Question

Math question

Answer 1

hint:
\[\sin \theta = \pm \sqrt {1 - {{\left( {\cos \theta } \right)}^2}} \]

Answer 2

hint:
\[\large \begin{gathered}
\sin \theta = \pm \sqrt {1 - {{\left( {\cos \theta } \right)}^2}} = \pm \sqrt {1 - \frac{{16}}{{49}}} = \pm \sqrt {\frac{{33}}{{49}}} = ...? \hfill \\
\hfill \\
\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\left( { \pm \sqrt {\frac{{33}}{{49}}} } \right)}}{{ - \frac{4}{7}}} = ...? \hfill \\
\end{gathered} \]

Answer 3

I'm not sure...

Answer 4

hint:
\[ \pm \sqrt {\frac{{33}}{{49}}} = \pm \frac{{\sqrt {33} }}{7}\]

Answer 5

@Michele_Laino Okay, so sin0 = (sqrt)33/7 ?

Answer 6

not to butt in but this is somewhat easier if you draw a triangle

Answer 7

there is picture of an angle whose cosine is \(\frac{4}{7}\)
find the "opposite" side via pythagoras then you can take any trig ratio you like

Answer 8

since we have 2 square roots, then:
\[\sin \theta = \pm \frac{{\sqrt {33} }}{7}\]

Answer 9

How do we find tanθ Michele?

Answer 10

it is simple, you have to compute this ratio:
\[\large \tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\left( { \pm \frac{{\sqrt {33} }}{7}} \right)}}{{ - \frac{4}{7}}} = \left( { \pm \frac{{\sqrt {33} }}{7}} \right) \times \left( { - \frac{7}{4}} \right) = ...?\]

Answer 11

-7(sqrt)33/28 @Michele_Laino ?

Answer 12

memorize the trig ratios, look at the triangle and you can find any trig ratio you like

Answer 13

for example
\[\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}=\pm\frac{\sqrt{33}}{7}\]

Answer 14

\[\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=?\]

Answer 15

there is really very very little work to solve these
just find the missing side of the triangle and use the trig ratios you already know

Answer 16

tanθ = (sqrt)33/4 right?

Answer 17

yes of course

Answer 18

Thank you.

Answer 19

yw