# Trigonometry Question help (sine and cosine stuff)

#### davidngg8

Hello, I have 2 questions that I have been stuck on and trouble getting started and need help.

Questions
1. In an electrical circuit, voltages are in the form of a sine or cosine wave. Two alternating current waves V1 = 10 sin (50(pi)t) and V2 = 10 cos(50(pi)t) interfere with each other to produce a third wave in the form of V(t) = V1(t) + V2(t) . Find the exact values of the positive number A and for theta in [0, 2pi] such that V(t)=A sin(50(pi)t + theta) . Formula for this is A sin x + B cos x = C sin (x+ theta)
2. The horizontal range, R(theta), of a projectile fired with an initial velocity of 60 meters per second at an angle theta is given by R(theta) = 60^2 sin(theta)cos(theta) / 4.9 , where R(theta) is in Quadrant 1. At what angle must the projectile be fired so that the horizontal range is 280 meters? - Use double-angle identity.

These are the questions, I just need help getting started or If possible you can provide an example of how to solve it.

thank you

#### Cervesa

for number (1)

$A\sin{x}+B\cos{x} = C\sin(x+\theta) = C[\sin{x}\cos{\theta}+\cos{x}\sin{\theta}] \implies A = C\cos{\theta} \text{ and } B = C\sin{\theta}$

therefore:

$A^2 = C^2\cos^2{\theta}$
$B^2 = C^2\sin^2{\theta}$
$A^2+B^2 = C^2(\cos^2{\theta}+\sin^2{\theta}) = C^2 (1) = C^2$

$C = \sqrt{A^2+B^2}$

also

$\dfrac{B^2}{A^2} = \tan^2{\theta} \implies \theta = \tan^{-1}(B/A)$

for number (2)

$280 = \dfrac{(60^2/2) \cdot 2\sin{\theta}\cos{\theta}}{4.9} = \dfrac{60^2 \cdot \sin(2\theta)}{9.8} \implies \sin(2\theta) = \dfrac{9.8 \cdot 280}{60^2}$

$2\theta = \sin^{-1}\left(\dfrac{9.8 \cdot 280}{60^2}\right)$

take it from here?

1 person

#### davidngg8

I figured out number 2 , but number 1 is still really confusing

#### Cervesa

Two alternating current waves V1 = 10 sin (50(pi)t) and V2 = 10 cos(50(pi)t) interfere with each other to produce a third wave in the form of V(t) = V1(t) + V2(t) . Find the exact values of the positive number A and for theta in [0, 2pi] such that V(t)=A sin(50(pi)t + theta)
let C be the coefficient (instead of A) to conform with your given formula

Formula for this is A sin x + B cos x = C sin (x+ theta)
$10\sin(50\pi t) + 10\cos(50\pi t) = {\color{red}{C}}\sin(50\pi t + \theta)$

$C = \sqrt{10^2+10^2}$

$\theta = \tan^{-1}(10/10)$

also, $\theta$ should be in radians