Find minimum value of $8 \cos x + 4 \sin x $ and corresponding value of $x$

I used the R method to simplify it -

$ \sqrt{80} \cos (x-26.565) $

Minimum value of that =

$ \sqrt{80} \cos (x-26.565) = - \sqrt{80}$

$ \cos (x-26.565) = -1$

This cosine value lies in the 2nd and 3rd quadrant

letting $x-26.565 = y$

y reference angle = $ \cos^-1 (-1) = 180$

2nd quadrant - $180 - y (ref) = 0 $

3rd quadrant - $180 + y(ref) = 360$

Therefore , $x = 26.565, 386.565$

Why am I wrong ? The minimum value is $206.6$


Your method is perfectly fine, you just made a mistake at the end.

$\cos(u)=-1\iff u\equiv 180°\pmod{360°}$

Here $u=x-x_0$ so you should get $x\equiv 180°+x_0\pmod{360°}$

Applying to $x_0=26.565°$ you get $x = 206.565°$


Since $$ 8^2+4^2=80 $$ you know that $8\cos x+4\sin x=\sqrt{80}\cos(x-\alpha)$, for some angle that can be determined by setting $x=0$ and $x=\pi/2$: \begin{align} 8&=\sqrt{80}\cos\alpha\\ 4&=\sqrt{80}\sin\alpha \end{align} Thus the angle $\alpha$ is in the first quadrant and so $$ \alpha=\arcsin\frac{4}{\sqrt{80}}=\arcsin\frac{1}{\sqrt{5}} $$ In degrees this is $26.565$. In radians it is $0.464$ (rounding to three decimal digits).

The point where the minimum value $-\sqrt{80}$ is reached is when $x-\alpha$ is the straight angle. In degrees the value of $x$ is $180+26.565=206.565$.

In radians it is $\pi+0.464=3.605$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.