Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ  = 5$$
Find 
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
 A: If you recognize the Pythagorean relation $3^2+4^2=5^2$, then the equation
$${4\over5}\sin\theta+{3\over5}\cos\theta=1$$
is satisfied if $\sin\theta={4\over5}$ and $\cos\theta={3\over5}$, which makes
$$4\cos\theta-3\sin\theta=4\cdot{3\over5}-3\cdot{4\over5}=0$$
It's not obvious that this approach produces the only possible answer, but a little extra thought shows that it does.  (For one thing, the statement of the problem suggests the answer is unique.)
A: HINT:
using Brahmagupta–Fibonacci identity
$$(4\sin\theta+3\cos\theta)^2+(4\cos\theta-3\sin\theta)^2=(4^2+3^2)(\sin^2\theta+\cos^2\theta)$$
A: The 3, 4, 5 should perhaps make one think of a a right angled triangle with sides 3, 4, 5.
Express the sines and cosines in terms of ratios of sides of a triangle h (hypotenuse), o (opposite) and a (adjacent), so that $\sin(\theta) = o/h$ and $\cos(\theta) = a/h$ then the given condition is that $4.o/h + 3.a/h = 5$. 
By observation this is satisfied by $o = 4; a=3; h=5$, so that $\sin(\theta) = 4/5$ and $\cos(\theta) = 3/5$.
Now plug this into your second expression so that $4\cos(\theta) - 3\sin(\theta) = 12/5 - 12/5 = 0$
A: Another approach:
Rewrite
$$
\begin{align}
4\sin\theta +3\cos\theta  &= 5\\
4\sin\theta&=5-3\cos\theta.\tag1
\end{align}
$$
Square $(1)$, yields
$$
16\sin^2\theta=25-30\cos\theta+9\cos^2\theta.\tag2
$$
Use identity $\sin^2\theta=1-\cos^2\theta$ and substitute to $(2)$.
$$
\begin{align}
16(1-\cos^2\theta)&=25-30\cos\theta+9\cos^2\theta\\
25\cos^2\theta-30\cos\theta+9&=0\\
(5\cos\theta-3)^2&=0\\
\cos\theta&=\frac35.
\end{align}
$$
Consequently, $\sin\theta=\dfrac45$. Thus,
$$
4\cos\theta-3\sin\theta=4\left(\frac35\right)-3\left(\frac45\right)=\Large\color{blue}0.
$$
A: If $a\sin\theta+b\cos\theta=c$
Squaring we get $$a^2\sin^2\theta+b^2\cos^2\theta+2ab\cos\theta\sin\theta=c^2$$
$$\iff a^2(1-\cos^2\theta)+b^2(1-\sin^2\theta)+2ab\cos\theta\sin\theta=c^2$$
$$\iff c^2-a^2-b^2=(a\cos\theta-b\sin\theta)^2$$
Here $\displaystyle a=4,b=3,c=5\implies c^2-a^2-b^2=?$

Alternatively, 
Let $a=r\cos\phi, b=r\sin\phi\  \ \ \ (1)$ where $r\ge0$
So, $\displaystyle a\sin\theta+b\cos\theta=c\implies r\sin(\theta+\phi)=c$
$\displaystyle a\cos\theta-b\sin\theta=r\cos(\theta+\phi)=r\cdot\pm\sqrt{1-\frac{c^2}{r^2}}=\pm\sqrt{r^2-c^2}$
Now, squaring & adding $(1)$  we get $\displaystyle r^2=a^2+b^2$
A: This might be slightly more advanced, a solution using vectors: let $${\bf v}=(3,4),$$ for each $\theta$ let $${\bf w}(\theta)=(\cos\theta,\sin\theta)$$ and let $${\bf u}(\theta)=(-\sin\theta,\cos\theta).$$
Note that ${\bf u}(\theta)$ and ${\bf w}(\theta)$ are perpendicular for all $\theta$. Your equation tells you something about the dot product of ${\bf v}$ and ${\bf w}(\theta)$, namely that $${\bf v}\cdot{\bf w}(\theta)=5.$$ Since the norms of these two vectors are $1$ and $5$, respectively, we have: $$|{\bf v}\cdot{\bf w}(\theta)| = \|{\bf v}\| \|{\bf w}(\theta)\| = 5,$$ i.e. equality holds in the Cauchy-Schwarz inequality. But this means that ${\bf v}$ and ${\bf w}(\theta)$ are parallel, so ${\bf v}$ and ${\bf u}(\theta)$ are perpendicular, which means precisely that $${\bf v}\cdot {\bf u}(\theta)=0,$$ or in other words that $$-3\sin\theta+4\cos\theta=0.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$$
\left.\begin{array}{rcrcl}
4\sin\pars{\theta} & + & 3\cos\pars{\theta} & = & 5
\\
-3\sin\pars{\theta} & + & 4\cos\pars{\theta} & \equiv & \mu:\ {\large ?}
\end{array}\right\rbrace\ \imp\
\begin{array}{|rcl}
\ \sin\pars{\theta}& = & {20 - 3\mu \over 25}
\\[3mm]
\cos\pars{\theta}& = & {4\mu + 15 \over 25}
\end{array}
$$

Then,
  \begin{align}
1&=\sin^{2}\pars{\theta} + \cos^{2}\pars{\theta}
=\pars{20 - 3\mu \over 25}^{2} + \pars{4\mu + 15 \over 25}^{2}
={\mu^{2} \over 25} + 1\ \imp\ \boxed{\quad\mu = 0\quad}
\end{align}

$$
\imp\quad\color{#66f}{\large 4\cos\pars{\theta} - 3\sin\pars{\theta} = 0}
$$
A: I found this post and I decided to refresh my math skills. Below you can see my solution.
Let's say:
$$s = \sinθ$$
$$c = \cosθ$$
We have got:
$$4s +3c  = 5 [ X ] $$ 
$$4c - 3s = x [ XX ]$$
A) Let's multiply [ X ] by 3 and [ XX ] by 4 and add both:
$$12s +9c  = 15$$
$$16c - 12s = 4x$$
so: $$25c= 15 + 4x [XXX]$$
B) Let's multiply [ X ] by 4 and [ XX ] by 3 take a difference:
$$16s + 12c  = 20 $$ 
$$12c - 9s = 3x $$
so: $$25s= 20 - 3x [XXXX]$$
Let's do ^2 on both [ XXX ] [ XXXX ] and add:
$$(25c)^2 + (25s)^2 = (15 + 4x)^2 + (20 -3x)^2 $$
but $$(25c)^2 + (25s)^2 = 25^2 $$
so we have got:
$$25^2 = (15 + 4x)^2 + (20 -3x)^2 $$
which gives us easily:
$$x = 0 $$
A: Furthermore, another approach.
We can show that $4 \sin \theta + 3 \cos \theta$ has its maximum at $y = 5$. Or, more generally, $a \sin \theta + b \cos \theta$ has a maximum at $y = \sqrt{a^2 + b^2}$. 
We can manipulate the expression $a \sin \theta + b \cos \theta$ to become $\sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2 + b^2}} \sin \theta + \frac{b}{\sqrt{a^2 + b^2}}  \cos \theta\right)$
Notice that $\frac{a}{\sqrt{a^2 + b^2}}$ and $\frac{b}{\sqrt{a^2 + b^2}}$ are equivalent to $\cos \alpha$ and $\sin \alpha$ for $\alpha = \tan^{-1} \frac ba$.So now we have $\sqrt{a^2 + b^2} \left(\cos \alpha \sin \theta + \sin \alpha \cos \theta \right) = \sqrt{a^2+b^2} \sin(\alpha + \theta)$, which is maximized when $\alpha + \theta = \pi/2 \rightarrow \theta = \pi / 2 - \alpha$
Thus, we know that $4 \sin \theta + 3 \cos \theta$ is maximized since it equals $5$. Continuing to work generally, we should substitute $\theta = \pi/2 - \alpha$ into $a \cos \theta - b \sin \theta: a \cos(\pi/2 - \alpha) - b \sin (\pi/2 - \alpha) \rightarrow a \sin \alpha - b \cos \alpha$. Recall that $\alpha = \tan^{-1} \frac{b}{a}$, so it follows that $\alpha = \sin^{-1} \frac{b}{\sqrt{a^2 + b^2}} = \cos^{-1} \frac{a}{\sqrt{a^2 + b^2}}$. So finally we have $a \sin \alpha - b \cos \alpha = \frac{ab}{\sqrt{a^2 + b^2}} - \frac{ba}{\sqrt{a^2+b^2}} = \boxed{0}$
A: I learnt yet another way in High School (in the Netherlands).
$$4 \sin\theta + 3 \cos \theta = 5$$
Can be thought of as a vector dot product:
$$\binom{\cos\theta}{\sin\theta}\cdot\binom{3}{4}=5$$
Another way to write a dot product is
$$\sqrt{\sin^2\theta+\cos^2\theta}\cdot\sqrt{4^2+3^2}\cdot \cos\phi = 5$$
where $\phi$ is the angle between the two vectors that make up the dot product.
This simplifies to $$\cos\phi=1$$
So $\phi=0$ or $\phi=\pi$.
Thus we know that the vector $\binom{\cos\theta}{\sin\theta}$ lies parallel to, or in the opposite direction of, $\binom{3}{4}$. In other words, we can write$$\tan\theta=\frac43$$
from which it follows that$$\theta = \arctan\left(\frac43\right) \mod \pi$$
Obviously this works for cases where $\phi\ne0$ - you just include whatever $\phi$ is.
A: The solutions involving algebraic manipulation are preferred in this case, but if you were to run across an equation that doesn't lend itself so cleanly to the other methods, these are alternatives.
If you have a graphing calculator, you could graph $4\sin\theta+3\cos\theta$ and $5$, and look for their intersection point(s).
You could also use Newton's method (or some other approach) by hand, which is similar to what a graphing calculator would do.
A: Very important fact of physics and harmonics and digital signal processing:

The sum of two sinusoids of equal frequency is another sinusoid of that same frequency, regardless of amplitude or phase.

Or stated mathematically:

If 
  $$f(x) = A_1\cos\left(\omega\,x + p_1\right) + A_2\cos\left(\omega\,x + p_2\right)$$
  then
  $$f(x) = A_3\cos\left(\omega\,x + p_3\right)$$
  and
  $$A_1 \angle p_1 + A_2 \angle p_2 = A_3 \angle p_3$$
  where $M \angle \theta$ is polar notation.  The same holds is $\cos$ is replaced with $\sin$.

So your problem is:
$$4\sin(\theta) + 3\cos(\theta) = 5$$
$$4\sin(\theta) + 3\sin(\theta + \pi/2) = 5$$
$$M\sin(\theta + p) = 5$$
$$\theta = \arcsin(5/M) - p$$
And you find $M$ and $p$ by adding:
$$M\angle p = 4 \angle 0 + 3 \angle \pi/2$$
$$\begin{cases} M = 5 \\ p = \arctan(3/4)\end{cases}$$
So
$$\theta = \arcsin(1) - \arctan(3/4)$$
$$\theta = \pi/2 + Z\cdot 2\pi - \arctan(3/4)$$
So now
$$y = 4\cos(\theta) - 3\sin(\theta)$$
$$y = 4\cos(\theta) + 3\cos(\theta + \pi/2)$$
$$y = 5\cos(\theta + \arctan(3/4))$$
So
$$y = 5\cos(\pi/2 + Z\cdot 2\pi - \arctan(3/4) + \arctan(3/4))$$
$$y = 5\cos(\pi/2 + Z\cdot 2\pi)$$
$$y = 0$$
A: Its simple:
Given
$$4\sin\theta + 3\cos\theta=5$$
Squaring on both sides gives
$$16\sin^2\theta +24\sin\theta \cos\theta+9\cos^2\theta=25$$
$$16-16\cos^2\theta+24\sin\theta \cos\theta+9-9\sin^2\theta=25$$
$$16\cos^2\theta-24\sin\theta \cos\theta+9\sin^2\theta=0$$
$$(4\cos\theta-3\sin\theta)^2=0$$
$$4\cos\theta-3\sin\theta=0$$
Hope it helps!.
A: A very fast solution
Maximum value of $\displaystyle4\sin\theta+3\cos\theta$ is $\displaystyle{5}$. This value is achieved when $\displaystyle\tan\theta=\dfrac{4}{3}$ by differentiating.So $\cos\theta=\dfrac{3}{5}$ and $\sin\theta=\dfrac{4}{5}$.
Hence $4\cos\theta-3\sin\theta=0$. 
