Answer:
cosine theta equals plus or minus two times square root of two over three, tangent theta equals plus or minus square root of two over four
Step-by-step explanation:
Given:
sinθ=1/3
θ=19.47 degrees
then
cosθ= cos(19.47)=0.942 = 2(√2/3)
tanθ=tan(19.47)=0.35= √2/4
Hence option two is correct:cosine theta equals plus or minus two times square root of two over three, tangent theta equals plus or minus square root of two over four!
Answer:
So second choice.
[tex]\cos(\theta)=\pm \frac{2\sqrt{2}}{3}[/tex]
[tex]\tan(\theta)=\pm \frac{\sqrt{2}}{4}[/tex]
Step-by-step explanation:
I'm going to use a Pythagorean Identity, name the one that says:
[tex]\sin^2(\theta)+\cos^2(\theta)=1[/tex].
We are given: [tex]\sin(\theta)=\frac{1}{3}[/tex].
Inserting this into our identity above gives us:
[tex](\frac{1}{3})^2+\cos^2(\theta)=1[/tex]
Time to solve this for the cosine value:
[tex]\frac{1}{9}+\cos^2(\theta)=1[/tex]
Subtract 1/9 on both sides:
[tex]\cos^2(\theta)=1-\frac{1}{9}[/tex]
[tex]\cos^2(\theta)=\frac{8}{9}[/tex]
Square root both sides:
[tex]\cos(\theta)=\pm \sqrt{\frac{8}{9}}[/tex]
9 is a perfect square but 8 is not.
8 does contain a factor that is a perfect square which is 4.
So time for a rewrite:
[tex]\cos(\theta)=\pm \frac{\sqrt{4}\sqrt{2}}{3}[/tex]
[tex]\cos(\theta)=\pm \frac{2\sqrt{2}}{3}[/tex]
So without any other give information we can't know if cosine is positive or negative.
Now time for the tangent value.
You can find tangent value by using a quotient identity:
[tex]\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}[/tex]
[tex]\tan(\theta)= \frac{\frac{1}{3}}{\pm \frac{2\sqrt{2}}{3}}[/tex]
Multiply top and bottom by 3 get's rid of the 3's on the bottom of each mini-fraction:
[tex]\tan(\theta)=\pm \frac{1}{2 \sqrt{2}}[/tex]
Multiply top and bottom by sqrt(2) to get rid of the square root on bottom:
[tex]\tan(\theta)=\pm \frac{1(\sqrt{2})}{2\sqrt{2}(\sqrt{2})}[/tex]
Simplifying:
[tex]\tan(\theta)=\pm \frac{\sqrt{2}}{2(2)}[/tex]
[tex]\tan(\theta)=\pm \frac{\sqrt{2}}{4}[/tex]
