If $3 \sin B + 5 \cos B =5$, prove that:

Question

If $3 \sin B + 5 \cos B =5$ then prove that $3\cos B-5\sin B=\pm 3$.

My Attempt

Given: $$3\sin B+5\cos B=5$$ Dividing both sides by $\cos B$: $$3\tan B+5=5\sec B$$ Squaring on both sides: $$9\tan^2 B+30\tan B+ 25=25 \sec^2 B$$

Now, what should I do next?

Use formula $1+\tan^2B=\sec^2B$ on the RHS. Like this you have a second degree equation in $\tan B$. — Jean Marie, Mar 12 '16 at 16:33

score 3 · Answer 1 · 2016-03-12T17:06:07.723

3

Hint:

$$\begin{align}\color{blue}{(3\sin B+5\cos B)^2}&+\color{green}{(3\cos B-5\sin B)^2}\\ =3^2\sin^2B+2\cdot5\cdot3\sin B\cos B+5^2\cos^2B&+3^2\cos^2B-2\cdot3\cdot5\cos B\sin B+5^2\sin^2B\\ =\color{blue}{5^2}&+\color{green}{3^2}.\end{align}$$

edited Mar 12 '16 at 17:06

answered Mar 12 '16 at 16:54

Could you explain a bit, I could not understand? – Iaamuser user Mar 12 '16 at 17:06
@Iaamuseruser: what don't you understand, this is elementary development ? – Mar 12 '16 at 17:07
@ Yves Daoust, How does it complete the proof? – Iaamuser user Mar 12 '16 at 17:09
@Iaamuseruser: do you agree that the third line equals the first line ? – Mar 12 '16 at 17:10
@YvesDaoust hey you can't solve the question this way!! i think so. – user5954246 Mar 12 '16 at 17:12
@DeNiSkA: there is nothing wrong with this proof (titled as a hint by the way, but a valid proof indeed). Can you elaborate ? – Mar 12 '16 at 17:13
@YvesDaoust sorry you are correct!!!!!! – user5954246 Mar 12 '16 at 17:15
@DeNiSkA: it is not presented in the usual order, but it is so much nicer this way :) – Mar 12 '16 at 17:18
yes i will give +1 for this – user5954246 Mar 12 '16 at 17:18

egreg · Answer 2 · 2016-03-12T17:10:05.663

2

It's not restrictive to assume $0\le B<2\pi$, since only $\sin B$ and $\cos B$ appear. Note that $B\ne\pi$, because $3\sin\pi+5\cos\pi=-5\ne5$.

Let $t=\tan(B/2)$; then $$ \frac{6t}{1+t^2}+\frac{5-5t^2}{1+t^2}=5 $$ so $$ 10t^2-6t=0 $$ Therefore either $t=0$ or $t=3/5$.

Now $$ 3\cos B-5\sin B=\frac{3-3t^2}{1+t^2}-\frac{10t}{1+t^2} $$ and you can substitute the values above.

edited Mar 12 '16 at 17:10

answered Mar 12 '16 at 16:49

egreg

238,574

$B\neq \pi+2\pi k$, $k\in\mathbb Z$. – user236182 Mar 12 '16 at 16:52
@egreg, but how its $B\ne\pi$? – Iaamuser user Mar 12 '16 at 17:00
@Iaamuseruser $3\sin\pi+5\cos\pi=-5\ne5$. – egreg Mar 12 '16 at 17:03
@user236182 It's not restrictive to assume $0\le B<2\pi$, because only $\sin B$ and $\cos B$ appear. – egreg Mar 12 '16 at 17:06
@egreg That assumption isn't really universal and should be mentioned. – user236182 Mar 12 '16 at 17:07

user5954246 · Answer 3 · 2016-03-12T17:01:03.177

1

i can write the first equation as $$6\sin \frac{B}{2}\cos \frac{B}{2}=5(1-cos B)$$ i have used $\sin 2x=2sin\frac{x}{2}cos\frac{x}{2}$ formula $$1-\cos B=2\sin^2 \frac{B}{2}$$ $$\implies 6\cos \frac{B}2= 5*2\sin \frac{B}{2}$$ here, i have assumed $\sin x$ not equal to $0$, because if $sin x =0$ the answer is straight forward

$$\implies \tan \frac{B}{2}=3/5$$

then you know $\tan x=\frac{opp}{hyp}$ and by pythagoras theorem you can find $sinx$ and $cosx$

and $$\sin B=15/17 , \cos B=8/17$$ on substituting value in the second equation it is proved!!

i hope this method helped

note:

conversion of ratios is very helpful

edited Mar 12 '16 at 17:01

answered Mar 12 '16 at 16:54

user5954246

1,135

without $sinx=0$, how should I complete the proof? – Iaamuser user Mar 13 '16 at 02:03
@Iaamuseruser hey i didn't get you sorry? – user5954246 Mar 13 '16 at 04:30
if you mean if $sin x=0$ then how will this be proved then the answer is... see the first equation if $sin x =0$ then $cos x$ must be 1, so substitute the value in the second equation and then identity is proved!! – user5954246 Mar 13 '16 at 04:33

jim · Answer 4 · 2016-03-14T20:42:37.113

Write $X = 3 \sin B + 5 \cos B = \sqrt{34} \sin (B + \alpha)$ (with $\tan \ \alpha = \frac{5}{3}$ but not needed) then $Y = 3\cos B - 5\sin B = \sqrt{34} \cos( B + \alpha)$, then use $\sin^2 \theta + \cos^2 \theta = 1$.

The details of the calculation are: You can write $3 \sin B + 5 \cos B$ as $\sqrt{34} (\frac{3}{\sqrt{34}} \sin B + \frac{5}{\sqrt{34}} \cos B)$.

This is OK and it hasn't changed the result, but why divide and multiply by $\sqrt{34}$? That is a bit of experience, but notice that $3^2 + 5^2 = 9 + 25 = 34$.

But now you can put $\cos \alpha = \frac{3}{\sqrt{34}}$ and $\sin \alpha = \frac{5}{\sqrt{34}}$ (you have $\sin^2 \alpha + \cos^2 \alpha = 1$ so that this is admissible) and the final result for $X$ is $\sqrt{34} \sin ( B + \alpha)$.

After that, you can then look at $Y$ and divide and multiply by 6 again and this time you have $Y = 3 \cos B - 5 \sin B = \sqrt{34} (\frac{3}{\sqrt{34}} \cos B - \frac{5}{\sqrt{34}} \sin B) = \sqrt{34} (\cos \alpha \cos B - \sin \alpha \sin B) = \sqrt{34} \cos( B + \alpha)$.

Because $X^2 + Y^2 = 34$ (from $\sin^2 + \cos^2 = 1)$ then $Y^2 = 9, Y = \pm 3$ since you are given $X = 5$.

@http://math.stackexchange.com/users/310149/iaamuser-user have put details in. — jim, Mar 13 '16 at 20:46

If $3 \sin B + 5 \cos B =5$, prove that:

4 Answers4