Author Topic: Trig question: symmetry and exact values (Read 1289 times) Tweet Share

Ematuro · « **on:** April 15, 2011, 11:50:41 am »

If sin(x)=0.3, cos(a)=0.5, tan(b)=2.4 and x,a,b are all in the first quadrant, find the value of:
$cos(\pi-x)$

Couldn't do this question and looked at the worked solutions.. the next step they had was
$cos(\pi-x) = -cos(x) = -\sqrt{1-sin^2x}$

Could someone explain how they become each other

xZero · « **Reply #1 on:** April 15, 2011, 12:08:18 pm »

If x is in the first quadrant, then -x is in the fourth, the +pi will shift it to the 2nd quadrant, hence the -cos(x). Now cos^2+sin^2=1. Rearrange it to make cos(x) the subject and multiply both side by -1 to get -cos(x)

luffy · « **Reply #2 on:** April 15, 2011, 12:26:53 pm »

Its a simple property of the unit circle that $\cos{(\pi - x)} = -\cos{(x)}$

There is also a rule (can be derived from the unit circle as well) that states $\sin^2{(x)} + \cos^2{(x)} = 1$

Thus, $\cos^2{(x)} = 1 - \sin^2{(x)}$ (i.e. $\cos{(x)} = \sqrt{1 - \sin^2{(x)}}$

$\therefore \cos^2{(x)} = 1 - (0.3)^2$

$\therefore \cos^2{(x)} = 0.91$

$\therefore \cos{(x)} = \sqrt{0.91}$

$\therefore \cos{(\pi - x)} = -\cos{(x)} = -\sqrt{0.91}$

Hope I helped.

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Author Topic: Trig question: symmetry and exact values (Read 1289 times) Tweet Share

Ematuro

Trig question: symmetry and exact values

xZero

Re: Trig question: symmetry and exact values

luffy

Re: Trig question: symmetry and exact values

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