Trig Identities Prob

• April 15th 2008, 01:10 PM
hermannjp
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
• April 15th 2008, 01:48 PM
TKHunny
My favorite:

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

It's a lot easier with only one function.
• April 15th 2008, 02:07 PM
Jhevon
Quote:

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? :p
• April 15th 2008, 04:38 PM
hermannjp
so... is my answer right or wrong?
if wrong, what's right? lol
• April 15th 2008, 07:18 PM
Soroban
Hello, Herman!

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

Quote:

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: . $16 \;=\;9\sec^2\!x - 6\sec x\tan x + \tan^2\!x$

• April 16th 2008, 07:08 AM
TKHunny
Quote:

Originally Posted by Jhevon
why would you even know such an identity? :p

It's a philosophy, "Derive, Don't Memorize." (Wink)