SOLUTION: I need to find the value of tan(theta) but the only value I have is sin(theta)=1/4 and it is in quadrant I.
I know that (sin)(sin)theta + (cos)(cos)theta=1 so if sin(theta)= 1/
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Question 211542: I need to find the value of tan(theta) but the only value I have is sin(theta)=1/4 and it is in quadrant I.
I know that (sin)(sin)theta + (cos)(cos)theta=1 so if sin(theta)= 1/4 then sin(sin)theta =1/16... so cos(cos)theta=square root of 15/16.. and this is positive because the values are in quadrant I which is always positive. From there I know that sin(theta) over cos(theta) will give me tan(theta) but I dont know how to simplify from there. Can you help me with that?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
+\cos^2(\theta)=1)
... Start with the first Pythagorean identity.
^2+\cos^2(\theta)=1)
... Plug in =\frac{1}{4})
=1)
... Square to get
=1-\frac{1}{16})
... Subtract from both sides.
=\frac{16}{16}-\frac{1}{16})
... Multiply 1 by
=\frac{16-1}{16})
... Combine the fractions.
=\frac{15}{16})
... Subtract
=\sqrt{\frac{15}{16}})
... Take the square root of both sides. Note: because 
is in the first quadrant, this means that >0)
(ie cosine is positive). So we don't need to worry about the negative square root.
=\frac{\sqrt{15}}{\sqrt{16}})
... Break up the square root.
=\frac{\sqrt{15}}{4})
... Take the square root of 16 to get 4.
-------------------------
=\frac{\sin(\theta)}{\cos(\theta)})
... Move onto the tangent identity.
=\frac{\frac{1}{4}}{\frac{\sqrt{15}}{4}})
... Plug in and
=\frac{1}{4}\cdot\frac{4}{\sqrt{15}})
... Multiply the first fraction by the reciprocal of the second fraction.
=\frac{1}{\sqrt{15}})
... Cancel out the common terms and simplify.
=\frac{1}{\sqrt{15}}\cdot\frac{\sqrt{15}}{\sqrt{15}})
... Multiply the fraction by (this will rationalize the denominator).
=\frac{1\cdot\sqrt{15}}{\sqrt{15}\cdot\sqrt{15}})
... Combine the fractions.
=\frac{\sqrt{15}}{15})
... Multiply
So if , then
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