# Thread: Trigonometry problem , easy for someone who knows math :)

1. ## Trigonometry problem , easy for someone who knows math :)

2. Is this problem for studying for the test? Or is it a take-home test?

Alpha=x.. k?
cos x=5/13
x is between 270,360 so x is negative of the form (360-x).

cos x= cos -x
therefore, we can take cos x as acute.
but sin x is negative
sin360-x = - sin x

sin(60-x)=sin60*cosX -sin(-X)cos60
=root3 *5/13 + 12/13*1/2
=5root3 + 12
------------
26

cos(45+x)=cos45cosX-sin45sin(-X)
=1/root2*5/13 + 1/root2*12/13
= 17
------
13root2

4. Your addition of angles formula is the right way to go about this. The restriction on x just tells you the sign of sin(x), where you use Pythagorean theorem to find sin(x) without actually needing to solve for x. I would double-check your trig there, especially the sin(60).

5. Hello, TheGame!

$\cos\theta \:=\:\frac{5}{13},\;\;\theta \in Q4$

Find: . $(a)\;\sin(60^o-\theta)$
. . . . . $(b)\;\cos(\theta + 45^o)$

$\cos\theta \:=\:\dfrac{5}{13} \:=\:\dfrac{adj}{hyp}$

$\theta$ is in a right triangle with: $adj = 5,\;hyp = 13$

Pythagorus tells us that: $opp = \pm 12$

Since $\theta$ is in $Q4,\;\sin\theta \:=\:\text{-}\frac{12}{13}$

$(a)\;\sin(60^o - \theta) \;=\;\sin60^o\cos\theta - \cos60^o\sin\theta$

. . . . . . . . . . . . $=\;\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{5} {13}\right) - \left(\dfrac{1}{2}\right)\left(\text{-}\dfrac{12}{13}\right)$

. . . . . . . . . . . . $=\;\dfrac{5\sqrt{3}+12}{26}$

$(b)\;\cos(\theta + 45^o) \;=\;\cos\theta\cos45^o - \sin\theta\sin45^o$

. . . . . . . . . . . . $=\;\left(\dfrac{5}{13}\right)\left(\dfrac{1}{\sqrt {2}}\right) - \left(\text{-}\dfrac{12}{13}\right)\left(\dfrac{1}{\sqrt{2}}\ri ght)$

. . . . . . . . . . . . $=\;\dfrac{17}{13\sqrt{2}}$

6. Also,i use sin^2 x + cos^2 x=1.. another form of pythagoras.. but straight to the point..