Trigonometric Identities

[ParveshD.Singh]

Hi there I'm currently stuck on one of the questions from my math assignment... I'm hoping some kind and smart souls on the Internet can help me with it. Here's the question.
Rewrite 3 cos(x) - 4 sin(x) as r cos(x + alpha ), where r > 0 and alpha (0, pi/2).
(hint, cos(a + b) = cos(a) cos(b) - sin(a) sin(b))

 

[Cubist]

Try to manipulate the "hint" identity so that the RHS looks like "3 cos(x) - 4 sin(x)". Start by multiplying both sides of the hint by "r". Then substitute the "b" variable for an "x" (let b=x)...

r*cos(a + x) = r*cos(a) cos(x) - r*sin(a) sin(x)

Can you find an "a" and "r" to make the RHS look like "3 cos(x) - 4 sin(x)"?

 

[Deleted member 4993]

Hi there I'm currently stuck on one of the questions from my math assignment... I'm hoping some kind and smart souls on the Internet can help me with it. Here's the question.
Rewrite 3 cos(x) - 4 sin(x) as r cos(x + alpha ), where r > 0 and alpha (0, pi/2).
(hint, cos(a + b) = cos(a) cos(b) - sin(a) sin(b))

You need to transform:

A * cos(Θ) + B * sin(Θ) = R * cos(Θ + Φ), where,

R = √[A² + B²]

cos(Φ) = A/√(A² + B²)

sin(Φ) = -B/√(A² + B²) ........ continue

 

[HallsofIvy]

Looking at the formula you are given for cos(x+ y), you want to associate the "cos(a)" with the "3" multiplying cos(x) and "sin(a)" with the "4" multiplying sin(x). But you realize that cosine and sine are never larger than 1 or less than -1, right?So you need reduce the coefficients by dividing by something. We also need to recall that sin^2(a)+ cos^2(a). That is where the "3^2+ 4^2 (= 16)= 5²" comes from 3cos(x)- 4sin(x)= \(\displaystyle 5(\frac{3}{5}cos(x) - \frac{4}{5}sin(x)).\)

Now you should see that you need to find alpha so that cos(alpha)= 3/5 and sin(alpha)= 4/5