Cool Trig Equation

Here a different way to look at the problem that has the advantage of being extremely easy. Add $\displaystyle 4\sin(A)$ to both sides and you get $\cos(A)=1+4\sin(A).$ Since $-1\leqslant\cos A\leqslant 1\;\forall A\in\Bbb R$ one of the ways this could be possible is if $\sin A=0\implies A=0 \text{ or }A=\pi$ . Plug in those values in $\sin(A)+4\cos(A)$ gives us the answer: 4 and -4.

let $t = tan( \frac{A}{2} )$

then $sin(A) = \frac{2t}{1+t^{2}}$ and $cos(A) = \frac{1-t^{2}}{1+t^{2}}$

$cos(A) - 4 sin(A) = 1$ .. we subtract two sides by 1

$cos(A) - 4 sin(A) - 1 = 0$

$\frac{1-t^{2}}{1+t^{2}} -4\frac{2t}{1+t^{2}} - \frac{1+t^{2}}{1+t^{2}} = 0$

Bring them together and they will be equal to

$\frac{2t^{2} + 8t } { -1-t^2} = 0$ .. and we multiply both sides by $-1-t^2$

$2t^2 + 8t = 0$ .. and we take $2t$ common factor

$2t(t+4) = 0$

so there is two solution $t=0$ ... and $t=-4$

when $t = 0$ .. $A=0$ .. and .. $sin(A)+4cos(A) = 4$

when $t = -4$ .. $A=208.0724869$ .. and ... $sin(A)+4cos(A) = -4$

\sin{A}+ 4 \cos{A}= x.

Square both equations and add. Use (sine)^{2} + (cos)^{2} = 1 Get ans.

$(sinA+4cosA)^2+(cosA-4sinA)^2=1+16$ or, $(sinA+4cosA)^2 = 16$ or, $(sinA+4cosA) = 4, -4$

why everyone is complicating the problem just take $sin(A)+4cos(A)=x$ and now notice that ${ x }^{ 2 }+1={ (sin(A)+4cos(A)) }^{ 2 }+{ (cos(A)-4sin(A)) }^{ 2 }\\ \quad \quad \quad \quad ={ (sin }^{ 2 }(A)+16{ cos }^{ 2 }(A)+8sin(A)cos(A))+({ cos }^{ 2 }(A)+16{ sin }^{ 2 }(A)-8sin(A)cos(A))\\ \quad \quad \quad \quad =({ sin }^{ 2 }(A)+{ cos }^{ 2 }(A))+16({ sin }^{ 2 }(A)+{ cos }^{ 2 }(A))\\ \quad \quad \quad \quad =17\quad$ and we are done.

Some of these "solutions" don't actually work. Others assume the correct answer because it contains the one trivial solution. Others which actually generate two correct answers are inordinately complicated. Here is mine.

Given cos(A) - 4 sin(A) = 1, rearrange the terms to cos(A) - 1 = 4 sin(A).

Then square both sides of the equation, cos(A)^2 - 2 cos(A) + 1 = 16 sin(A)^2 = 16 (1 - cos(A)^2).

Collect terms, 17 cos(A)^2 - 2 cos(A) - 15 = 0.

This is a simple quadratic in cos(A), and the 2 roots are [-(-2) +/- ((-2)^2 - 4 (17) (-15))^(1/2)] / 2 (17).

After the arithmetic, cos(A) = (2 +/- 32) / 34 = {1, -15/17}.

Checking our work, the first root is the trivial solution, A = 0, cos(A) = 1 and sin(A) = 0, and result is 4.

The second root is slightly more complicated. First, we observe that A is an acute angle of an 8, 15, 17 right triangle, and |sin(A)| = 8/17. If cos(A) is negative, to satisfy the given equation, sin(A) cannot be positive.

Plugging this data back into the second equation yields (-8/17) + 4 (-15/17) = -68/17 = -4.

Hence, the solution set is {4, -4}.

abvious set of sol-cosA,sinA=(1,0)or (0,-1)now (1,0) fits for A=0deg but (0,-1) has no suc h solution set so A=0 hence ans 4 ,-4 achieved

I did this in a somewhat different way, as well, but it worked, so here goes:

cos(A) – 4sin(A) = 1

Since cos^2(A) + sin^2(A) = 1, then

cos(A) – 4sin(A) = cos^2(A) + sin^2(A)

cos(A) – 4sin(A) = cos(A)cos(A) + sin(A)sin(A)

Looking at the coefficients, we can see that

cos(A) = 1

sin(A) = - 4

We can rule out the second option, since sin(A) must be within -1 and 1, then

A = πn

sin(A) + 4cos(A) = ± 4

$cos \theta - 4 sin\theta = 1$ $\implies cos^2 \theta + sin^2 \theta - 8sin\theta cos \theta + 15 sin\theta = 1$ $\implies sin\theta (15sin\theta - 8cos\theta) = 0$ $\implies sin\theta = 0$ $\implies \theta = 0^{\circ}$ $\implies sin\theta + 4 cos\theta = 0 + 4*1 = 4$

Only the options contains $+4$ is $(+4, -4)$ . So, the answer is $(+4, -4)$

1.Square first eqn. \newline 2.Expand sin^2 and cos^2 using the formula (cos(A))^2=1-(sin(A))^2 \newline 3.You will get an expression which nothing but square of second eqn.\newline \cos{A}^{2} + 16\sin{A}^{2} - 8\cos{A} \sin{A} = 1 \newline 1 - \sin{A}^{2} + 16 - 16\cos{A}^{2} - 8\cos{A} \sin{A} = 1 \newline -[\sin{A}^{2} + 16\cos{A}^{2} + 8\cos{A} \sin{A} ] = -16 \newline or (\sin{A} + 4\cos{A})^2 = 16 \newline \therefore \sin{A} + 4\cos{A} = 4 or -4