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November 22, 2014

November 22, 2014

Posted by **jen** on Thursday, January 8, 2009 at 9:49pm.

I got tan(theta)=[sqrt(294)]/42

Is that right?

if sin(theta)=-a where 0<a<1, and theta is in quadrant 3, find the exact algebraic expressionm for cos(theta)

- revised problem, ignore the original post -
**jen**, Thursday, January 8, 2009 at 9:49pmIf sin(theta)=[sqrt(7)]/7 and theta is in Quadrant two, find the exact numerical value of tan theta without using a calculator.

I got tan(theta)=[sqrt(294)]/42

Is that right?

if sin(theta)=-a where 0<a<1, and theta is in quadrant 3, find the exact algebraic expressionm for cos(theta)

- easy trig -
**Reiny**, Thursday, January 8, 2009 at 9:58pmYou posted the same question earlier today, and I told you .....

http://www.jiskha.com/display.cgi?id=1231449736

- easy trig -
**jen**, Thursday, January 8, 2009 at 10:22pmyes, but i changed my mistake from

sin(theta)=[sqrt(70)]/7 to sin(theta)=[sqrt(7)]/7.

so it's sin(theta)=.378

- easy trig -
**Reiny**, Thursday, January 8, 2009 at 10:54pmok then, that's better

recall that sine(angle) = opposite/hypotenus

so we need a right-angled triangle in the II quadrant with a height of √7 and a hypotenuse of 7

let the base be x

x^2 + (√7)^2 = 7^2

x^2 = 42

x = ±√42, but we are in the second quadrant so x = -√42

then for yours

tan(theta) = √7/-√42

= -1/√6

notice all steps were done without a calculator.

that does not mean we couldn't use a calculator to check our answer

enter the following

√7/7 =

2nd function sin

-180=

± key (to make our answer positive in the second quadrant)(on some calculators you might have to multiply by -1 to get it to a positive, do whatever your calc needs done)

tan =

store or write down that number

now do

-1/√6 =

compare the two results, they are the same.

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