# Math Help - Trig Identities Prob

1. ## Trig Identities Prob

'Solve sin x + 4 cos x = 3'
Can anyone help me with the above problem? Trigonometric identities and formulae confuse me.

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I have tried doing this:
sin x = 3 - 4 cos x

divide all by cos x
then tan x = 3/cos x - 4
rearrange so 4 = 3/cos x - tan x
which is same as 4 = 3 sec x - tan x

square both sides
so 16 = 9 sec^2 x - tan^2 x
this is 16 = 9(tan^2x + 1) - tan^2 x
and 16 = 9tan^2x - tan^2x + 9

rearrange so 7 = 8 tan^2 x
7/8 = tan^2 x
so tan x= sqrt(7/8)
and x = arctan sqrt(7/8)

Is this correct?
If anyone could provide any help or solution to this problem, I'll be so grateful.

Thank you!
Herman

2. My favorite:

$\sin(x) + 4\cos(x) = \sqrt{17}\sin(x+atan(4))$

It's a lot easier with only one function.

3. Originally Posted by TKHunny
My favorite:

$\sin(x) + 4\cos(x) = \sqrt{17}\sin(x+atan(4))$

It's a lot easier with only one function.
why would you even know such an identity?

4. so... is my answer right or wrong?
if wrong, what's right? lol

5. Hello, Herman!

You started off well, but you squared incorrectly . . .

Solve: . $\sin x + 4\cos x \:= \:3$

I have tried doing this: . $\sin x \:= \:3 - 4\cos x$

Divide all by $\cos x\!:\;\;\;\tan x \:= \:3\sec x - 4 \quad\Rightarrow\quad 4 \:=\:3\sec x - \tan x$

Square both sides: . $16 \:= \:9\sec^2\!x - \tan^2\!x$ . . . . no
You should have: . $16 \;=\;9\sec^2\!x - 6\sec x\tan x + \tan^2\!x$

6. Originally Posted by Jhevon
why would you even know such an identity?
It's a philosophy, "Derive, Don't Memorize."