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

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The discussion focuses on solving the trigonometric equation sin(theta) + 5cos(theta) = 4 to find the smallest angle. Participants suggest using the quadratic formula and identities to transform the equation into a solvable form. A key approach involves rewriting the equation as a quadratic in terms of tan(theta) after manipulating the original equation. The final form leads to a quadratic equation that can be solved for sin(theta). The conversation emphasizes the importance of correctly applying trigonometric identities and algebraic manipulation to find the solution.
wadcock
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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.
 
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I would likely use a linear combination identity to rewrite the equation as:

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

Can you proceed?
 
Alternatively,

$$\sin\!\theta+5\cos\!\theta=4$$

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

$$\tan\!\theta+5=4\sec\!\theta$$

Square both sides:

$$\tan^2\!\theta+10\tan\!\theta+25=16\sec^2\!\theta$$

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

$$15\tan^2\!\theta-10\tan\!\theta-9=0$$
 
greg1313 said:
Alternatively,

$$\sin\!\theta+5\cos\!\theta=4$$

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

$$\tan\!\theta+5=4\sec\!\theta$$

Square both sides:

$$\tan^2\!\theta+10\tan\!\theta+25=16\sec^2\!\theta$$

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

$$15\tan^2\!\theta-10\tan\!\theta-9=0$$

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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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