# Math Help - trigonometric equation?

1. ## trigonometric equation?

hello i would like to find the smallest angle in degrees from the equation

3sinθ+2cosθ=3

i think that i need the identity sin^2θ=cos^2θ=1 to solve this and i was then unsure where to head to next either squaring the equation to match the identity or because there is no powers on either side working it out as sin^2θ=1-cos^2θ which i put like
3(1-cosθ)+2cosθ=3.

could someone please tell me if i am going right in any way and what the next steps are that i have to take....without to many calculations so i can have another go to understand thanks olie

2. ## Re: trigonometric equation?

Originally Posted by ibiza007
hello i would like to find the smallest angle in degrees from the equation

3sinθ+2cosθ=3

i think that i need the identity sin^2θ=cos^2θ=1 to solve this and i was then unsure where to head to next either squaring the equation to match the identity or because there is no powers on either side working it out as sin^2θ=1-cos^2θ which i put like
3(1-cosθ)+2cosθ=3.

could someone please tell me if i am going right in any way and what the next steps are that i have to take....without to many calculations so i can have another go to understand thanks olie
first off , $\sin{t} \ne 1 - \cos{t}$ ... so get that idea out of your head.

note that ...

$a\sin{t} + b\sin{t} = R\sin(x + \phi)$

$R = \sqrt{a^2+b^2}$

$\tan{\phi} = \frac{b}{a}$

$3\sin{t} + 2\sin{t} = 3 = \sqrt{13}\sin(x + \phi)$

$\sin(x + \phi) = \frac{3}{\sqrt{13}}$

$x + \phi = \arcsin\left(\frac{3}{\sqrt{13}}\right)$

$x = \arcsin\left(\frac{3}{\sqrt{13}}\right) - \arctan\left(\frac{2}{3}\right) \approx 22.6^\circ$

3. ## Re: trigonometric equation?

You can also try another approach.
$3\sin{\theta}+2\cos{\theta}=3$

$\Rightarrow (3\sin{\theta}+2\cos{\theta})^2=9$...........[Square both the sides]

$\Rightarrow 9\sin^2{\theta}+4\cos^2{\theta}+12\sin{\theta}\cos {\theta}=9$

$\Rightarrow 5\sin^2{\theta}+4(\sin^2{\theta}+\cos^2{\theta})+1 2\sin{\theta}\cos{\theta}=9$

$\Rightarrow 5\sin^2{\theta}+4+12\sin{\theta}\cos{\theta}=9$

$\Rightarrow 5\sin^2{\theta}+12\sin{\theta}\cos{\theta}=5$

$\Rightarrow 12\sin{\theta}\cos{\theta}=5-5\sin^2{\theta}$

$\Rightarrow 12\sin{\theta}\cos{\theta}=5(1-\sin^2{\theta})$

$\Rightarrow 12\sin{\theta}\cos{\theta}=5(\cos^2{\theta})$

$\Rightarrow 12\sin{\theta}=5(\cos{\theta})$

$\Rightarrow \frac{\sin{\theta}}{\cos{\theta}}=\frac{5}{12}$

$\Rightarrow \tan{\theta}=\frac{5}{12}$

$\Rightarrow \theta=\tan^{-1}(\frac{5}{12})$

$\Rightarrow \theta \approx 22.62^{\circ}$

4. ## Re: trigonometric equation?

thankyou for your help. you use t in your formula what does it represent? also i have read up on this method but i keep getting stuck im trying to get two angles within the range 0-360 degrees i looked at a formula close to this which is to write the formula as Rcos(x-a)(or Rsin(x-a) i get the almost there when
x-a = arcsin(3/square root 13) then the results don't match yours, i also don't understand why you subtracted arctan(2/3) at the end.
in the example i have read they get an angle closer to 0 and an angle closer to 360 please can you help me understand a bit further please i thought i was nearly there a few times but don't quite make it each time.

5. ## Re: trigonometric equation?

Originally Posted by ibiza007
thankyou for your help. you use t in your formula what does it represent? also i have read up on this method but i keep getting stuck im trying to get two angles within the range 0-360 degrees i looked at a formula close to this which is to write the formula as Rcos(x-a)(or Rsin(x-a) i get the almost there when
x-a = arcsin(3/square root 13) then the results don't match yours, i also don't understand why you subtracted arctan(2/3) at the end.
in the example i have read they get an angle closer to 0 and an angle closer to 360 please can you help me understand a bit further please i thought i was nearly there a few times but don't quite make it each time.
t is just short for $\theta$ ... cause I'm too lazy to type it out all the time.

go back and look at my solution again ... if $\tan{\phi} = \frac{b}{a} = \frac{2}{3}$ , then $\phi = \arctan\left(\frac{2}{3}\right)$