22
$\begingroup$

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)

$\endgroup$
3
  • 5
    $\begingroup$ Why does this have so many views? This is a standard trig question. $\endgroup$
    – user85798
    Apr 18 '14 at 4:04
  • 2
    $\begingroup$ @Oliver Maths is always amazing no matter how tough or easy the question are 'cz there exist unimaginably countless ways you could reach the same answer. $\endgroup$
    – Anz Joy
    Apr 18 '14 at 5:36
  • $\begingroup$ @user1488 probably because standard trig questions are useful and interesting to many people doing maths. $\endgroup$ Apr 20 '18 at 17:49

14 Answers 14

50
$\begingroup$

HINT:

using Brahmagupta–Fibonacci identity

$$(4\sin\theta+3\cos\theta)^2+(4\cos\theta-3\sin\theta)^2=(4^2+3^2)(\sin^2\theta+\cos^2\theta)$$

$\endgroup$
7
  • $\begingroup$ A bit more explanation would suffice $\endgroup$
    – Anz Joy
    Apr 17 '14 at 8:36
  • $\begingroup$ @AnzJoy, Sorry misreading . Please find the edited version $\endgroup$ Apr 17 '14 at 8:39
  • 14
    $\begingroup$ Expand $(a\sin\theta+b\cos\theta)^2+(a\cos\theta-b\sin\theta)^2$ and simplify to get $a^2+b^2$. What lab bhattacharjee gave as an answer is really elegant. $\endgroup$ Apr 17 '14 at 9:05
  • 1
    $\begingroup$ @AnzJoy, Please let me know if you have any confusion? $\endgroup$ Apr 17 '14 at 9:21
  • 1
    $\begingroup$ @Downvoter, May I request you to pinpoint the mistake $\endgroup$ Apr 18 '14 at 5:40
37
$\begingroup$

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, $$ 4\cos\theta-3\sin\theta=4\left(\frac35\right)-3\left(\frac45\right)=\Large\color{blue}0. $$

$\endgroup$
3
  • $\begingroup$ Easy to follow, and I can imagine thinking of this method for a future problem. Good stuff. $\endgroup$
    – Almo
    Apr 17 '14 at 15:21
  • 7
    $\begingroup$ Warning: Squaring equations introduces spurious solutions; all solutions must be checked in the equations before one is done. In this particular case, the "other" square root gives $\sin \theta = - \frac{4}{5}$, which is not mentioned, which (unsurprisingly) when checked is not a solution of the equation. (If you reason you put $\cos \theta = 3/5$ into (1) and found only one sine case, you reason incorrectly. You must substitute into (2), the equation it actually came from. Ignoring signs causes the solution set of (2) to differ from that of (1) and you must respect this difference.) $\endgroup$ Apr 17 '14 at 16:21
  • $\begingroup$ @EricTowers Thanks for your warning Sir... $\endgroup$
    – Tunk-Fey
    Apr 17 '14 at 21:33
11
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.)

$\endgroup$
4
$\begingroup$

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 $4\cos(\theta) - 3\sin(\theta) = 12/5 - 12/5 = 0$

$\endgroup$
2
  • $\begingroup$ Hummmm ! Not so sure. $\endgroup$ Apr 17 '14 at 9:04
  • $\begingroup$ @Claude Leibovici: yes you shouldn't be: I've fixed it now though (got the sines and cosines the wrong way round initially). $\endgroup$ Apr 17 '14 at 9:12
3
$\begingroup$

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 $\displaystyle r^2=a^2+b^2$

$\endgroup$
2
  • 1
    $\begingroup$ @AnzJoy, How about this? Observe that we don't need $\displaystyle\cos\theta,\sin\theta$ separately $\endgroup$ Apr 17 '14 at 14:44
  • $\begingroup$ Wew..!Maths amazes me..Thank you so much.. $\endgroup$
    – Anz Joy
    Apr 18 '14 at 5:31
3
$\begingroup$

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 $$-3\sin\theta+4\cos\theta=0.$$

$\endgroup$
3
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \left.\begin{array}{rcrcl} 4\sin\pars{\theta} & + & 3\cos\pars{\theta} & = & 5 \\ -3\sin\pars{\theta} & + & 4\cos\pars{\theta} & \equiv & \mu:\ {\large ?} \end{array}\right\rbrace\ \imp\ \begin{array}{|rcl} \ \sin\pars{\theta}& = & {20 - 3\mu \over 25} \\[3mm] \cos\pars{\theta}& = & {4\mu + 15 \over 25} \end{array} $$

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}

$$ \imp\quad\color{#66f}{\large 4\cos\pars{\theta} - 3\sin\pars{\theta} = 0} $$

$\endgroup$
2
  • 1
    $\begingroup$ I came here to see my OP and I saw your answer. +1 $\endgroup$
    – Tunk-Fey
    Sep 2 '14 at 16:47
  • $\begingroup$ @Tunk-Fey Thanks. $\endgroup$ Sep 2 '14 at 18:09
2
$\begingroup$

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:

$$x = 0 $$

$\endgroup$
1
  • $\begingroup$ The last step could be made a bit better by expanding the RHS to get $625 = 25x^2 + 625 \rightarrow 25x^2 = 0$ $\endgroup$
    – MCT
    Apr 17 '14 at 15:17
2
$\begingroup$

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 $a \sin \alpha - b \cos \alpha = \frac{ab}{\sqrt{a^2 + b^2}} - \frac{ba}{\sqrt{a^2+b^2}} = \boxed{0}$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$$y = 5\cos(\pi/2 + Z\cdot 2\pi - \arctan(3/4) + \arctan(3/4))$$ $$y = 5\cos(\pi/2 + Z\cdot 2\pi)$$ $$y = 0$$

$\endgroup$
1
$\begingroup$

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!.

$\endgroup$
-1
$\begingroup$

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 $4\cos\theta-3\sin\theta=0$.

$\endgroup$

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