Find cos theta and tan theta using sin theta

[mathlearn]

If sin \(\displaystyle \theta\) =\(\displaystyle \frac{4}{5}\) , find cos \(\displaystyle \theta\) and tan \(\displaystyle \theta\)

Can you help me to solve. :)

Many thanks :)

 

[I like Serena]

Re: Find cos theta and tan theta using sin thetha

If sin \(\displaystyle \theta\) =\(\displaystyle \frac{4}{5}\) , find cos \(\displaystyle \theta\) and tan \(\displaystyle \theta\)

Can you help me to solve. :)

Many thanks :)

Hey mathlearn! ;)

The sine is the opposite side divided by the hypotenuse.
What would be the length of the adjacent side, knowing we have a right triangle?
And what would then be the cosine respectively the tangent?

 

[mathlearn]

Re: Find cos theta and tan theta using sin thetha

:)

If sin \(\displaystyle \theta\) =\(\displaystyle \frac{4}{5}\) , find cos \(\displaystyle \theta\) and tan \(\displaystyle \theta\)

sin \(\displaystyle \theta\) = \(\displaystyle \frac{opposite side}{hypotenuese}\)

\(\displaystyle \therefore sin \) \(\displaystyle \theta\) =\(\displaystyle \frac{4}{5}\)

So applying Pythagoras theorem

Hypotenuse² = opposite side ² + adjacent side²

\(\displaystyle 5^{2}\) = \(\displaystyle 4^{2}\) + \(\displaystyle adjacent side^{2}\)

\(\displaystyle 25\) = \(\displaystyle 16\) + \(\displaystyle adjacent side^{2}\)

\(\displaystyle 25\) - \(\displaystyle 16\)= \(\displaystyle adjacent side^{2}\)

\(\displaystyle 9\)= \(\displaystyle adjacent side^{2}\)

\(\displaystyle \sqrt{9}\)= \(\displaystyle \sqrt{adjacent side^{2}}\)

\(\displaystyle 3\)= \(\displaystyle adjacent side\)

\(\displaystyle \therefore cos \theta\) = \(\displaystyle \frac{adjacent side}{hypotenuse }\)\(\displaystyle \therefore cos \theta\) = \(\displaystyle \frac{3}{5}\)

and

\(\displaystyle \therefore tan \theta\) = \(\displaystyle \frac{opposite side}{adjacent side}\)

\(\displaystyle \therefore tan \theta\) = \(\displaystyle \frac{4}{3}\)

Correct I Guess?

Many Thanks :)

 

[I like Serena]

Yep. All correct. (Nod)

 

[Prove It]

Yep. All correct. (Nod)

Actually, it's possible that the angle could be in the second quadrant, in which case the cosine and tangent values would be negative.

Without any information about which quadrant the angle lies, you would need to write both the positive and negative answers.