typo

2nd line clearly should have said

let Rsin(x+a) = 5cosx + 12sinx

Answer 4

To express each of the given expressions in the form Rsin(x+a), where r>0 and 0< a <2pi, you can use the trigonometric identities.

(a) 5cosx + 12sinx:
To rewrite this expression in the desired form, we'll use the identity sin(x + a) = sinxcos a + cosxsin a.

In this case:
5cosx + 12sinx = Rsin(x+a)

Let's express 5 and 12 as the hypotenuse and the opposite side of a right-angled triangle with angle a:

5 = Rcos a
12 = Rsin a

To find R, we can use the Pythagorean theorem:
(Rcos a)^2 + (Rsin a)^2 = R^2[(cos a)^2 + (sin a)^2] = R^2 = (5^2 + 12^2) = 169
R^2 = 169
R = ±√169
Note: Since R > 0, we take the positive square root: R = 13

Now, substitute the values of Rcos a and Rsin a into the original expression:
5cosx + 12sinx = 13(sinxcos a + cosxsin a)

Comparing this expression with the desired form, we have:
R = 13
sin(x+a) = sinxcos a + cosxsin a

Therefore, by comparing coefficients:
r = 13
a = arctan(12/5)

So, the expression 5cosx + 12sinx can be written in the form Rsin(x+a) as 13sin(x + arctan(12/5)).

(b) 12cosx + 5sinx:

Using the same method as above, we have:
12cosx + 5sinx = Rsin(x+a)

Let's express 12 and 5 as the hypotenuse and the opposite side of a right-angled triangle with angle a:

12 = Rcos a
5 = Rsin a

Again, using the Pythagorean theorem:
(Rcos a)^2 + (Rsin a)^2 = R^2[(cos a)^2 + (sin a)^2] = R^2 = (12^2 + 5^2) = 169
R^2 = 169
R = ±√169
Since R > 0, we take the positive square root: R = 13

Now, substitute the values of Rcos a and Rsin a into the original expression:
12cosx + 5sinx = 13(sinxcos a + cosxsin a)

Comparing this expression with the desired form, we have:
R = 13
sin(x+a) = sinxcos a + cosxsin a

Therefore, by comparing coefficients:
r = 13
a = arctan(5/12)

So, the expression 12cosx + 5sinx can be written in the form Rsin(x+a) as 13sin(x + arctan(5/12)).