Question

can someone help me with pre cal

Answer 1

what type?

Answer 2

Trig

Answer 3

Let theta be an acute angle such that sin(theta)=(6/7). evaluate the other five trigonometric functions of theta.

Answer 4

instead of theta let us use x here

Answer 5

given sinx =6/7 ,cosecx =1/sinx
cosx =sqrt( 1-(sinx)^2 )
tanx = sinx / (sqrt( 1-(sinx)^2 ))
and cotx = (sqrt( 1-(sinx)^2 ))/sinx

Answer 6

\[\boxed{\sin\theta=\frac{\text{opposite side}}{\text{hypotenuse}}},\\\boxed{\cos\theta=\frac{\text{adjacent side}}{\text{hypotenuse}}},\\\boxed{\tan\theta=\frac{\text{opposite side}}{\text{adjacent side}}},\]

\[\boxed{\csc\theta=\frac1{\sin\theta}=\frac{\text{hypotenuse}}{\text{opposite side}}},\\\boxed{\sec\theta=\frac1{\cos\theta}=\frac{\text{hypotenuse}}{\text{adjacent side}}},\\\boxed{\cot\theta =\frac1{\tan\theta}=\frac{\text{adjacent side}}{\text{opposite side}}}.\]

Answer 7

matricked , i am confused

Answer 8

thank you unkleRhaukus

Answer 9

Both have told you essentially the same thing, although @UnkleRhaukus' formulation may appear simpler.

\[\sin \theta = \frac{6}{7}\]This implies that the opposite side is 6, and the hypotenuse is 7 (or multiples thereof). To find the adjacent side, we'll have to use the Pythagorean theorem: \[a^2 + 6^2 = 7^2\]\[a^2=49-36\]\[a=\sqrt{13}\]
Continuing on, \[\cos \theta = \frac{\sqrt{13}}{7}\]\[\sec \theta = \frac{1}{\cos \theta} = \frac{7}{\sqrt{13}} = \frac{7\sqrt{13}}{13}\]\[\tan\theta = \frac{6}{\sqrt{13}} = \frac{6\sqrt{13}}{13}\]\[\csc\theta = \frac{7}{6}\]\[\cot\theta = \frac{\sqrt{13}}{6}\]

Now let's do them the @matricked way:

\[\sin\theta = \frac{6}{7}\]\[\cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-{(6/7)^2}} = \frac{\sqrt{13}}{7}\]\[\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{6}{\sqrt{13}} = \frac{6\sqrt{13}}{13}\]and so on. Exactly the same answers, just looks a bit more frightening because he writes out the taking of a square root to find the unknown side each time he needs it.