How do you use #sintheta=1/3# to find #tantheta#?

Answer 1

#+- sqrt2/4#

First find cos t. #cos^2 t = 1 - sin^2 t# #cos ^2 t = 1 - 1/9 = 8/9# #cos t = +- (2sqrt2)/3# #tan t = sin t/(cos t) = +- (1/3)(3/(2sqrt2)) = +- 1/(2sqrt2) = +- sqrt2/4# Note about the signs of tan t. On the unit circle: sin t = 1/3 --> cos t either + or - , there for --> tan t either + or - Check by calculator. #sin t = 1/3# --> #t = 19^@47# and #t = 180 - 19.47 = 160^@53# tan 19.47 = 0.35 and tan 160.53 = - 0.35 #+- sqrt2/4 = +- 1.414/5 = +- 0.35# OK

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Answer 2

Anna Andrews

You can use the relationship between sine and tangent to find $\tan(\theta)$ using the given value of $\sin(\theta)$ . First, recognize that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ . Since $\sin(\theta) = \frac{1}{3}$ , you need to find $\cos(\theta)$ to compute $\tan(\theta)$ . To find $\cos(\theta)$ , use the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$ . Substitute $\sin(\theta) = \frac{1}{3}$ into this equation and solve for $\cos(\theta)$ . Once you have $\cos(\theta)$ , you can compute $\tan(\theta)$ using the formula $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.