Find the Smallest Angle for sin theta+5cos theta=4 | Trigonometric Equation Help

[wadcock]

Can anyone help with this trig question,

Determine the smallest angle in degrees such that sin theta+5cos theta=4

Iknow i need to use the quadratic formula but really stuck on it.

 

[MarkFL]

I would likely use a linear combination identity to rewrite the equation as:

\(\displaystyle \sqrt{26}\sin\left(\theta+\arctan(5)\right)=4\)

Can you proceed?

 

[Greg]

Alternatively,

\(\displaystyle \sin\!\theta+5\cos\!\theta=4\)

There is no solution when \(\displaystyle \cos\!\theta=0\) so we may divide both sides by \(\displaystyle \cos\!\theta\):

\(\displaystyle \tan\!\theta+5=4\sec\!\theta\)

Square both sides:

\(\displaystyle \tan^2\!\theta+10\tan\!\theta+25=16\sec^2\!\theta\)

Using the identity \(\displaystyle \sec^2\!\theta=\tan^2\!\theta+1\) and rearranging we have

\(\displaystyle 15\tan^2\!\theta-10\tan\!\theta-9=0\)

 

[Prove It]

Alternatively,

\(\displaystyle \sin\!\theta+5\cos\!\theta=4\)

There is no solution when \(\displaystyle \cos\!\theta=0\) so we may divide both sides by \(\displaystyle \cos\!\theta\):

\(\displaystyle \tan\!\theta+5=4\sec\!\theta\)

Square both sides:

\(\displaystyle \tan^2\!\theta+10\tan\!\theta+25=16\sec^2\!\theta\)

Using the identity \(\displaystyle \sec^2\!\theta=\tan^2\!\theta+1\) and rearranging we have

\(\displaystyle 15\tan^2\!\theta-10\tan\!\theta-9=0\)

You can use this approach without dividing...

$\displaystyle \begin{align*} \sin{ \left( \theta \right) } + 5\cos{ \left( \theta \right) } &= 4 \\ 5\cos{ \left( \theta \right) } &= 4 - \sin{ \left( \theta \right) } \\ \left[ 5\cos{ \left( \theta \right) } \right] ^2 &= \left[ 4 - \sin{ \left( \theta \right) } \right] ^2 \\ 25\cos^2{ \left( \theta \right) } &= 16 - 8\sin{ \left( \theta \right) } + \sin^2{ \left( \theta \right) } \\ 25 \left[ 1 - \sin^2{ \left( \theta \right) } \right] &= 16 - 8\sin{ \left( \theta \right) } + \sin^2{ \left( \theta \right) } \\ 25 - 25\sin^2{ \left( \theta \right)} &= 16 - 8\sin{ \left( \theta \right) } + \sin^2{ \left( \theta \right) } \\ 0 &= 26\sin^2{ \left( \theta \right) } - 8\sin{ \left( \theta \right) } - 9 \end{align*}$

and you can now solve this quadratic equation.